Allocation as default initialization

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












16















Say I have a struct (or class) with a dynamic array, its length, and a constructor:



struct array 
int l;
int* t;
array(int length);
;

array::array(int length)
l=length;
t=new int[l];



I assume everything so far is legal: this is how I would write it (although it maybe not be the only way to do things), but I have seen the following code, that somewhat seems to work:



struct array 
int l;
int* t = new int[l];
array(int length);


array::array(int length)
l=length;



It looks bad, so I wonder if this works out of sheer luck and undefined behaviors, or if there is some internal rule that makes this code work fine.










share|improve this question

















  • 7





    don't use 'l' as a variable name, It's easily confused with the number '1',

    – regomodo
    Jan 28 at 13:31











  • @regomodo: I'd rather tell people not to use fonts (for coding) that make it easy to confuse I, l and 1. :-)

    – Heinzi
    Jan 28 at 14:37







  • 5





    id rather people read "Clean Code" instead

    – regomodo
    Jan 28 at 14:54






  • 2





    @Heinzi Or maybe stop using single letter variables damnit.

    – Bakuriu
    Jan 28 at 20:09















16















Say I have a struct (or class) with a dynamic array, its length, and a constructor:



struct array 
int l;
int* t;
array(int length);
;

array::array(int length)
l=length;
t=new int[l];



I assume everything so far is legal: this is how I would write it (although it maybe not be the only way to do things), but I have seen the following code, that somewhat seems to work:



struct array 
int l;
int* t = new int[l];
array(int length);


array::array(int length)
l=length;



It looks bad, so I wonder if this works out of sheer luck and undefined behaviors, or if there is some internal rule that makes this code work fine.










share|improve this question

















  • 7





    don't use 'l' as a variable name, It's easily confused with the number '1',

    – regomodo
    Jan 28 at 13:31











  • @regomodo: I'd rather tell people not to use fonts (for coding) that make it easy to confuse I, l and 1. :-)

    – Heinzi
    Jan 28 at 14:37







  • 5





    id rather people read "Clean Code" instead

    – regomodo
    Jan 28 at 14:54






  • 2





    @Heinzi Or maybe stop using single letter variables damnit.

    – Bakuriu
    Jan 28 at 20:09













16












16








16








Say I have a struct (or class) with a dynamic array, its length, and a constructor:



struct array 
int l;
int* t;
array(int length);
;

array::array(int length)
l=length;
t=new int[l];



I assume everything so far is legal: this is how I would write it (although it maybe not be the only way to do things), but I have seen the following code, that somewhat seems to work:



struct array 
int l;
int* t = new int[l];
array(int length);


array::array(int length)
l=length;



It looks bad, so I wonder if this works out of sheer luck and undefined behaviors, or if there is some internal rule that makes this code work fine.










share|improve this question














Say I have a struct (or class) with a dynamic array, its length, and a constructor:



struct array 
int l;
int* t;
array(int length);
;

array::array(int length)
l=length;
t=new int[l];



I assume everything so far is legal: this is how I would write it (although it maybe not be the only way to do things), but I have seen the following code, that somewhat seems to work:



struct array 
int l;
int* t = new int[l];
array(int length);


array::array(int length)
l=length;



It looks bad, so I wonder if this works out of sheer luck and undefined behaviors, or if there is some internal rule that makes this code work fine.







c++ constructor






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Jan 28 at 10:24









tarulentarulen

1,798413




1,798413







  • 7





    don't use 'l' as a variable name, It's easily confused with the number '1',

    – regomodo
    Jan 28 at 13:31











  • @regomodo: I'd rather tell people not to use fonts (for coding) that make it easy to confuse I, l and 1. :-)

    – Heinzi
    Jan 28 at 14:37







  • 5





    id rather people read "Clean Code" instead

    – regomodo
    Jan 28 at 14:54






  • 2





    @Heinzi Or maybe stop using single letter variables damnit.

    – Bakuriu
    Jan 28 at 20:09












  • 7





    don't use 'l' as a variable name, It's easily confused with the number '1',

    – regomodo
    Jan 28 at 13:31











  • @regomodo: I'd rather tell people not to use fonts (for coding) that make it easy to confuse I, l and 1. :-)

    – Heinzi
    Jan 28 at 14:37







  • 5





    id rather people read "Clean Code" instead

    – regomodo
    Jan 28 at 14:54






  • 2





    @Heinzi Or maybe stop using single letter variables damnit.

