FizzBuzz in Javascript

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP












16












$begingroup$


I'm sure everyone here knows what FizzBuzz is. I would like constructive criticism for my solution.



I'm a beginner to programming as a whole and this isn't my first solution, but it's what I think is my best solution. I'd like to see if this is a reasonable solution, or is just plain terrible.



for(var i=1;i<=100;i++) 
var output = "";

if(i % 15 == 0) output = "FizzBuzz"
else if(i % 3 == 0) output = "Fizz"
else if(i % 5 == 0) output = "Buzz"
else output = i;

console.log(output);



The output is correct. It's your standard FizzBuzz output.










share|improve this question











$endgroup$



migrated from stackoverflow.com Jan 8 at 15:27


This question came from our site for professional and enthusiast programmers.










  • 4




    $begingroup$
    It's hard to give constructive criticism on such a small, simple function. It's a reasonable solution and the only things I could pick out being wrong would be me being extremely picky or a personal preference.
    $endgroup$
    – George
    Jan 8 at 15:36







  • 4




    $begingroup$
    You could go with var output = i; and then remove that final else block.
    $endgroup$
    – user189829
    Jan 8 at 15:37










  • $begingroup$
    One possible issue is you are relying on a standard fizzbuzz using 3 and 5 (hence checking for 3 * 5 as 15). Someone might give you a test using 2 and 4 to see if you fell into the trap of checking for 8.
    $endgroup$
    – Kickstart
    Jan 9 at 10:39










  • $begingroup$
    It would be better to include the actual spec you are following. The FizzBuzz solution looks fine. Any extra points goes to how you follow the nuance of the spec. If you really wanted to get fancy, you could do it without the explicit "FizzBuzz" part of the algorithm.
    $endgroup$
    – Tombo
    Jan 9 at 14:39















16












$begingroup$


I'm sure everyone here knows what FizzBuzz is. I would like constructive criticism for my solution.



I'm a beginner to programming as a whole and this isn't my first solution, but it's what I think is my best solution. I'd like to see if this is a reasonable solution, or is just plain terrible.



for(var i=1;i<=100;i++) 
var output = "";

if(i % 15 == 0) output = "FizzBuzz"
else if(i % 3 == 0) output = "Fizz"
else if(i % 5 == 0) output = "Buzz"
else output = i;

console.log(output);



The output is correct. It's your standard FizzBuzz output.










share|improve this question











$endgroup$



migrated from stackoverflow.com Jan 8 at 15:27


This question came from our site for professional and enthusiast programmers.










  • 4




    $begingroup$
    It's hard to give constructive criticism on such a small, simple function. It's a reasonable solution and the only things I could pick out being wrong would be me being extremely picky or a personal preference.
    $endgroup$
    – George
    Jan 8 at 15:36







  • 4




    $begingroup$
    You could go with var output = i; and then remove that final else block.
    $endgroup$
    – user189829
    Jan 8 at 15:37










  • $begingroup$
    One possible issue is you are relying on a standard fizzbuzz using 3 and 5 (hence checking for 3 * 5 as 15). Someone might give you a test using 2 and 4 to see if you fell into the trap of checking for 8.
    $endgroup$
    – Kickstart
    Jan 9 at 10:39










  • $begingroup$
    It would be better to include the actual spec you are following. The FizzBuzz solution looks fine. Any extra points goes to how you follow the nuance of the spec. If you really wanted to get fancy, you could do it without the explicit "FizzBuzz" part of the algorithm.
    $endgroup$
    – Tombo
    Jan 9 at 14:39













16












16








16


5



$begingroup$


I'm sure everyone here knows what FizzBuzz is. I would like constructive criticism for my solution.



I'm a beginner to programming as a whole and this isn't my first solution, but it's what I think is my best solution. I'd like to see if this is a reasonable solution, or is just plain terrible.



for(var i=1;i<=100;i++) 
var output = "";

if(i % 15 == 0) output = "FizzBuzz"
else if(i % 3 == 0) output = "Fizz"
else if(i % 5 == 0) output = "Buzz"
else output = i;

console.log(output);



The output is correct. It's your standard FizzBuzz output.










share|improve this question











$endgroup$




I'm sure everyone here knows what FizzBuzz is. I would like constructive criticism for my solution.



I'm a beginner to programming as a whole and this isn't my first solution, but it's what I think is my best solution. I'd like to see if this is a reasonable solution, or is just plain terrible.



for(var i=1;i<=100;i++) 
var output = "";

if(i % 15 == 0) output = "FizzBuzz"
else if(i % 3 == 0) output = "Fizz"
else if(i % 5 == 0) output = "Buzz"
else output = i;

console.log(output);



The output is correct. It's your standard FizzBuzz output.







javascript beginner fizzbuzz iteration






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Jan 10 at 14:34









Pedro A

1334




1334










asked Jan 8 at 15:26









Zachary WoodsZachary Woods

8915




8915




migrated from stackoverflow.com Jan 8 at 15:27


This question came from our site for professional and enthusiast programmers.






migrated from stackoverflow.com Jan 8 at 15:27


This question came from our site for professional and enthusiast programmers.









  • 4




    $begingroup$
    It's hard to give constructive criticism on such a small, simple function. It's a reasonable solution and the only things I could pick out being wrong would be me being extremely picky or a personal preference.
    $endgroup$
    – George
    Jan 8 at 15:36







  • 4




    $begingroup$
    You could go with var output = i; and then remove that final else block.
    $endgroup$
    – user189829
    Jan 8 at 15:37










  • $begingroup$
    One possible issue is you are relying on a standard fizzbuzz using 3 and 5 (hence checking for 3 * 5 as 15). Someone might give you a test using 2 and 4 to see if you fell into the trap of checking for 8.
    $endgroup$
    – Kickstart
    Jan 9 at 10:39










  • $begingroup$
    It would be better to include the actual spec you are following. The FizzBuzz solution looks fine. Any extra points goes to how you follow the nuance of the spec. If you really wanted to get fancy, you could do it without the explicit "FizzBuzz" part of the algorithm.
    $endgroup$
    – Tombo
    Jan 9 at 14:39












  • 4




    $begingroup$
    It's hard to give constructive criticism on such a small, simple function. It's a reasonable solution and the only things I could pick out being wrong would be me being extremely picky or a personal preference.
    $endgroup$
    – George
    Jan 8 at 15:36







  • 4




    $begingroup$
    You could go with var output = i; and then remove that final else block.
    $endgroup$
    – user189829
    Jan 8 at 15:37










  • $begingroup$
    One possible issue is you are relying on a standard fizzbuzz using 3 and 5 (hence checking for 3 * 5 as 15). Someone might give you a test using 2 and 4 to see if you fell into the trap of checking for 8.
    $endgroup$
    – Kickstart
    Jan 9 at 10:39










  • $begingroup$
    It would be better to include the actual spec you are following. The FizzBuzz solution looks fine. Any extra points goes to how you follow the nuance of the spec. If you really wanted to get fancy, you could do it without the explicit "FizzBuzz" part of the algorithm.
    $endgroup$
    – Tombo
    Jan 9 at 14:39







4




4




$begingroup$
It's hard to give constructive criticism on such a small, simple function. It's a reasonable solution and the only things I could pick out being wrong would be me being extremely picky or a personal preference.
$endgroup$
– George
Jan 8 at 15:36





$begingroup$
It's hard to give constructive criticism on such a small, simple function. It's a reasonable solution and the only things I could pick out being wrong would be me being extremely picky or a personal preference.
$endgroup$
– George
Jan 8 at 15:36





4




4




$begingroup$
You could go with var output = i; and then remove that final else block.
$endgroup$
– user189829
Jan 8 at 15:37




$begingroup$
You could go with var output = i; and then remove that final else block.
$endgroup$
– user189829
Jan 8 at 15:37












$begingroup$
One possible issue is you are relying on a standard fizzbuzz using 3 and 5 (hence checking for 3 * 5 as 15). Someone might give you a test using 2 and 4 to see if you fell into the trap of checking for 8.
$endgroup$
– Kickstart
Jan 9 at 10:39




$begingroup$
One possible issue is you are relying on a standard fizzbuzz using 3 and 5 (hence checking for 3 * 5 as 15). Someone might give you a test using 2 and 4 to see if you fell into the trap of checking for 8.
$endgroup$
– Kickstart
Jan 9 at 10:39












$begingroup$
It would be better to include the actual spec you are following. The FizzBuzz solution looks fine. Any extra points goes to how you follow the nuance of the spec. If you really wanted to get fancy, you could do it without the explicit "FizzBuzz" part of the algorithm.
$endgroup$
– Tombo
Jan 9 at 14:39




$begingroup$
It would be better to include the actual spec you are following. The FizzBuzz solution looks fine. Any extra points goes to how you follow the nuance of the spec. If you really wanted to get fancy, you could do it without the explicit "FizzBuzz" part of the algorithm.
$endgroup$
– Tombo
Jan 9 at 14:39










7 Answers
7






active

oldest

votes


















47












$begingroup$


I'd like to see if this is a reasonable solution, or is just plain terrible.




I wouldn't say it is "terrible" - mostly because it works and doesn't appear to be very inefficient. However, there are some improvements that can be made.




  • use strict equality comparison - i.e. === when comparing values. That way it won't need to convert the types.


  • Style Guide Consider following a style guide. Many common style guides advise separating keywords with a space - e.g. if ( instead of if(.


  • Use consistent indentation The first and last lines within the for loop are not indented, while the other lines between them are, though maybe it was the case that your code was indented properly but when you pasted here it was thrown off because of the markdown formatting...


