Statements in Java
Clash Royale CLAN TAG#URR8PPP
Each code block translates to a specific statement in English. There is a theme behind the statements. As code puzzles tend to be "easier" (relative to other PSE puzzles) than expected, it will not be revealed unless people are unable to make any progress on the phrases.
1:
public class Statement()
void Riddle1()
Citizen you = new Citizen();
Government() g = new Government();
\g.getService(you)
\after a code revision, I've determined the following is better:
you.getService(g)
2:
System.out.println(listofconcerns.length); \returns 1
System.out.println(listofconcerns[0]); \returns the string "concerns"
3:
(house has been declared and initialized at some point before this line)
int quotient = house/house;
Boolean b = quotient == 1;
System.out.println(b); \returns false, with a compiler specific error on the result not holding up
4:
int anglevel = anger();
if(anglevel > threshold1)
wait(10000);
if( anglevel > threshold2)
wait(90000);
english computer-puzzle
asked Dec 23 '18 at 1:22
Display name
860216
add a comment |
Each code block translates to a specific statement in English. There is a theme behind the statements. As code puzzles tend to be "easier" (relative to other PSE puzzles) than expected, it will not be revealed unless people are unable to make any progress on the phrases.
1:
public class Statement()
void Riddle1()
Citizen you = new Citizen();
Government() g = new Government();
\g.getService(you)
\after a code revision, I've determined the following is better:
you.getService(g)
2:
System.out.println(listofconcerns.length); \returns 1
System.out.println(listofconcerns[0]); \returns the string "concerns"
3:
(house has been declared and initialized at some point before this line)
int quotient = house/house;
Boolean b = quotient == 1;
System.out.println(b); \returns false, with a compiler specific error on the result not holding up
4:
int anglevel = anger();
if(anglevel > threshold1)
wait(10000);
if( anglevel > threshold2)
wait(90000);
english computer-puzzle
asked Dec 23 '18 at 1:22
Display name
860216
add a comment |
Each code block translates to a specific statement in English. There is a theme behind the statements. As code puzzles tend to be "easier" (relative to other PSE puzzles) than expected, it will not be revealed unless people are unable to make any progress on the phrases.
1:
public class Statement()
void Riddle1()
Citizen you = new Citizen();
Government() g = new Government();
\g.getService(you)
\after a code revision, I've determined the following is better:
you.getService(g)
2:
System.out.println(listofconcerns.length); \returns 1
System.out.println(listofconcerns[0]); \returns the string "concerns"
3:
(house has been declared and initialized at some point before this line)
int quotient = house/house;
Boolean b = quotient == 1;
System.out.println(b); \returns false, with a compiler specific error on the result not holding up
4:
int anglevel = anger();
if(anglevel > threshold1)
wait(10000);
if( anglevel > threshold2)
wait(90000);
english computer-puzzle
asked Dec 23 '18 at 1:22
Display name
860216
Each code block translates to a specific statement in English. There is a theme behind the statements. As code puzzles tend to be "easier" (relative to other PSE puzzles) than expected, it will not be revealed unless people are unable to make any progress on the phrases.
1:
public class Statement()
void Riddle1()
Citizen you = new Citizen();
Government() g = new Government();
\g.getService(you)
\after a code revision, I've determined the following is better:
you.getService(g)
2:
System.out.println(listofconcerns.length); \returns 1
System.out.println(listofconcerns[0]); \returns the string "concerns"
3:
(house has been declared and initialized at some point before this line)
int quotient = house/house;
Boolean b = quotient == 1;
System.out.println(b); \returns false, with a compiler specific error on the result not holding up
4:
int anglevel = anger();
if(anglevel > threshold1)
wait(10000);
if( anglevel > threshold2)
wait(90000);
english computer-puzzle
english computer-puzzle
asked Dec 23 '18 at 1:22
Display name
860216
asked Dec 23 '18 at 1:22
Display name
860216
edited Dec 23 '18 at 1:55
asked Dec 23 '18 at 1:22
Display name
860216
asked Dec 23 '18 at 1:22
Display name
860216
asked Dec 23 '18 at 1:22
Display name
860216
860216
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
public class Statement()
void Riddle1()
Citizen you = new Citizen();
Government() g = new Government();
\g.getService(you)
\after a code revision, I've determined the following is better:
Ask not what your country can do for you...
you.getService(g)
...ask what you can do for your country.
