Statements in Java
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 4 hours ago
Display name
577216
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 4 hours ago
Display name
577216
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 4 hours ago
Display name
577216
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 4 hours ago
Display name
577216
asked 4 hours ago
Display name
577216
edited 3 hours ago
asked 4 hours ago
Display name
577216
asked 4 hours ago
Display name
577216
asked 4 hours ago
Display name
577216
577216
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
Our only fear ("concern") is...
System.out.println(listofconcerns[0]); \returns the string "concerns"
...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 4 hours ago
deep thought
2,4361633
Correct. I suppose there's no way that code puzzles like these will ever be too difficult.
– Display name
4 hours ago
@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
4 hours ago
@Displayname If you like code puzzles that are apparently not easy enough try <plug>this</plug>!
– WAF
13 mins ago
add a comment |
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1 Answer
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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
Our only fear ("concern") is...
System.out.println(listofconcerns[0]); \returns the string "concerns"
...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 4 hours ago
deep thought
2,4361633
Correct. I suppose there's no way that code puzzles like these will ever be too difficult.
– Display name
4 hours ago
@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
4 hours ago
@Displayname If you like code puzzles that are apparently not easy enough try <plug>this</plug>!
– WAF
13 mins ago
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
Our only fear ("concern") is...
System.out.println(listofconcerns[0]); \returns the string "concerns"
...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 4 hours ago
deep thought
2,4361633
Correct. I suppose there's no way that code puzzles like these will ever be too difficult.
– Display name
4 hours ago
@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
4 hours ago
@Displayname If you like code puzzles that are apparently not easy enough try <plug>this</plug>!
– WAF
13 mins ago
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
Our only fear ("concern") is...
System.out.println(listofconcerns[0]); \returns the string "concerns"
...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 4 hours ago
deep thought
2,4361633
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
Our only fear ("concern") is...
System.out.println(listofconcerns[0]); \returns the string "concerns"
...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 4 hours ago
deep thought
2,4361633
answered 4 hours ago
deep thought
2,4361633
answered 4 hours ago
deep thought
2,4361633
answered 4 hours ago
deep thought
2,4361633
2,4361633
Correct. I suppose there's no way that code puzzles like these will ever be too difficult.
– Display name
4 hours ago
@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
4 hours ago
@Displayname If you like code puzzles that are apparently not easy enough try <plug>this</plug>!
– WAF
13 mins ago
add a comment |
Correct. I suppose there's no way that code puzzles like these will ever be too difficult.
– Display name
4 hours ago
@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
4 hours ago
@Displayname If you like code puzzles that are apparently not easy enough try <plug>this</plug>!
– WAF
13 mins ago
Correct. I suppose there's no way that code puzzles like these will ever be too difficult.
– Display name
4 hours ago
Correct. I suppose there's no way that code puzzles like these will ever be too difficult.
– Display name
4 hours ago
@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
4 hours ago
@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
4 hours ago
@Displayname If you like code puzzles that are apparently not easy enough try <plug>this</plug>!
– WAF
13 mins ago
@Displayname If you like code puzzles that are apparently not easy enough try <plug>this</plug>!
– WAF
13 mins ago
add a comment |
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