  • Statements and blocks - If you are going to use block statements then expand them so each statement is on a separate line. Otherwise, just put the single statement without a block. Also use consistent statement termination. The first three conditional blocks don't have a semi-colon after the statement within the block, while the last one (in the else block) does.


  • Abstract logic into a function As Paul's answer suggests: you can put the core logic into a function that returns the output but doesn't handle outputting the value (e.g. to the console). This allows such code to be atomic and testable - congruent with the Single Responsibility Principle. Also, the return statement can eliminate the need for else keywords within a function. One drawback is that calling a function on each iteration may slow down operations but should be negligible for a set of numbers 1 through 100.

Updated Code



Consider the modified code below, utilizing the feedback above.



Note: the inline console in the snippets is truncated to ~50 lines, but the complete console log should be visible in your browser console.






function fizzBuzz(value) 
if (value % 15 === 0) return "FizzBuzz";
if (value % 3 === 0) return "Fizz";
if (value % 5 === 0) return "Buzz";
return value;

for (var i = 1; i <= 100; i++)
console.log(fizzBuzz(i));





One option I did consider is to minimize the number of modulo operations, append to the output string instead of outputting it. If you are trying to optimize for speed, this might not be the approach to take because appending to the string may be much slower than doing an extra couple modulo operations. Try a comparison in this jsPerf test.






function fizzBuzz(value) 
for (var i = 1; i <= 100; i++)
console.log(fizzBuzz(i));








share|improve this answer











$endgroup$








  • 3




    $begingroup$
    There is also the "never use semicolons" school. blog.izs.me/2010/12/… feross.org/never-use-semicolons
    $endgroup$
    – Džuris
    Jan 8 at 22:01






  • 4




    $begingroup$
    @CaseyB I don't think arguing about micro-optimizations like this when OP is a beginner is a good idea at all.
    $endgroup$
    – Voile
    Jan 9 at 9:43






  • 4




    $begingroup$
    @Voile One million percent. Even if this were a production-ready, enterprise-grade, cloud-based microservice™, readability and maintainability is always the number one concern. The concept of micro-optimizing a JavaScript FizzBuzz is honestly hilarious.
    $endgroup$
    – Michael
    Jan 9 at 10:34






  • 7




    $begingroup$
    @Michael FizzBuzz Enterprise Edition begs to differ. :P But FWIW, I believe Casey was merely providing a counterargument for the microoptimization that the answer itself first suggested.
    $endgroup$
    – Mateen Ulhaq
    Jan 9 at 12:05






  • 2




    $begingroup$
    @MateenUlhaq 260 issues... man, they really need to refactor it, arrange UATs, and use Kanban.
    $endgroup$
    – Juha Untinen
    Jan 9 at 13:10


















25












$begingroup$

One of the problems is that the case where you check i % 15 (i.e. i is a multiple of 3 and 5) is unnecessary. You have the concept of 3 and 5 repeated, and the concept of Fizz and Buzz repeated.



This is not currently much of a problem but suppose someone asks you to extend your program to print "Jazz" when i is a multiple of 7. Using your current strategy we now need a case where i is a multiple of:



  1. 3

  2. 5

  3. 7

  4. 3 and 5

  5. 3 and 7

  6. 5 and 7

  7. 3 and 5 and 7

It would look something like this:






for(var i = 1;i <= 100; i++) 
var output = "";

if(i % 105 == 0) output = "FizzBuzzJazz"
else if(i % 15 == 0) output = "FizzBuzz"
else if(i % 35 == 0) output = "BuzzJazz"
else if(i % 21 == 0) output = "FizzJazz"
else if(i % 3 == 0) output = "Fizz"
else if(i % 5 == 0) output = "Buzz"
else if(i % 7 == 0) output = "Jazz"
else output = i;

console.log(output);





See how quickly that got out of hand? If we add a fourth word it becomes even worse.



If we use a different strategy by appending text to the output variable, we can get away with having as few conditions as we have words.






for(var i = 1; i <= 100; i++) 
var output = "";
if (i % 3 === 0) output += "Fizz";
if (i % 5 === 0) output += "Buzz";
if (i % 7 === 0) output += "Jazz";

console.log(output === "" ? i : output);





(I've fixed a few other things as suggested in Sam's answer)



One thing that might be new to you here, if you're a beginner, is that the expression used as the argument for console.log or called the conditional or ternary operator. Ours says that if the output is blank (i.e. not a multiple of 3, 5 or 7) then print i, else print the string that we've compounded.



The ternary operator can always be replaced by an if-statement if you're not yet comfortable with it.






share|improve this answer











$endgroup$












  • $begingroup$
    @PatrickRoberts True. I wanted to focus more on the approach to solving the problem than those things. Buuut I did correct everything else (I couldn't help myself) so yeah I've made that change as well. Thanks
    $endgroup$
    – Michael
    Jan 10 at 16:15






  • 1




    $begingroup$
    You forgot the equality in the ternary expression. I tried to edit myself but I can only make edits of at least 6 characters :|
    $endgroup$
    – Patrick Roberts
    Jan 10 at 16:23


















15












$begingroup$

I think it's fine as is, though some folks like to return early instead of using a series of if..else. For example:



function calc(i) 
if(i % 15 == 0) return "FizzBuzz";
if(i % 3 == 0) return "Fizz";
if(i % 5 == 0) return "Buzz";
return i;


for(var i=1;i<=100;i++)
console.log(calc(i));






share|improve this answer











$endgroup$








  • 1




    $begingroup$
    "Returning early" is often called "multiple exit points" and is sometimes considered poor practice.
    $endgroup$
    – Moby Disk
    Jan 9 at 18:59






  • 6




    $begingroup$
    "Returning early" is often called "validation" and is often considered good practice.
    $endgroup$
    – nurettin
    Jan 10 at 11:22






  • 3




    $begingroup$
    Returning early has many names and people get very worked up about whether their way is best.
    $endgroup$
    – Tim B
    Jan 10 at 17:25






  • 1




    $begingroup$
    @MobyDisk That advice is originally from Assembly, and referred to "places to exit to", not "places to exit from". That removes the "it's always been done this way, so there must've been a reason" argument. Before applying your revised version of this advice to a particular scenario, check whether it actually makes sense. In this case, it doesn't.
    $endgroup$
    – wizzwizz4
    Jan 12 at 12:44


















5












$begingroup$

Expanding on Michael's answer, you could also create an object with all the words you want, with the key being the number your input must be divisible by, to make this more dynamic and a bit more future proof (A lot easier to add a new entry to the object versus adding more lines of code), and the object itself can also be dynamically generated.






divisions = 3: "Fizz", 5: "Buzz", 7: "Jazz";

for(var i = 1; i <= 100; i++)
var output = "";

for (var x in divisions)
if(i % x == 0) output += divisions[x];


console.log(output == "" ? i : output);








share|improve this answer











$endgroup$




















    2












    $begingroup$

    From a maintenance stand point, I think it's better that you check for 3 and 5 instead of checking for 15. The issue with checking for 15 is that when you come back to your code 6 months from now, you won't realize that you were just printing fizzbuzz when something was divisible by both 3 and 5. That might matter in the future, but I think that's something worth talking about in an interview.






    share|improve this answer











    $endgroup$




















      1












      $begingroup$

      Though your code is fine, another thing to look is amount of division happening the code, Division is computationally expensive operation and as @Michael has suggested that it is redundant. So always try to minimise the multiplication and division operation in your code.



      You have mentioned that you are a beginner programmer. I believe it is a good practice to start familiarising your self with topics like Computational Complexity.



      Look here for computational complexity of mathematical functions






      share|improve this answer











      $endgroup$












      • $begingroup$
        While I think it's a bit early for a beginning programmer to be learning about things such as memory management (assuming this is one of their FIRST actual programs), I think this is a good answer!
        $endgroup$
        – nostalgk
        Jan 9 at 20:10






      • 3




        $begingroup$
        This advice screams "premature optimization". Division is not an expensive operation. It's a bit more expensive that addition but it's not really expensive on today's computers. Your mobile phone can do millions of divisions in a fraction of a second. String concatenation is much more expensive, by far. The most important metric in code quality is readability. Also, if you are talking about computational complexity, you should note that your change won't change complexity at all.
        $endgroup$
        – Sulthan
        Jan 9 at 23:37







      • 1




        $begingroup$
        The theoretical computational complexity of the underlying division algorithm is practically irrelevant.
        $endgroup$
        – crasic
        Jan 10 at 3:05











      • $begingroup$
        @Sulthan The subject computational complexity doesn't take in to account how many operations a machine can perform, it takes into account how many operations/steps are required to solve a problem. While CC of addition i linear, division is quadratic. Dividing only by 3 and 5 would change the complexity and results of those can be saved in a bool variable which can be evaluated in constant time complexity.
        $endgroup$
        – Hemanshu
        Jan 10 at 13:58







      • 1




        $begingroup$
        On a modern CPU integer division takes 1 CPU cycle, the JS version of % a little longer. JS will use 32bit signed ints when it can so the operation is insignificant in comparison to just calling a function. BTW the linked computational complexity page has nothing to do with how CPUs ALU and FLU process the basic math operations.
        $endgroup$
        – Blindman67
        Jan 10 at 14:08


















      1












      $begingroup$

      I tend to be somewhat "literalist", at least at first, when implementing algorithms. So I would start with this, and then optimize from there. I like that it keeps things very clear.