System.out.println(listofconcerns.length); \returns 1
The only thing we have to fear... ("listofconcerns" has just one item)
System.out.println(listofconcerns[0]); \returns the string "concerns"
...is fear itself
(house has been declared and initialized at some point before this line)
int quotient = house/house;
A house divided against itself...
Boolean b = quotient == 1;
System.out.println(b); \returns false, with a compiler specific error on the result not holding up
...cannot stand (does not hold up)
int anglevel = anger();
if(anglevel > threshold1)
When angry...
wait(10000);
count to ten before you speak. (wait() is in milliseconds)
if( anglevel > threshold2)
If very angry...
wait(90000);
...count to one hundred. (we've already counted to 10 at this point, so 10+90)
Theme:
Things said by dead american presidents: Kennedy, Roosevelt, Lincoln, and Jefferson respectively.
answered Dec 23 '18 at 1:41
deep thought
2,7971734
Correct. I suppose there's no way that code puzzles like these will ever be too difficult.
– Display name
Dec 23 '18 at 1:42
@Displayname Putting them together makes them easier, because once you get one, you get a hint to the others. However, easy doesn't mean bad, I think theme makes it better.
– deep thought
Dec 23 '18 at 1:45
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
public class Statement()
void Riddle1()
Citizen you = new Citizen();
Government() g = new Government();
\g.getService(you)
\after a code revision, I've determined the following is better:
Ask not what your country can do for you...
you.getService(g)
...ask what you can do for your country.
System.out.println(listofconcerns.length); \returns 1
The only thing we have to fear... ("listofconcerns" has just one item)
System.out.println(listofconcerns[0]); \returns the string "concerns"
...is fear itself
(house has been declared and initialized at some point before this line)
int quotient = house/house;
A house divided against itself...
Boolean b = quotient == 1;
System.out.println(b); \returns false, with a compiler specific error on the result not holding up
...cannot stand (does not hold up)
int anglevel = anger();
if(anglevel > threshold1)
When angry...
wait(10000);
count to ten before you speak. (wait() is in milliseconds)
if( anglevel > threshold2)
If very angry...
wait(90000);
...count to one hundred. (we've already counted to 10 at this point, so 10+90)
Theme:
Things said by dead american presidents: Kennedy, Roosevelt, Lincoln, and Jefferson respectively.
answered Dec 23 '18 at 1:41
deep thought
2,7971734
Correct. I suppose there's no way that code puzzles like these will ever be too difficult.
– Display name
Dec 23 '18 at 1:42
@Displayname Putting them together makes them easier, because once you get one, you get a hint to the others. However, easy doesn't mean bad, I think theme makes it better.
– deep thought
Dec 23 '18 at 1:45
add a comment |
public class Statement()
void Riddle1()
Citizen you = new Citizen();
Government() g = new Government();
\g.getService(you)
\after a code revision, I've determined the following is better:
Ask not what your country can do for you...
you.getService(g)
...ask what you can do for your country.
System.out.println(listofconcerns.length); \returns 1
The only thing we have to fear... ("listofconcerns" has just one item)
System.out.println(listofconcerns[0]); \returns the string "concerns"
...is fear itself
(house has been declared and initialized at some point before this line)
int quotient = house/house;
A house divided against itself...
Boolean b = quotient == 1;
System.out.println(b); \returns false, with a compiler specific error on the result not holding up
...cannot stand (does not hold up)
int anglevel = anger();
if(anglevel > threshold1)
When angry...
wait(10000);
count to ten before you speak. (wait() is in milliseconds)
if( anglevel > threshold2)
If very angry...
wait(90000);
...count to one hundred. (we've already counted to 10 at this point, so 10+90)
Theme:
Things said by dead american presidents: Kennedy, Roosevelt, Lincoln, and Jefferson respectively.
answered Dec 23 '18 at 1:41
deep thought
2,7971734
Correct. I suppose there's no way that code puzzles like these will ever be too difficult.