    – Bakuriu
    Jan 28 at 20:09







7




7





don't use 'l' as a variable name, It's easily confused with the number '1',

– regomodo
Jan 28 at 13:31





don't use 'l' as a variable name, It's easily confused with the number '1',

– regomodo
Jan 28 at 13:31













@regomodo: I'd rather tell people not to use fonts (for coding) that make it easy to confuse I, l and 1. :-)

– Heinzi
Jan 28 at 14:37






@regomodo: I'd rather tell people not to use fonts (for coding) that make it easy to confuse I, l and 1. :-)

– Heinzi
Jan 28 at 14:37





5




5





id rather people read "Clean Code" instead

– regomodo
Jan 28 at 14:54





id rather people read "Clean Code" instead

– regomodo
Jan 28 at 14:54




2




2





@Heinzi Or maybe stop using single letter variables damnit.

– Bakuriu
Jan 28 at 20:09





@Heinzi Or maybe stop using single letter variables damnit.

– Bakuriu
Jan 28 at 20:09












2 Answers
2






active

oldest

votes


















27














This code is not correct.



int* t = new int[l]; will happen before l=length;, thus reading the uninitialized variable l. Member initializers are handled before the constructor's body runs.



array::array(int length) : llength 


instead would work because l is declared before t.



However, doing this "by hand" is a bad idea to begin with. You should be using std::vector.






share|improve this answer






























    9














    The 2nd code snippet might have undefined behavior.



    The data members are initialized at the order of how they're declared. For class array, when t is initialized l is not initialized yet. For objects with automatic and dynamic storage duration l will be initialized to indeterminate value, then the usage of l (i.e. new int[l]) leads to UB.



    Note that l=length; inside the body of the constructor is just assignment; the initialization of data members has been finished before that.



    BTW: With member initializer list the 1st code snippet chould be rewritten as



    array::array(int length) : l(length), t(new int[l]) 






    share|improve this answer




















    • 3





      Why do you say "might"? It does have UB, unconditionally.

      – Ruslan
      Jan 28 at 16:29






    • 1





      @Ruslan Because for static or thread-local objects l will get zero initialized at first.

      – songyuanyao
      Jan 29 at 1:13












    • @Ruslan I suppose, since a program with undefined behaviour can make anything happen, we could argue that one possible outcome of this program is making the code well-defined ;)

      – Lightness Races in Orbit
      Jan 30 at 16:33










    Your Answer






    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: "1"
    ;
    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: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    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%2fstackoverflow.com%2fquestions%2f54399900%2fallocation-as-default-initialization%23new-answer', 'question_page');

    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    27














    This code is not correct.



    int* t = new int[l]; will happen before l=length;, thus reading the uninitialized variable l. Member initializers are handled before the constructor's body runs.



    array::array(int length) : llength 


    instead would work because l is declared before t.



    However, doing this "by hand" is a bad idea to begin with. You should be using std::vector.






    share|improve this answer



























      27














      This code is not correct.



      int* t = new int[l]; will happen before l=length;, thus reading the uninitialized variable l. Member initializers are handled before the constructor's body runs.



      array::array(int length) : llength 


      instead would work because l is declared before t.



      However, doing this "by hand" is a bad idea to begin with. You should be using std::vector.






      share|improve this answer

























        27












        27








        27







        This code is not correct.



        int* t = new int[l]; will happen before l=length;, thus reading the uninitialized variable l. Member initializers are handled before the constructor's body runs.



        array::array(int length) : llength 


        instead would work because l is declared before t.



        However, doing this "by hand" is a bad idea to begin with. You should be using std::vector.






        share|improve this answer













        This code is not correct.



        int* t = new int[l]; will happen before l=length;, thus reading the uninitialized variable l. Member initializers are handled before the constructor's body runs.



        array::array(int length) : llength 


        instead would work because l is declared before t.



        However, doing this "by hand" is a bad idea to begin with. You should be using std::vector.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Jan 28 at 10:29









        Baum mit AugenBaum mit Augen

        41.2k12118154




        41.2k12118154























            9














            The 2nd code snippet might have undefined behavior.



            The data members are initialized at the order of how they're declared. For class array, when t is initialized l is not initialized yet. For objects with automatic and dynamic storage duration l will be initialized to indeterminate value, then the usage of l (i.e. new int[l]) leads to UB.



            Note that l=length; inside the body of the constructor is just assignment; the initialization of data members has been finished before that.



            BTW: With member initializer list the 1st code snippet chould be rewritten as



            array::array(int length) : l(length), t(new int[l]) 






            share|improve this answer




















            • 3





              Why do you say "might"? It does have UB, unconditionally.

              – Ruslan
              Jan 28 at 16:29






            • 1





              @Ruslan Because for static or thread-local objects l will get zero initialized at first.