      const calcFizzBuzz = function(num) 
      let result = "";
      if (num % 3 === 0)
      result += "Fizz";
      if (num % 5 === 0)
      result += "Buzz";


      else if (num % 5 === 0)
      result += "Buzz";

      else
      result = num;

      return result;


      for (let i = 0; i < 100; i++)
      console.log(calcFizzBuzz(i));






      share|improve this answer









      $endgroup$












        Your Answer





        StackExchange.ifUsing("editor", function ()
        return StackExchange.using("mathjaxEditing", function ()
        StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
        StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["\$", "\$"]]);
        );
        );
        , "mathjax-editing");

        StackExchange.ifUsing("editor", function ()
        StackExchange.using("externalEditor", function ()
        StackExchange.using("snippets", function ()
        StackExchange.snippets.init();
        );
        );
        , "code-snippets");

        StackExchange.ready(function()
        var channelOptions =
        tags: "".split(" "),
        id: "196"
        ;
        initTagRenderer("".split(" "), "".split(" "), channelOptions);

        StackExchange.using("externalEditor", function()
        // Have to fire editor after snippets, if snippets enabled
        if (StackExchange.settings.snippets.snippetsEnabled)
        StackExchange.using("snippets", function()
        createEditor();
        );

        else
        createEditor();

        );

        function createEditor()
        StackExchange.prepareEditor(
        heartbeatType: 'answer',
        autoActivateHeartbeat: false,
        convertImagesToLinks: false,
        noModals: true,
        showLowRepImageUploadWarning: true,
        reputationToPostImages: null,
        bindNavPrevention: true,
        postfix: "",
        imageUploader:
        brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
        contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
        allowUrls: true
        ,
        onDemand: true,
        discardSelector: ".discard-answer"
        ,immediatelyShowMarkdownHelp:true
        );



        );













        draft saved

        draft discarded


















        StackExchange.ready(
        function ()
        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcodereview.stackexchange.com%2fquestions%2f211113%2ffizzbuzz-in-javascript%23new-answer', 'question_page');

        );

        Post as a guest















        Required, but never shown

























        7 Answers
        7






        active

        oldest

        votes








        7 Answers
        7






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        47












        $begingroup$


        I'd like to see if this is a reasonable solution, or is just plain terrible.




        I wouldn't say it is "terrible" - mostly because it works and doesn't appear to be very inefficient. However, there are some improvements that can be made.




        • use strict equality comparison - i.e. === when comparing values. That way it won't need to convert the types.


        • Style Guide Consider following a style guide. Many common style guides advise separating keywords with a space - e.g. if ( instead of if(.


        • Use consistent indentation The first and last lines within the for loop are not indented, while the other lines between them are, though maybe it was the case that your code was indented properly but when you pasted here it was thrown off because of the markdown formatting...


        • Statements and blocks - If you are going to use block statements then expand them so each statement is on a separate line. Otherwise, just put the single statement without a block. Also use consistent statement termination. The first three conditional blocks don't have a semi-colon after the statement within the block, while the last one (in the else block) does.


        • Abstract logic into a function As Paul's answer suggests: you can put the core logic into a function that returns the output but doesn't handle outputting the value (e.g. to the console). This allows such code to be atomic and testable - congruent with the Single Responsibility Principle. Also, the return statement can eliminate the need for else keywords within a function. One drawback is that calling a function on each iteration may slow down operations but should be negligible for a set of numbers 1 through 100.

        Updated Code



        Consider the modified code below, utilizing the feedback above.



        Note: the inline console in the snippets is truncated to ~50 lines, but the complete console log should be visible in your browser console.






        function fizzBuzz(value) 
        if (value % 15 === 0) return "FizzBuzz";
        if (value % 3 === 0) return "Fizz";
        if (value % 5 === 0) return "Buzz";
        return value;

        for (var i = 1; i <= 100; i++)
        console.log(fizzBuzz(i));





        One option I did consider is to minimize the number of modulo operations, append to the output string instead of outputting it. If you are trying to optimize for speed, this might not be the approach to take because appending to the string may be much slower than doing an extra couple modulo operations. Try a comparison in this jsPerf test.






        function fizzBuzz(value) 
        for (var i = 1; i <= 100; i++)
        console.log(fizzBuzz(i));








        share|improve this answer











        $endgroup$








        • 3




          $begingroup$
          There is also the "never use semicolons" school. blog.izs.me/2010/12/… feross.org/never-use-semicolons
          $endgroup$
          – Džuris
          Jan 8 at 22:01






        • 4




          $begingroup$
          @CaseyB I don't think arguing about micro-optimizations like this when OP is a beginner is a good idea at all.
          $endgroup$
          – Voile
          Jan 9 at 9:43






        • 4




          $begingroup$
          @Voile One million percent. Even if this were a production-ready, enterprise-grade, cloud-based microservice™, readability and maintainability is always the number one concern. The concept of micro-optimizing a JavaScript FizzBuzz is honestly hilarious.
          $endgroup$
          – Michael
          Jan 9 at 10:34






        • 7




          $begingroup$
          @Michael FizzBuzz Enterprise Edition begs to differ. :P But FWIW, I believe Casey was merely providing a counterargument for the microoptimization that the answer itself first suggested.
          $endgroup$
          – Mateen Ulhaq
          Jan 9 at 12:05






        • 2




          $begingroup$
          @MateenUlhaq 260 issues... man, they really need to refactor it, arrange UATs, and use Kanban.
          $endgroup$
          – Juha Untinen
          Jan 9 at 13:10















        47












        $begingroup$


        I'd like to see if this is a reasonable solution, or is just plain terrible.




        I wouldn't say it is "terrible" - mostly because it works and doesn't appear to be very inefficient. However, there are some improvements that can be made.




        • use strict equality comparison - i.e. === when comparing values. That way it won't need to convert the types.


        • Style Guide Consider following a style guide. Many common style guides advise separating keywords with a space - e.g. if ( instead of if(.


        • Use consistent indentation The first and last lines within the for loop are not indented, while the other lines between them are, though maybe it was the case that your code was indented properly but when you pasted here it was thrown off because of the markdown formatting...


        • Statements and blocks - If you are going to use block statements then expand them so each statement is on a separate line. Otherwise, just put the single statement without a block. Also use consistent statement termination. The first three conditional blocks don't have a semi-colon after the statement within the block, while the last one (in the else block) does.


        • Abstract logic into a function As Paul's answer suggests: you can put the core logic into a function that returns the output but doesn't handle outputting the value (e.g. to the console). This allows such code to be atomic and testable - congruent with the Single Responsibility Principle. Also, the return statement can eliminate the need for else keywords within a function. One drawback is that calling a function on each iteration may slow down operations but should be negligible for a set of numbers 1 through 100.

        Updated Code



        Consider the modified code below, utilizing the feedback above.



        Note: the inline console in the snippets is truncated to ~50 lines, but the complete console log should be visible in your browser console.






        function fizzBuzz(value) 
        if (value % 15 === 0) return "FizzBuzz";
        if (value % 3 === 0) return "Fizz";
        if (value % 5 === 0) return "Buzz";
        return value;

        for (var i = 1; i <= 100; i++)
        console.log(fizzBuzz(i));





        One option I did consider is to minimize the number of modulo operations, append to the output string instead of outputting it. If you are trying to optimize for speed, this might not be the approach to take because appending to the string may be much slower than doing an extra couple modulo operations. Try a comparison in this jsPerf test.






        function fizzBuzz(value) 
        for (var i = 1; i <= 100; i++)
        console.log(fizzBuzz(i));








        share|improve this answer











        $endgroup$








        • 3




          $begingroup$
          There is also the "never use semicolons" school. blog.izs.me/2010/12/… feross.org/never-use-semicolons
          $endgroup$
          – Džuris
          Jan 8 at 22:01






        • 4




          $begingroup$
          @CaseyB I don't think arguing about micro-optimizations like this when OP is a beginner is a good idea at all.
          $endgroup$
          – Voile
          Jan 9 at 9:43






        • 4




          $begingroup$
          @Voile One million percent. Even if this were a production-ready, enterprise-grade, cloud-based microservice™, readability and maintainability is always the number one concern. The concept of micro-optimizing a JavaScript FizzBuzz is honestly hilarious.
          $endgroup$
          – Michael
          Jan 9 at 10:34






        • 7




          $begingroup$
          @Michael FizzBuzz Enterprise Edition begs to differ. :P But FWIW, I believe Casey was merely providing a counterargument for the microoptimization that the answer itself first suggested.
          $endgroup$
          – Mateen Ulhaq
          Jan 9 at 12:05






        • 2




          $begingroup$
          @MateenUlhaq 260 issues... man, they really need to refactor it, arrange UATs, and use Kanban.
          $endgroup$
          – Juha Untinen
          Jan 9 at 13:10













        47












        47








        47





        $begingroup$


        I'd like to see if this is a reasonable solution, or is just plain terrible.




        I wouldn't say it is "terrible" - mostly because it works and doesn't appear to be very inefficient. However, there are some improvements that can be made.




        • use strict equality comparison - i.e. === when comparing values. That way it won't need to convert the types.


        • Style Guide Consider following a style guide. Many common style guides advise separating keywords with a space - e.g. if ( instead of if(.


        • Use consistent indentation The first and last lines within the for loop are not indented, while the other lines between them are, though maybe it was the case that your code was indented properly but when you pasted here it was thrown off because of the markdown formatting...


        • Statements and blocks - If you are going to use block statements then expand them so each statement is on a separate line. Otherwise, just put the single statement without a block. Also use consistent statement termination. The first three conditional blocks don't have a semi-colon after the statement within the block, while the last one (in the else block) does.


        • Abstract logic into a function As Paul's answer suggests: you can put the core logic into a function that returns the output but doesn't handle outputting the value (e.g. to the console). This allows such code to be atomic and testable - congruent with the Single Responsibility Principle. Also, the return statement can eliminate the need for else keywords within a function. One drawback is that calling a function on each iteration may slow down operations but should be negligible for a set of numbers 1 through 100.

