/prog/ - fizz buzz

Name: Anonymous 2011-10-15 16:47

I have achieved my lifelong dream. Finally a fizz buzz program with 0 comparison operator.

#include <stdio.h>

#include <stdlib.h>



void main(int j){

  const char* s[] = {"%d\n", "fizz\n", "buzz\n", "fizzbuzz\n"};

  printf(s[3-!!(j%3)-2*!!(j%5)], j);

  (main+((exit-main)*(j/100)))((1-j/100)*j+1);

}

Name: Anonymous 2011-10-15 18:23

See:

http://codepad.org/T1grqp7Z

Final code:

#include <stdio.h>

#include <stdlib.h>

#include <stdint.h>

typedef uintptr_t p;

char *s[] = {"%d\n", "fizz\n", "buzz\n", "fizzbuzz\n"};

int e(int i, char **j) {

    int k;

    char **l;

    k = i;

    l = j;

    exit(0);

}

int main(int i, char **j) {

    char **k;

    k = j;

    printf(s

[(!(i % 3)) | ((!(i % 5)) << 1)]

, i);

    (*((int (*)(int, char **))((p)(main) + ((p)(e) -

    (p)(main)) * (i / 100))))(i + 1, NULL);

    return 0;

}

Compiles, warning-free, with:

gcc -ansi -Wall -Werror -Wextra -pedantic

fizz buzz

1 Name: Anonymous 2011-10-15 16:47

8 Name: Anonymous 2011-10-15 18:23

Name: Anonymous 2011-10-15 16:47

Name: Anonymous 2011-10-15 18:23