Thread: A C Puzzler...

The program shall display to stdout using printf the string "Hello C"

The program shall not contain any or a single semi-colon.

The program should compile on gcc without errors using switches -Wall -Pedantic -std=c99

The switch -std=c99 is optional and c89 may display a warning,
that control reaches end of non-void function. This is acceptable.

Compilation on any other C compiler system that results in an executable
that adheres to the first and second directives is also acceptable.

First correct entry wins.

This is old hat. There's even a FAQ on it.

Quzah.

This would be better placed in the Contests forum.

Code:

#include <stdio.h>

int 
main(void)
{
     while( printf("Hello World\n"), 0)
     {
     }
}

There would be different versions based on the different loops. Is there an essentially different method?

Plus stdio.h contains semi-colons, so I guess that shouldn't be allowed.

If you can #include then:

noco.c

Code:

#include <stdio.h>

int main()
{
  ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
  ;;     ;printf("Hello C");     ;;
  ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;

  ;;;;;;;;;;;;;;;;
  ;;  return 0; ;;
  ;;;;;;;;;;;;;;;;
}

noco_wrap.c

Code:

#include "noco.c"

gcc -ansi -pedantic -Wall -std=c99 noco_wrap.c

Code:

while( printf("Hello World\n"))
     {
     }

printf returns the number of characters outputted. As such it would be in an infinate loop.

edit: Made the sentance clearer

Originally posted by wudmx
[B]Originally posted by pinko_liberal

Code:

#include <stdio.h>

int 
main(void)
{
     while( printf("Hello World\n"), 0)
     {
     }
}

i don't have here any compiler - but i think this won't work! shouldn't it be

It works due to the use of the comma operator. It evaluates each argument, and in doing so, the printf is executed. It also evaluates zero, which terminates the loop.

Quzah.

A simpler variation:

Code:

 int main()
{
     while (printf("Hello World!") && 0){}
}

Originally posted by quzah
It works due to the use of the comma operator. It evaluates each argument, and in doing so, the printf is executed. It also evaluates zero, which terminates the loop.

It's also note-worthy to add that the comma operator is evaluated
from left to right meaning that the printf in pinko's while loop was
evaluated first, then the expression on the right side of the comma was evaluated.

pinko_liberal was the first to satisfy all conditions, pinko wins.

Code:

 while (printf("Hello C"), 0) {}

wudmx also supplied a correct entry,

Code:

if (printf("Hello C")) {}

Sebastiani also supplied a correct entry,

Code:

 while (printf("Hello C") && 0) {}

Originally posted by Thantos
Its also note-worthy to add that EVERY expression
is evaluated from left to right not just ones using the comma.

Now, at which point did I say that the comma expression
evaluates from left to right?

I did however state that the comma operator is evaluated from left to right.

It's understandable that you are a little confused. Oh! the P
well if you didn't get that then you're just out of the loop.

Thantos.

Worked it out yet?

We're talking about operator associativity, not expressional evaluation.

Not everything evaluates from left to right. Take this as an exampple:

Code:

a = b = c = d = e = 3;

That evaluates from right to left.