        Updated Code



        Consider the modified code below, utilizing the feedback above.



        Note: the inline console in the snippets is truncated to ~50 lines, but the complete console log should be visible in your browser console.






        function fizzBuzz(value) 
        if (value % 15 === 0) return "FizzBuzz";
        if (value % 3 === 0) return "Fizz";
        if (value % 5 === 0) return "Buzz";
        return value;

        for (var i = 1; i <= 100; i++)
        console.log(fizzBuzz(i));





        One option I did consider is to minimize the number of modulo operations, append to the output string instead of outputting it. If you are trying to optimize for speed, this might not be the approach to take because appending to the string may be much slower than doing an extra couple modulo operations. Try a comparison in this jsPerf test.






        function fizzBuzz(value) 
        for (var i = 1; i <= 100; i++)
        console.log(fizzBuzz(i));








        share|improve this answer











        $endgroup$




        I'd like to see if this is a reasonable solution, or is just plain terrible.




        I wouldn't say it is "terrible" - mostly because it works and doesn't appear to be very inefficient. However, there are some improvements that can be made.




        • use strict equality comparison - i.e. === when comparing values. That way it won't need to convert the types.


        • Style Guide Consider following a style guide. Many common style guides advise separating keywords with a space - e.g. if ( instead of if(.


        • Use consistent indentation The first and last lines within the for loop are not indented, while the other lines between them are, though maybe it was the case that your code was indented properly but when you pasted here it was thrown off because of the markdown formatting...


        • Statements and blocks - If you are going to use block statements then expand them so each statement is on a separate line. Otherwise, just put the single statement without a block. Also use consistent statement termination. The first three conditional blocks don't have a semi-colon after the statement within the block, while the last one (in the else block) does.


        • Abstract logic into a function As Paul's answer suggests: you can put the core logic into a function that returns the output but doesn't handle outputting the value (e.g. to the console). This allows such code to be atomic and testable - congruent with the Single Responsibility Principle. Also, the return statement can eliminate the need for else keywords within a function. One drawback is that calling a function on each iteration may slow down operations but should be negligible for a set of numbers 1 through 100.

        Updated Code



        Consider the modified code below, utilizing the feedback above.



        Note: the inline console in the snippets is truncated to ~50 lines, but the complete console log should be visible in your browser console.






        function fizzBuzz(value) 
        if (value % 15 === 0) return "FizzBuzz";
        if (value % 3 === 0) return "Fizz";
        if (value % 5 === 0) return "Buzz";
        return value;

        for (var i = 1; i <= 100; i++)
        console.log(fizzBuzz(i));





        One option I did consider is to minimize the number of modulo operations, append to the output string instead of outputting it. If you are trying to optimize for speed, this might not be the approach to take because appending to the string may be much slower than doing an extra couple modulo operations. Try a comparison in this jsPerf test.






        function fizzBuzz(value) 
        for (var i = 1; i <= 100; i++)
        console.log(fizzBuzz(i));








        function fizzBuzz(value) 
        if (value % 15 === 0) return "FizzBuzz";
        if (value % 3 === 0) return "Fizz";
        if (value % 5 === 0) return "Buzz";
        return value;

        for (var i = 1; i <= 100; i++)
        console.log(fizzBuzz(i));





        function fizzBuzz(value) 
        if (value % 15 === 0) return "FizzBuzz";
        if (value % 3 === 0) return "Fizz";
        if (value % 5 === 0) return "Buzz";
        return value;

        for (var i = 1; i <= 100; i++)
        console.log(fizzBuzz(i));





        function fizzBuzz(value) 
        for (var i = 1; i <= 100; i++)
        console.log(fizzBuzz(i));





        function fizzBuzz(value) 
        for (var i = 1; i <= 100; i++)
        console.log(fizzBuzz(i));






        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Jan 11 at 16:19

























        answered Jan 8 at 16:54









        Sᴀᴍ OnᴇᴌᴀSᴀᴍ Onᴇᴌᴀ

        9,32262161




        9,32262161







        • 3




          $begingroup$
          There is also the "never use semicolons" school. blog.izs.me/2010/12/… feross.org/never-use-semicolons
          $endgroup$
          – Džuris
          Jan 8 at 22:01






        • 4




          $begingroup$
          @CaseyB I don't think arguing about micro-optimizations like this when OP is a beginner is a good idea at all.
          $endgroup$
          – Voile
          Jan 9 at 9:43






        • 4




          $begingroup$
          @Voile One million percent. Even if this were a production-ready, enterprise-grade, cloud-based microservice™, readability and maintainability is always the number one concern. The concept of micro-optimizing a JavaScript FizzBuzz is honestly hilarious.
          $endgroup$
          – Michael
          Jan 9 at 10:34






        • 7




          $begingroup$
          @Michael FizzBuzz Enterprise Edition begs to differ. :P But FWIW, I believe Casey was merely providing a counterargument for the microoptimization that the answer itself first suggested.
          $endgroup$
          – Mateen Ulhaq
          Jan 9 at 12:05






        • 2




          $begingroup$
          @MateenUlhaq 260 issues... man, they really need to refactor it, arrange UATs, and use Kanban.
          $endgroup$
          – Juha Untinen
          Jan 9 at 13:10












        • 3




          $begingroup$
          There is also the "never use semicolons" school. blog.izs.me/2010/12/… feross.org/never-use-semicolons
          $endgroup$
          – Džuris
          Jan 8 at 22:01






        • 4




          $begingroup$
          @CaseyB I don't think arguing about micro-optimizations like this when OP is a beginner is a good idea at all.
          $endgroup$
          – Voile
          Jan 9 at 9:43






        • 4




          $begingroup$
          @Voile One million percent. Even if this were a production-ready, enterprise-grade, cloud-based microservice™, readability and maintainability is always the number one concern. The concept of micro-optimizing a JavaScript FizzBuzz is honestly hilarious.
          $endgroup$
          – Michael
          Jan 9 at 10:34






        • 7




          $begingroup$
          @Michael FizzBuzz Enterprise Edition begs to differ. :P But FWIW, I believe Casey was merely providing a counterargument for the microoptimization that the answer itself first suggested.
          $endgroup$
          – Mateen Ulhaq
          Jan 9 at 12:05






        • 2




          $begingroup$
          @MateenUlhaq 260 issues... man, they really need to refactor it, arrange UATs, and use Kanban.
          $endgroup$
          – Juha Untinen
          Jan 9 at 13:10







        3




        3




        $begingroup$
        There is also the "never use semicolons" school. blog.izs.me/2010/12/… feross.org/never-use-semicolons
        $endgroup$
        – Džuris
        Jan 8 at 22:01




        $begingroup$
        There is also the "never use semicolons" school. blog.izs.me/2010/12/… feross.org/never-use-semicolons
        $endgroup$
        – Džuris
        Jan 8 at 22:01




        4




        4




        $begingroup$
        @CaseyB I don't think arguing about micro-optimizations like this when OP is a beginner is a good idea at all.
        $endgroup$
        – Voile
        Jan 9 at 9:43




        $begingroup$
        @CaseyB I don't think arguing about micro-optimizations like this when OP is a beginner is a good idea at all.
        $endgroup$
        – Voile
        Jan 9 at 9:43




        4




        4




        $begingroup$
        @Voile One million percent. Even if this were a production-ready, enterprise-grade, cloud-based microservice™, readability and maintainability is always the number one concern. The concept of micro-optimizing a JavaScript FizzBuzz is honestly hilarious.
        $endgroup$
        – Michael
        Jan 9 at 10:34




        $begingroup$
        @Voile One million percent. Even if this were a production-ready, enterprise-grade, cloud-based microservice™, readability and maintainability is always the number one concern. The concept of micro-optimizing a JavaScript FizzBuzz is honestly hilarious.
        $endgroup$
        – Michael
        Jan 9 at 10:34




        7




        7




        $begingroup$
        @Michael FizzBuzz Enterprise Edition begs to differ. :P But FWIW, I believe Casey was merely providing a counterargument for the microoptimization that the answer itself first suggested.
        $endgroup$
        – Mateen Ulhaq
        Jan 9 at 12:05




        $begingroup$
        @Michael FizzBuzz Enterprise Edition begs to differ. :P But FWIW, I believe Casey was merely providing a counterargument for the microoptimization that the answer itself first suggested.
        $endgroup$
        – Mateen Ulhaq
        Jan 9 at 12:05




        2




        2




        $begingroup$
        @MateenUlhaq 260 issues... man, they really need to refactor it, arrange UATs, and use Kanban.
        $endgroup$
        – Juha Untinen
        Jan 9 at 13:10




        $begingroup$
        @MateenUlhaq 260 issues... man, they really need to refactor it, arrange UATs, and use Kanban.
        $endgroup$
        – Juha Untinen
        Jan 9 at 13:10













        25












        $begingroup$

        One of the problems is that the case where you check i % 15 (i.e. i is a multiple of 3 and 5) is unnecessary. You have the concept of 3 and 5 repeated, and the concept of Fizz and Buzz repeated.



        This is not currently much of a problem but suppose someone asks you to extend your program to print "Jazz" when i is a multiple of 7. Using your current strategy we now need a case where i is a multiple of:



        1. 3

        2. 5

        3. 7

        4. 3 and 5

        5. 3 and 7

        6. 5 and 7

        7. 3 and 5 and 7

        It would look something like this:






        for(var i = 1;i <= 100; i++) 
        var output = "";

        if(i % 105 == 0) output = "FizzBuzzJazz"
        else if(i % 15 == 0) output = "FizzBuzz"
        else if(i % 35 == 0) output = "BuzzJazz"
        else if(i % 21 == 0) output = "FizzJazz"
        else if(i % 3 == 0) output = "Fizz"
        else if(i % 5 == 0) output = "Buzz"
        else if(i % 7 == 0) output = "Jazz"
        else output = i;

        console.log(output);





        See how quickly that got out of hand? If we add a fourth word it becomes even worse.