              – songyuanyao
              Jan 29 at 1:13












            • @Ruslan I suppose, since a program with undefined behaviour can make anything happen, we could argue that one possible outcome of this program is making the code well-defined ;)

              – Lightness Races in Orbit
              Jan 30 at 16:33















            9














            The 2nd code snippet might have undefined behavior.



            The data members are initialized at the order of how they're declared. For class array, when t is initialized l is not initialized yet. For objects with automatic and dynamic storage duration l will be initialized to indeterminate value, then the usage of l (i.e. new int[l]) leads to UB.



            Note that l=length; inside the body of the constructor is just assignment; the initialization of data members has been finished before that.



            BTW: With member initializer list the 1st code snippet chould be rewritten as



            array::array(int length) : l(length), t(new int[l]) 






            share|improve this answer




















            • 3





              Why do you say "might"? It does have UB, unconditionally.

              – Ruslan
              Jan 28 at 16:29






            • 1





              @Ruslan Because for static or thread-local objects l will get zero initialized at first.

              – songyuanyao
              Jan 29 at 1:13












            • @Ruslan I suppose, since a program with undefined behaviour can make anything happen, we could argue that one possible outcome of this program is making the code well-defined ;)

              – Lightness Races in Orbit
              Jan 30 at 16:33













            9












            9








            9







            The 2nd code snippet might have undefined behavior.



            The data members are initialized at the order of how they're declared. For class array, when t is initialized l is not initialized yet. For objects with automatic and dynamic storage duration l will be initialized to indeterminate value, then the usage of l (i.e. new int[l]) leads to UB.



            Note that l=length; inside the body of the constructor is just assignment; the initialization of data members has been finished before that.



            BTW: With member initializer list the 1st code snippet chould be rewritten as



            array::array(int length) : l(length), t(new int[l]) 






            share|improve this answer















            The 2nd code snippet might have undefined behavior.



            The data members are initialized at the order of how they're declared. For class array, when t is initialized l is not initialized yet. For objects with automatic and dynamic storage duration l will be initialized to indeterminate value, then the usage of l (i.e. new int[l]) leads to UB.



            Note that l=length; inside the body of the constructor is just assignment; the initialization of data members has been finished before that.



            BTW: With member initializer list the 1st code snippet chould be rewritten as



            array::array(int length) : l(length), t(new int[l]) 







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited Jan 28 at 10:33

























            answered Jan 28 at 10:28









            songyuanyaosongyuanyao

            91.6k11175238




            91.6k11175238







            • 3





              Why do you say "might"? It does have UB, unconditionally.

              – Ruslan
              Jan 28 at 16:29






            • 1





              @Ruslan Because for static or thread-local objects l will get zero initialized at first.

              – songyuanyao
              Jan 29 at 1:13












            • @Ruslan I suppose, since a program with undefined behaviour can make anything happen, we could argue that one possible outcome of this program is making the code well-defined ;)

              – Lightness Races in Orbit
              Jan 30 at 16:33












            • 3





              Why do you say "might"? It does have UB, unconditionally.

              – Ruslan
              Jan 28 at 16:29






            • 1





              @Ruslan Because for static or thread-local objects l will get zero initialized at first.

              – songyuanyao
              Jan 29 at 1:13












            • @Ruslan I suppose, since a program with undefined behaviour can make anything happen, we could argue that one possible outcome of this program is making the code well-defined ;)

              – Lightness Races in Orbit
              Jan 30 at 16:33







            3




            3





            Why do you say "might"? It does have UB, unconditionally.

            – Ruslan
            Jan 28 at 16:29





            Why do you say "might"? It does have UB, unconditionally.

            – Ruslan
            Jan 28 at 16:29




            1




            1





            @Ruslan Because for static or thread-local objects l will get zero initialized at first.

            – songyuanyao
            Jan 29 at 1:13






            @Ruslan Because for static or thread-local objects l will get zero initialized at first.

            – songyuanyao
            Jan 29 at 1:13














            @Ruslan I suppose, since a program with undefined behaviour can make anything happen, we could argue that one possible outcome of this program is making the code well-defined ;)

            – Lightness Races in Orbit
            Jan 30 at 16:33





            @Ruslan I suppose, since a program with undefined behaviour can make anything happen, we could argue that one possible outcome of this program is making the code well-defined ;)

            – Lightness Races in Orbit
            Jan 30 at 16:33

















            draft saved

            draft discarded
















































            Thanks for contributing an answer to Stack Overflow!


            • 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.

            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%2fstackoverflow.com%2fquestions%2f54399900%2fallocation-as-default-initialization%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?

            Bahrain

            Postfix configuration issue with fips on centos 7; mailgun relay