– Display name
Dec 23 '18 at 1:42
@Displayname Putting them together makes them easier, because once you get one, you get a hint to the others. However, easy doesn't mean bad, I think theme makes it better.
– deep thought
Dec 23 '18 at 1:45
add a comment |
public class Statement()
void Riddle1()
Citizen you = new Citizen();
Government() g = new Government();
\g.getService(you)
\after a code revision, I've determined the following is better:
Ask not what your country can do for you...
you.getService(g)
...ask what you can do for your country.
System.out.println(listofconcerns.length); \returns 1
The only thing we have to fear... ("listofconcerns" has just one item)
System.out.println(listofconcerns[0]); \returns the string "concerns"
...is fear itself
(house has been declared and initialized at some point before this line)
int quotient = house/house;
A house divided against itself...
Boolean b = quotient == 1;
System.out.println(b); \returns false, with a compiler specific error on the result not holding up
...cannot stand (does not hold up)
int anglevel = anger();
if(anglevel > threshold1)
When angry...
wait(10000);
count to ten before you speak. (wait() is in milliseconds)
if( anglevel > threshold2)
If very angry...
wait(90000);
...count to one hundred. (we've already counted to 10 at this point, so 10+90)
Theme:
Things said by dead american presidents: Kennedy, Roosevelt, Lincoln, and Jefferson respectively.
answered Dec 23 '18 at 1:41
deep thought
2,7971734
public class Statement()
void Riddle1()
Citizen you = new Citizen();
Government() g = new Government();
\g.getService(you)
\after a code revision, I've determined the following is better:
Ask not what your country can do for you...
you.getService(g)
...ask what you can do for your country.
System.out.println(listofconcerns.length); \returns 1
The only thing we have to fear... ("listofconcerns" has just one item)
System.out.println(listofconcerns[0]); \returns the string "concerns"
...is fear itself
(house has been declared and initialized at some point before this line)
int quotient = house/house;
A house divided against itself...
Boolean b = quotient == 1;
System.out.println(b); \returns false, with a compiler specific error on the result not holding up
...cannot stand (does not hold up)
int anglevel = anger();
if(anglevel > threshold1)
When angry...
wait(10000);
count to ten before you speak. (wait() is in milliseconds)
if( anglevel > threshold2)
If very angry...
wait(90000);
...count to one hundred. (we've already counted to 10 at this point, so 10+90)
Theme:
Things said by dead american presidents: Kennedy, Roosevelt, Lincoln, and Jefferson respectively.
answered Dec 23 '18 at 1:41
deep thought
2,7971734
edited Dec 23 '18 at 15:38
answered Dec 23 '18 at 1:41
deep thought
2,7971734
answered Dec 23 '18 at 1:41
deep thought
2,7971734
answered Dec 23 '18 at 1:41
deep thought
2,7971734
2,7971734
Correct. I suppose there's no way that code puzzles like these will ever be too difficult.
– Display name
Dec 23 '18 at 1:42
@Displayname Putting them together makes them easier, because once you get one, you get a hint to the others. However, easy doesn't mean bad, I think theme makes it better.
– deep thought
Dec 23 '18 at 1:45
add a comment |
Correct. I suppose there's no way that code puzzles like these will ever be too difficult.
– Display name
Dec 23 '18 at 1:42
@Displayname Putting them together makes them easier, because once you get one, you get a hint to the others. However, easy doesn't mean bad, I think theme makes it better.
– deep thought
Dec 23 '18 at 1:45
Correct. I suppose there's no way that code puzzles like these will ever be too difficult.
– Display name
Dec 23 '18 at 1:42
Correct. I suppose there's no way that code puzzles like these will ever be too difficult.
– Display name
Dec 23 '18 at 1:42
@Displayname Putting them together makes them easier, because once you get one, you get a hint to the others. However, easy doesn't mean bad, I think theme makes it better.
– deep thought
Dec 23 '18 at 1:45
@Displayname Putting them together makes them easier, because once you get one, you get a hint to the others. However, easy doesn't mean bad, I think theme makes it better.
– deep thought
Dec 23 '18 at 1:45
add a comment |
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