        If we use a different strategy by appending text to the output variable, we can get away with having as few conditions as we have words.






        for(var i = 1; i <= 100; i++) 
        var output = "";
        if (i % 3 === 0) output += "Fizz";
        if (i % 5 === 0) output += "Buzz";
        if (i % 7 === 0) output += "Jazz";

        console.log(output === "" ? i : output);





        (I've fixed a few other things as suggested in Sam's answer)



        One thing that might be new to you here, if you're a beginner, is that the expression used as the argument for console.log or called the conditional or ternary operator. Ours says that if the output is blank (i.e. not a multiple of 3, 5 or 7) then print i, else print the string that we've compounded.



        The ternary operator can always be replaced by an if-statement if you're not yet comfortable with it.






        share|improve this answer











        $endgroup$












        • $begingroup$
          @PatrickRoberts True. I wanted to focus more on the approach to solving the problem than those things. Buuut I did correct everything else (I couldn't help myself) so yeah I've made that change as well. Thanks
          $endgroup$
          – Michael
          Jan 10 at 16:15






        • 1




          $begingroup$
          You forgot the equality in the ternary expression. I tried to edit myself but I can only make edits of at least 6 characters :|
          $endgroup$
          – Patrick Roberts
          Jan 10 at 16:23















        25












        $begingroup$

        One of the problems is that the case where you check i % 15 (i.e. i is a multiple of 3 and 5) is unnecessary. You have the concept of 3 and 5 repeated, and the concept of Fizz and Buzz repeated.



        This is not currently much of a problem but suppose someone asks you to extend your program to print "Jazz" when i is a multiple of 7. Using your current strategy we now need a case where i is a multiple of:



        1. 3

        2. 5

        3. 7

        4. 3 and 5

        5. 3 and 7

        6. 5 and 7

        7. 3 and 5 and 7

        It would look something like this:






        for(var i = 1;i <= 100; i++) 
        var output = "";

        if(i % 105 == 0) output = "FizzBuzzJazz"
        else if(i % 15 == 0) output = "FizzBuzz"
        else if(i % 35 == 0) output = "BuzzJazz"
        else if(i % 21 == 0) output = "FizzJazz"
        else if(i % 3 == 0) output = "Fizz"
        else if(i % 5 == 0) output = "Buzz"
        else if(i % 7 == 0) output = "Jazz"
        else output = i;

        console.log(output);





        See how quickly that got out of hand? If we add a fourth word it becomes even worse.



        If we use a different strategy by appending text to the output variable, we can get away with having as few conditions as we have words.






        for(var i = 1; i <= 100; i++) 
        var output = "";
        if (i % 3 === 0) output += "Fizz";
        if (i % 5 === 0) output += "Buzz";
        if (i % 7 === 0) output += "Jazz";

        console.log(output === "" ? i : output);





        (I've fixed a few other things as suggested in Sam's answer)



        One thing that might be new to you here, if you're a beginner, is that the expression used as the argument for console.log or called the conditional or ternary operator. Ours says that if the output is blank (i.e. not a multiple of 3, 5 or 7) then print i, else print the string that we've compounded.



        The ternary operator can always be replaced by an if-statement if you're not yet comfortable with it.






        share|improve this answer











        $endgroup$












        • $begingroup$
          @PatrickRoberts True. I wanted to focus more on the approach to solving the problem than those things. Buuut I did correct everything else (I couldn't help myself) so yeah I've made that change as well. Thanks
          $endgroup$
          – Michael
          Jan 10 at 16:15






        • 1




          $begingroup$
          You forgot the equality in the ternary expression. I tried to edit myself but I can only make edits of at least 6 characters :|
          $endgroup$
          – Patrick Roberts
          Jan 10 at 16:23













        25












        25








        25





        $begingroup$

        One of the problems is that the case where you check i % 15 (i.e. i is a multiple of 3 and 5) is unnecessary. You have the concept of 3 and 5 repeated, and the concept of Fizz and Buzz repeated.



        This is not currently much of a problem but suppose someone asks you to extend your program to print "Jazz" when i is a multiple of 7. Using your current strategy we now need a case where i is a multiple of:



        1. 3

        2. 5

        3. 7

        4. 3 and 5

        5. 3 and 7

        6. 5 and 7

        7. 3 and 5 and 7

        It would look something like this:






        for(var i = 1;i <= 100; i++) 
        var output = "";

        if(i % 105 == 0) output = "FizzBuzzJazz"
        else if(i % 15 == 0) output = "FizzBuzz"
        else if(i % 35 == 0) output = "BuzzJazz"
        else if(i % 21 == 0) output = "FizzJazz"
        else if(i % 3 == 0) output = "Fizz"
        else if(i % 5 == 0) output = "Buzz"
        else if(i % 7 == 0) output = "Jazz"
        else output = i;

        console.log(output);





        See how quickly that got out of hand? If we add a fourth word it becomes even worse.



        If we use a different strategy by appending text to the output variable, we can get away with having as few conditions as we have words.






        for(var i = 1; i <= 100; i++) 
        var output = "";
        if (i % 3 === 0) output += "Fizz";
        if (i % 5 === 0) output += "Buzz";
        if (i % 7 === 0) output += "Jazz";

        console.log(output === "" ? i : output);





        (I've fixed a few other things as suggested in Sam's answer)



        One thing that might be new to you here, if you're a beginner, is that the expression used as the argument for console.log or called the conditional or ternary operator. Ours says that if the output is blank (i.e. not a multiple of 3, 5 or 7) then print i, else print the string that we've compounded.



        The ternary operator can always be replaced by an if-statement if you're not yet comfortable with it.






        share|improve this answer











        $endgroup$



        One of the problems is that the case where you check i % 15 (i.e. i is a multiple of 3 and 5) is unnecessary. You have the concept of 3 and 5 repeated, and the concept of Fizz and Buzz repeated.



        This is not currently much of a problem but suppose someone asks you to extend your program to print "Jazz" when i is a multiple of 7. Using your current strategy we now need a case where i is a multiple of:



        1. 3

        2. 5

        3. 7

        4. 3 and 5

        5. 3 and 7

        6. 5 and 7

        7. 3 and 5 and 7

        It would look something like this:






        for(var i = 1;i <= 100; i++) 
        var output = "";

        if(i % 105 == 0) output = "FizzBuzzJazz"
        else if(i % 15 == 0) output = "FizzBuzz"
        else if(i % 35 == 0) output = "BuzzJazz"
        else if(i % 21 == 0) output = "FizzJazz"
        else if(i % 3 == 0) output = "Fizz"
        else if(i % 5 == 0) output = "Buzz"
        else if(i % 7 == 0) output = "Jazz"
        else output = i;

        console.log(output);





        See how quickly that got out of hand? If we add a fourth word it becomes even worse.



        If we use a different strategy by appending text to the output variable, we can get away with having as few conditions as we have words.






        for(var i = 1; i <= 100; i++) 
        var output = "";
        if (i % 3 === 0) output += "Fizz";
        if (i % 5 === 0) output += "Buzz";
        if (i % 7 === 0) output += "Jazz";

        console.log(output === "" ? i : output);





        (I've fixed a few other things as suggested in Sam's answer)



        One thing that might be new to you here, if you're a beginner, is that the expression used as the argument for console.log or called the conditional or ternary operator. Ours says that if the output is blank (i.e. not a multiple of 3, 5 or 7) then print i, else print the string that we've compounded.



        The ternary operator can always be replaced by an if-statement if you're not yet comfortable with it.






        for(var i = 1;i <= 100; i++) 
        var output = "";

        if(i % 105 == 0) output = "FizzBuzzJazz"
        else if(i % 15 == 0) output = "FizzBuzz"
        else if(i % 35 == 0) output = "BuzzJazz"
        else if(i % 21 == 0) output = "FizzJazz"
        else if(i % 3 == 0) output = "Fizz"
        else if(i % 5 == 0) output = "Buzz"
        else if(i % 7 == 0) output = "Jazz"
        else output = i;

        console.log(output);





        for(var i = 1;i <= 100; i++) 
        var output = "";

        if(i % 105 == 0) output = "FizzBuzzJazz"
        else if(i % 15 == 0) output = "FizzBuzz"
        else if(i % 35 == 0) output = "BuzzJazz"
        else if(i % 21 == 0) output = "FizzJazz"
        else if(i % 3 == 0) output = "Fizz"
        else if(i % 5 == 0) output = "Buzz"
        else if(i % 7 == 0) output = "Jazz"
        else output = i;

        console.log(output);





        for(var i = 1; i <= 100; i++) 
        var output = "";
        if (i % 3 === 0) output += "Fizz";
        if (i % 5 === 0) output += "Buzz";
        if (i % 7 === 0) output += "Jazz";

        console.log(output === "" ? i : output);





        for(var i = 1; i <= 100; i++) 
        var output = "";
        if (i % 3 === 0) output += "Fizz";
        if (i % 5 === 0) output += "Buzz";
        if (i % 7 === 0) output += "Jazz";

        console.log(output === "" ? i : output);






        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Jan 10 at 16:40

























        answered Jan 9 at 10:22









        MichaelMichael

        698410




        698410











        • $begingroup$
          @PatrickRoberts True. I wanted to focus more on the approach to solving the problem than those things. Buuut I did correct everything else (I couldn't help myself) so yeah I've made that change as well. Thanks
          $endgroup$
          – Michael
          Jan 10 at 16:15






        • 1




          $begingroup$
          You forgot the equality in the ternary expression. I tried to edit myself but I can only make edits of at least 6 characters :|
          $endgroup$
          – Patrick Roberts
          Jan 10 at 16:23
















        • $begingroup$
          @PatrickRoberts True. I wanted to focus more on the approach to solving the problem than those things. Buuut I did correct everything else (I couldn't help myself) so yeah I've made that change as well. Thanks
          $endgroup$
          – Michael
          Jan 10 at 16:15






        • 1




          $begingroup$
          You forgot the equality in the ternary expression. I tried to edit myself but I can only make edits of at least 6 characters :|
          $endgroup$
          – Patrick Roberts
          Jan 10 at 16:23















        $begingroup$
        @PatrickRoberts True. I wanted to focus more on the approach to solving the problem than those things. Buuut I did correct everything else (I couldn't help myself) so yeah I've made that change as well. Thanks
        $endgroup$
        – Michael
        Jan 10 at 16:15




        $begingroup$
        @PatrickRoberts True. I wanted to focus more on the approach to solving the problem than those things. Buuut I did correct everything else (I couldn't help myself) so yeah I've made that change as well. Thanks
        $endgroup$
        – Michael
        Jan 10 at 16:15




        1




        1




        $begingroup$
        You forgot the equality in the ternary expression. I tried to edit myself but I can only make edits of at least 6 characters :|
        $endgroup$
        – Patrick Roberts
        Jan 10 at 16:23




        $begingroup$
        You forgot the equality in the ternary expression. I tried to edit myself but I can only make edits of at least 6 characters :|
        $endgroup$
        – Patrick Roberts
        Jan 10 at 16:23











        15












        $begingroup$

        I think it's fine as is, though some folks like to return early instead of using a series of if..else. For example:



        function calc(i) 
        if(i % 15 == 0) return "FizzBuzz";
        if(i % 3 == 0) return "Fizz";
        if(i % 5 == 0) return "Buzz";
        return i;


        for(var i=1;i<=100;i++)
        console.log(calc(i));






        share|improve this answer











        $endgroup$








        • 1




          $begingroup$
          "Returning early" is often called "multiple exit points" and is sometimes considered poor practice.
          $endgroup$
          – Moby Disk
          Jan 9 at 18:59






        • 6




          $begingroup$
          "Returning early" is often called "validation" and is often considered good practice.
          $endgroup$
          – nurettin
          Jan 10 at 11:22






        • 3




          $begingroup$
          Returning early has many names and people get very worked up about whether their way is best.
          $endgroup$
          – Tim B
          Jan 10 at 17:25






        • 1




          $begingroup$
          @MobyDisk That advice is originally from Assembly, and referred to "places to exit to", not "places to exit from". That removes the "it's always been done this way, so there must've been a reason" argument. Before applying your revised version of this advice to a particular scenario, check whether it actually makes sense. In this case, it doesn't.
          $endgroup$
          – wizzwizz4
          Jan 12 at 12:44















        15












        $begingroup$

        I think it's fine as is, though some folks like to return early instead of using a series of if..else. For example:



        function calc(i) 
        if(i % 15 == 0) return "FizzBuzz";
        if(i % 3 == 0) return "Fizz";
        if(i % 5 == 0) return "Buzz";
        return i;


        for(var i=1;i<=100;i++)
        console.log(calc(i));






        share|improve this answer











        $endgroup$








        • 1




          $begingroup$
          "Returning early" is often called "multiple exit points" and is sometimes considered poor practice.
          $endgroup$
          – Moby Disk
          Jan 9 at 18:59






        • 6




          $begingroup$
          "Returning early" is often called "validation" and is often considered good practice.
          $endgroup$
          – nurettin
          Jan 10 at 11:22






        • 3




          $begingroup$
          Returning early has many names and people get very worked up about whether their way is best.
          $endgroup$
          – Tim B
          Jan 10 at 17:25






        • 1




          $begingroup$
          @MobyDisk That advice is originally from Assembly, and referred to "places to exit to", not "places to exit from". That removes the "it's always been done this way, so there must've been a reason" argument. Before applying your revised version of this advice to a particular scenario, check whether it actually makes sense. In this case, it doesn't.
          $endgroup$
          – wizzwizz4
          Jan 12 at 12:44













        15












        15








        15





        $begingroup$

        I think it's fine as is, though some folks like to return early instead of using a series of if..else. For example:



        function calc(i) 
        if(i % 15 == 0) return "FizzBuzz";
        if(i % 3 == 0) return "Fizz";
        if(i % 5 == 0) return "Buzz";
        return i;


        for(var i=1;i<=100;i++)
        console.log(calc(i));






        share|improve this answer











        $endgroup$



        I think it's fine as is, though some folks like to return early instead of using a series of if..else. For example:



        function calc(i) 
        if(i % 15 == 0) return "FizzBuzz";
        if(i % 3 == 0) return "Fizz";
        if(i % 5 == 0) return "Buzz";
        return i;


        for(var i=1;i<=100;i++)
        console.log(calc(i));







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Jan 8 at 16:33

























        answered Jan 8 at 15:51









        PaulPaul

        31115




        31115







        • 1




          $begingroup$
          "Returning early" is often called "multiple exit points" and is sometimes considered poor practice.
          $endgroup$
          – Moby Disk
          Jan 9 at 18:59






        • 6




          $begingroup$
          "Returning early" is often called "validation" and is often considered good practice.
          $endgroup$
          – nurettin
          Jan 10 at 11:22






        • 3




          $begingroup$
          Returning early has many names and people get very worked up about whether their way is best.
          $endgroup$
          – Tim B
          Jan 10 at 17:25






        • 1




          $begingroup$
          @MobyDisk That advice is originally from Assembly, and referred to "places to exit to", not "places to exit from". That removes the "it's always been done this way, so there must've been a reason" argument. Before applying your revised version of this advice to a particular scenario, check whether it actually makes sense. In this case, it doesn't.
          $endgroup$
          – wizzwizz4
          Jan 12 at 12:44












        • 1




          $begingroup$
          "Returning early" is often called "multiple exit points" and is sometimes considered poor practice.
          $endgroup$
          – Moby Disk
          Jan 9 at 18:59






        • 6




          $begingroup$
          "Returning early" is often called "validation" and is often considered good practice.
          $endgroup$
          – nurettin
          Jan 10 at 11:22






        • 3




          $begingroup$
          Returning early has many names and people get very worked up about whether their way is best.
          $endgroup$
          – Tim B
          Jan 10 at 17:25






        • 1




          $begingroup$
          @MobyDisk That advice is originally from Assembly, and referred to "places to exit to", not "places to exit from". That removes the "it's always been done this way, so there must've been a reason" argument. Before applying your revised version of this advice to a particular scenario, check whether it actually makes sense. In this case, it doesn't.
          $endgroup$
          – wizzwizz4
          Jan 12 at 12:44







        1




        1




        $begingroup$
        "Returning early" is often called "multiple exit points" and is sometimes considered poor practice.
        $endgroup$
        – Moby Disk
        Jan 9 at 18:59




        $begingroup$
        "Returning early" is often called "multiple exit points" and is sometimes considered poor practice.
        $endgroup$
        – Moby Disk
        Jan 9 at 18:59




        6




        6




        $begingroup$
        "Returning early" is often called "validation" and is often considered good practice.
        $endgroup$
        – nurettin
        Jan 10 at 11:22




        $begingroup$
        "Returning early" is often called "validation" and is often considered good practice.
        $endgroup$
        – nurettin
        Jan 10 at 11:22




        3




        3




        $begingroup$
        Returning early has many names and people get very worked up about whether their way is best.
        $endgroup$
        – Tim B
        Jan 10 at 17:25




        $begingroup$
        Returning early has many names and people get very worked up about whether their way is best.
        $endgroup$
        – Tim B
        Jan 10 at 17:25




        1




        1




        $begingroup$
        @MobyDisk That advice is originally from Assembly, and referred to "places to exit to", not "places to exit from". That removes the "it's always been done this way, so there must've been a reason" argument. Before applying your revised version of this advice to a particular scenario, check whether it actually makes sense. In this case, it doesn't.
        $endgroup$
        – wizzwizz4
        Jan 12 at 12:44




        $begingroup$
        @MobyDisk That advice is originally from Assembly, and referred to "places to exit to", not "places to exit from". That removes the "it's always been done this way, so there must've been a reason" argument. Before applying your revised version of this advice to a particular scenario, check whether it actually makes sense. In this case, it doesn't.
        $endgroup$
        – wizzwizz4
        Jan 12 at 12:44











        5












        $begingroup$

        Expanding on Michael's answer, you could also create an object with all the words you want, with the key being the number your input must be divisible by, to make this more dynamic and a bit more future proof (A lot easier to add a new entry to the object versus adding more lines of code), and the object itself can also be dynamically generated.






        divisions = 3: "Fizz", 5: "Buzz", 7: "Jazz";

        for(var i = 1; i <= 100; i++)
        var output = "";

        for (var x in divisions)
        if(i % x == 0) output += divisions[x];


        console.log(output == "" ? i : output);








        share|improve this answer











        $endgroup$

















          5












          $begingroup$

          Expanding on Michael's answer, you could also create an object with all the words you want, with the key being the number your input must be divisible by, to make this more dynamic and a bit more future proof (A lot easier to add a new entry to the object versus adding more lines of code), and the object itself can also be dynamically generated.






          divisions = 3: "Fizz", 5: "Buzz", 7: "Jazz";

          for(var i = 1; i <= 100; i++)
          var output = "";

          for (var x in divisions)
          if(i % x == 0) output += divisions[x];


          console.log(output == "" ? i : output);








          share|improve this answer











          $endgroup$















            5












            5








            5





            $begingroup$

            Expanding on Michael's answer, you could also create an object with all the words you want, with the key being the number your input must be divisible by, to make this more dynamic and a bit more future proof (A lot easier to add a new entry to the object versus adding more lines of code), and the object itself can also be dynamically generated.






            divisions = 3: "Fizz", 5: "Buzz", 7: "Jazz";

            for(var i = 1; i <= 100; i++)
            var output = "";

            for (var x in divisions)
            if(i % x == 0) output += divisions[x];


            console.log(output == "" ? i : output);








            share|improve this answer











            $endgroup$



            Expanding on Michael's answer, you could also create an object with all the words you want, with the key being the number your input must be divisible by, to make this more dynamic and a bit more future proof (A lot easier to add a new entry to the object versus adding more lines of code), and the object itself can also be dynamically generated.






            divisions = 3: "Fizz", 5: "Buzz", 7: "Jazz";

            for(var i = 1; i <= 100; i++)
            var output = "";

            for (var x in divisions)
            if(i % x == 0) output += divisions[x];


            console.log(output == "" ? i : output);








            divisions = 3: "Fizz", 5: "Buzz", 7: "Jazz";

            for(var i = 1; i <= 100; i++)
            var output = "";

            for (var x in divisions)
            if(i % x == 0) output += divisions[x];


            console.log(output == "" ? i : output);





            divisions = 3: "Fizz", 5: "Buzz", 7: "Jazz";

            for(var i = 1; i <= 100; i++)
            var output = "";

            for (var x in divisions)
            if(i % x == 0) output += divisions[x];


            console.log(output == "" ? i : output);






            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited Jan 9 at 21:07

























            answered Jan 9 at 21:00









            GrumpyCroutonGrumpyCrouton

            261314




            261314





















                2












                $begingroup$

                From a maintenance stand point, I think it's better that you check for 3 and 5 instead of checking for 15. The issue with checking for 15 is that when you come back to your code 6 months from now, you won't realize that you were just printing fizzbuzz when something was divisible by both 3 and 5. That might matter in the future, but I think that's something worth talking about in an interview.






                share|improve this answer











                $endgroup$

















                  2












                  $begingroup$

                  From a maintenance stand point, I think it's better that you check for 3 and 5 instead of checking for 15. The issue with checking for 15 is that when you come back to your code 6 months from now, you won't realize that you were just printing fizzbuzz when something was divisible by both 3 and 5. That might matter in the future, but I think that's something worth talking about in an interview.






                  share|improve this answer











                  $endgroup$















                    2












                    2








                    2





                    $begingroup$

                    From a maintenance stand point, I think it's better that you check for 3 and 5 instead of checking for 15. The issue with checking for 15 is that when you come back to your code 6 months from now, you won't realize that you were just printing fizzbuzz when something was divisible by both 3 and 5. That might matter in the future, but I think that's something worth talking about in an interview.






                    share|improve this answer











                    $endgroup$



                    From a maintenance stand point, I think it's better that you check for 3 and 5 instead of checking for 15. The issue with checking for 15 is that when you come back to your code 6 months from now, you won't realize that you were just printing fizzbuzz when something was divisible by both 3 and 5. That might matter in the future, but I think that's something worth talking about in an interview.







                    share|improve this answer














                    share|improve this answer



                    share|improve this answer








                    edited Jan 9 at 22:18

























                    answered Jan 9 at 16:22









                    SteveSteve

                    9510




                    9510





















                        1












                        $begingroup$

                        Though your code is fine, another thing to look is amount of division happening the code, Division is computationally expensive operation and as @Michael has suggested that it is redundant. So always try to minimise the multiplication and division operation in your code.



                        You have mentioned that you are a beginner programmer. I believe it is a good practice to start familiarising your self with topics like Computational Complexity.



                        Look here for computational complexity of mathematical functions






                        share|improve this answer











                        $endgroup$












                        • $begingroup$
                          While I think it's a bit early for a beginning programmer to be learning about things such as memory management (assuming this is one of their FIRST actual programs), I think this is a good answer!
                          $endgroup$
                          – nostalgk
                          Jan 9 at 20:10






                        • 3




                          $begingroup$
                          This advice screams "premature optimization". Division is not an expensive operation. It's a bit more expensive that addition but it's not really expensive on today's computers. Your mobile phone can do millions of divisions in a fraction of a second. String concatenation is much more expensive, by far. The most important metric in code quality is readability. Also, if you are talking about computational complexity, you should note that your change won't change complexity at all.
                          $endgroup$
                          – Sulthan
                          Jan 9 at 23:37







                        • 1




                          $begingroup$
                          The theoretical computational complexity of the underlying division algorithm is practically irrelevant.
                          $endgroup$
                          – crasic
                          Jan 10 at 3:05











                        • $begingroup$
                          @Sulthan The subject computational complexity doesn't take in to account how many operations a machine can perform, it takes into account how many operations/steps are required to solve a problem. While CC of addition i linear, division is quadratic. Dividing only by 3 and 5 would change the complexity and results of those can be saved in a bool variable which can be evaluated in constant time complexity.
                          $endgroup$
                          – Hemanshu
                          Jan 10 at 13:58







                        • 1




                          $begingroup$
                          On a modern CPU integer division takes 1 CPU cycle, the JS version of % a little longer. JS will use 32bit signed ints when it can so the operation is insignificant in comparison to just calling a function. BTW the linked computational complexity page has nothing to do with how CPUs ALU and FLU process the basic math operations.
                          $endgroup$
                          – Blindman67
                          Jan 10 at 14:08















                        1












                        $begingroup$

                        Though your code is fine, another thing to look is amount of division happening the code, Division is computationally expensive operation and as @Michael has suggested that it is redundant. So always try to minimise the multiplication and division operation in your code.



                        You have mentioned that you are a beginner programmer. I believe it is a good practice to start familiarising your self with topics like Computational Complexity.



                        Look here for computational complexity of mathematical functions






                        share|improve this answer











                        $endgroup$












                        • $begingroup$
                          While I think it's a bit early for a beginning programmer to be learning about things such as memory management (assuming this is one of their FIRST actual programs), I think this is a good answer!
                          $endgroup$
                          – nostalgk
                          Jan 9 at 20:10






                        • 3




                          $begingroup$
                          This advice screams "premature optimization". Division is not an expensive operation. It's a bit more expensive that addition but it's not really expensive on today's computers. Your mobile phone can do millions of divisions in a fraction of a second. String concatenation is much more expensive, by far. The most important metric in code quality is readability. Also, if you are talking about computational complexity, you should note that your change won't change complexity at all.
                          $endgroup$
                          – Sulthan
                          Jan 9 at 23:37







                        • 1




                          $begingroup$
                          The theoretical computational complexity of the underlying division algorithm is practically irrelevant.
                          $endgroup$
                          – crasic
                          Jan 10 at 3:05











                        • $begingroup$
                          @Sulthan The subject computational complexity doesn't take in to account how many operations a machine can perform, it takes into account how many operations/steps are required to solve a problem. While CC of addition i linear, division is quadratic. Dividing only by 3 and 5 would change the complexity and results of those can be saved in a bool variable which can be evaluated in constant time complexity.
                          $endgroup$
                          – Hemanshu
                          Jan 10 at 13:58







                        • 1




                          $begingroup$
                          On a modern CPU integer division takes 1 CPU cycle, the JS version of % a little longer. JS will use 32bit signed ints when it can so the operation is insignificant in comparison to just calling a function. BTW the linked computational complexity page has nothing to do with how CPUs ALU and FLU process the basic math operations.
                          $endgroup$
                          – Blindman67
                          Jan 10 at 14:08













                        1












                        1








                        1





                        $begingroup$

                        Though your code is fine, another thing to look is amount of division happening the code, Division is computationally expensive operation and as @Michael has suggested that it is redundant. So always try to minimise the multiplication and division operation in your code.



                        You have mentioned that you are a beginner programmer. I believe it is a good practice to start familiarising your self with topics like Computational Complexity.



                        Look here for computational complexity of mathematical functions






                        share|improve this answer











                        $endgroup$



                        Though your code is fine, another thing to look is amount of division happening the code, Division is computationally expensive operation and as @Michael has suggested that it is redundant. So always try to minimise the multiplication and division operation in your code.



                        You have mentioned that you are a beginner programmer. I believe it is a good practice to start familiarising your self with topics like Computational Complexity.



                        Look here for computational complexity of mathematical functions







                        share|improve this answer














                        share|improve this answer



                        share|improve this answer








                        edited Jan 10 at 20:12









                        Sᴀᴍ Onᴇᴌᴀ

                        9,32262161




                        9,32262161










                        answered Jan 9 at 17:07









                        HemanshuHemanshu

                        314




                        314











                        • $begingroup$
                          While I think it's a bit early for a beginning programmer to be learning about things such as memory management (assuming this is one of their FIRST actual programs), I think this is a good answer!
                          $endgroup$
                          – nostalgk
                          Jan 9 at 20:10






                        • 3




                          $begingroup$
                          This advice screams "premature optimization". Division is not an expensive operation. It's a bit more expensive that addition but it's not really expensive on today's computers. Your mobile phone can do millions of divisions in a fraction of a second. String concatenation is much more expensive, by far. The most important metric in code quality is readability. Also, if you are talking about computational complexity, you should note that your change won't change complexity at all.
                          $endgroup$
                          – Sulthan
                          Jan 9 at 23:37







                        • 1




                          $begingroup$
                          The theoretical computational complexity of the underlying division algorithm is practically irrelevant.
                          $endgroup$
                          – crasic
                          Jan 10 at 3:05











                        • $begingroup$
                          @Sulthan The subject computational complexity doesn't take in to account how many operations a machine can perform, it takes into account how many operations/steps are required to solve a problem. While CC of addition i linear, division is quadratic. Dividing only by 3 and 5 would change the complexity and results of those can be saved in a bool variable which can be evaluated in constant time complexity.
                          $endgroup$
                          – Hemanshu
                          Jan 10 at 13:58







                        • 1




                          $begingroup$
                          On a modern CPU integer division takes 1 CPU cycle, the JS version of % a little longer. JS will use 32bit signed ints when it can so the operation is insignificant in comparison to just calling a function. BTW the linked computational complexity page has nothing to do with how CPUs ALU and FLU process the basic math operations.
                          $endgroup$
                          – Blindman67
                          Jan 10 at 14:08
















                        • $begingroup$
                          While I think it's a bit early for a beginning programmer to be learning about things such as memory management (assuming this is one of their FIRST actual programs), I think this is a good answer!
                          $endgroup$
                          – nostalgk
                          Jan 9 at 20:10






                        • 3




                          $begingroup$
                          This advice screams "premature optimization". Division is not an expensive operation. It's a bit more expensive that addition but it's not really expensive on today's computers. Your mobile phone can do millions of divisions in a fraction of a second. String concatenation is much more expensive, by far. The most important metric in code quality is readability. Also, if you are talking about computational complexity, you should note that your change won't change complexity at all.
                          $endgroup$
                          – Sulthan
                          Jan 9 at 23:37







                        • 1




                          $begingroup$
                          The theoretical computational complexity of the underlying division algorithm is practically irrelevant.
                          $endgroup$
                          – crasic
                          Jan 10 at 3:05











                        • $begingroup$
                          @Sulthan The subject computational complexity doesn't take in to account how many operations a machine can perform, it takes into account how many operations/steps are required to solve a problem. While CC of addition i linear, division is quadratic. Dividing only by 3 and 5 would change the complexity and results of those can be saved in a bool variable which can be evaluated in constant time complexity.
                          $endgroup$
                          – Hemanshu
                          Jan 10 at 13:58







                        • 1




                          $begingroup$
                          On a modern CPU integer division takes 1 CPU cycle, the JS version of % a little longer. JS will use 32bit signed ints when it can so the operation is insignificant in comparison to just calling a function. BTW the linked computational complexity page has nothing to do with how CPUs ALU and FLU process the basic math operations.
                          $endgroup$
                          – Blindman67
                          Jan 10 at 14:08















                        $begingroup$
                        While I think it's a bit early for a beginning programmer to be learning about things such as memory management (assuming this is one of their FIRST actual programs), I think this is a good answer!
                        $endgroup$
                        – nostalgk
                        Jan 9 at 20:10




                        $begingroup$
                        While I think it's a bit early for a beginning programmer to be learning about things such as memory management (assuming this is one of their FIRST actual programs), I think this is a good answer!
                        $endgroup$
                        – nostalgk
                        Jan 9 at 20:10




                        3




                        3




                        $begingroup$
                        This advice screams "premature optimization". Division is not an expensive operation. It's a bit more expensive that addition but it's not really expensive on today's computers. Your mobile phone can do millions of divisions in a fraction of a second. String concatenation is much more expensive, by far. The most important metric in code quality is readability. Also, if you are talking about computational complexity, you should note that your change won't change complexity at all.
                        $endgroup$
                        – Sulthan
                        Jan 9 at 23:37





                        $begingroup$
                        This advice screams "premature optimization". Division is not an expensive operation. It's a bit more expensive that addition but it's not really expensive on today's computers. Your mobile phone can do millions of divisions in a fraction of a second. String concatenation is much more expensive, by far. The most important metric in code quality is readability. Also, if you are talking about computational complexity, you should note that your change won't change complexity at all.
                        $endgroup$
                        – Sulthan
                        Jan 9 at 23:37





                        1




                        1




                        $begingroup$
                        The theoretical computational complexity of the underlying division algorithm is practically irrelevant.
                        $endgroup$
                        – crasic
                        Jan 10 at 3:05





                        $begingroup$
                        The theoretical computational complexity of the underlying division algorithm is practically irrelevant.
                        $endgroup$
                        – crasic
                        Jan 10 at 3:05













                        $begingroup$
                        @Sulthan The subject computational complexity doesn't take in to account how many operations a machine can perform, it takes into account how many operations/steps are required to solve a problem. While CC of addition i linear, division is quadratic. Dividing only by 3 and 5 would change the complexity and results of those can be saved in a bool variable which can be evaluated in constant time complexity.
                        $endgroup$
                        – Hemanshu
                        Jan 10 at 13:58





                        $begingroup$
                        @Sulthan The subject computational complexity doesn't take in to account how many operations a machine can perform, it takes into account how many operations/steps are required to solve a problem. While CC of addition i linear, division is quadratic. Dividing only by 3 and 5 would change the complexity and results of those can be saved in a bool variable which can be evaluated in constant time complexity.
                        $endgroup$
                        – Hemanshu
                        Jan 10 at 13:58





                        1




                        1




                        $begingroup$
                        On a modern CPU integer division takes 1 CPU cycle, the JS version of % a little longer. JS will use 32bit signed ints when it can so the operation is insignificant in comparison to just calling a function. BTW the linked computational complexity page has nothing to do with how CPUs ALU and FLU process the basic math operations.
                        $endgroup$
                        – Blindman67
                        Jan 10 at 14:08




                        $begingroup$
                        On a modern CPU integer division takes 1 CPU cycle, the JS version of % a little longer. JS will use 32bit signed ints when it can so the operation is insignificant in comparison to just calling a function. BTW the linked computational complexity page has nothing to do with how CPUs ALU and FLU process the basic math operations.
                        $endgroup$
                        – Blindman67
                        Jan 10 at 14:08











                        1












                        $begingroup$

                        I tend to be somewhat "literalist", at least at first, when implementing algorithms. So I would start with this, and then optimize from there. I like that it keeps things very clear.



                        const calcFizzBuzz = function(num) 
                        let result = "";
                        if (num % 3 === 0)
                        result += "Fizz";
                        if (num % 5 === 0)
                        result += "Buzz";


                        else if (num % 5 === 0)
                        result += "Buzz";

                        else
                        result = num;

                        return result;


                        for (let i = 0; i < 100; i++)
                        console.log(calcFizzBuzz(i));






                        share|improve this answer









                        $endgroup$

















                          1












                          $begingroup$

                          I tend to be somewhat "literalist", at least at first, when implementing algorithms. So I would start with this, and then optimize from there. I like that it keeps things very clear.



                          const calcFizzBuzz = function(num) 
                          let result = "";
                          if (num % 3 === 0)
                          result += "Fizz";
                          if (num % 5 === 0)
                          result += "Buzz";


                          else if (num % 5 === 0)
                          result += "Buzz";

                          else
                          result = num;

                          return result;


                          for (let i = 0; i < 100; i++)
                          console.log(calcFizzBuzz(i));






                          share|improve this answer









                          $endgroup$















                            1












                            1








                            1





                            $begingroup$

                            I tend to be somewhat "literalist", at least at first, when implementing algorithms. So I would start with this, and then optimize from there. I like that it keeps things very clear.



                            const calcFizzBuzz = function(num) 
                            let result = "";
                            if (num % 3 === 0)
                            result += "Fizz";
                            if (num % 5 === 0)
                            result += "Buzz";


                            else if (num % 5 === 0)
                            result += "Buzz";

                            else
                            result = num;

                            return result;


                            for (let i = 0; i < 100; i++)
                            console.log(calcFizzBuzz(i));






                            share|improve this answer









                            $endgroup$



                            I tend to be somewhat "literalist", at least at first, when implementing algorithms. So I would start with this, and then optimize from there. I like that it keeps things very clear.



                            const calcFizzBuzz = function(num) 
                            let result = "";
                            if (num % 3 === 0)
                            result += "Fizz";
                            if (num % 5 === 0)
                            result += "Buzz";


                            else if (num % 5 === 0)
                            result += "Buzz";

                            else
                            result = num;

                            return result;


                            for (let i = 0; i < 100; i++)
                            console.log(calcFizzBuzz(i));







                            share|improve this answer












                            share|improve this answer



                            share|improve this answer










                            answered Jan 11 at 9:34









                            PhrancisPhrancis

                            14.8k648141




                            14.8k648141



























                                draft saved

                                draft discarded
















































                                Thanks for contributing an answer to Code Review Stack Exchange!


                                • Please be sure to answer the question. Provide details and share your research!

                                But avoid


                                • Asking for help, clarification, or responding to other answers.

                                • Making statements based on opinion; back them up with references or personal experience.

                                Use MathJax to format equations. MathJax reference.


                                To learn more, see our tips on writing great answers.




                                draft saved


                                draft discarded














                                StackExchange.ready(
                                function ()
                                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcodereview.stackexchange.com%2fquestions%2f211113%2ffizzbuzz-in-javascript%23new-answer', 'question_page');

                                );

                                Post as a guest















                                Required, but never shown





















































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown

































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown






                                Popular posts from this blog

                                How to check contact read email or not when send email to Individual?

                                Displaying single band from multi-band raster using QGIS

                                How many registers does an x86_64 CPU actually have?