How does the below program output `C89` when compiled in C89 mode and `C99` when compiled in C99 mode?

问题

I've found this C program from the web:

#include <stdio.h>

int main(){

    printf("C%d\n",(int)(90-(-4.5//**/
    -4.5)));

    return 0;
}

The interesting thing with this program is that when it is compiled and run in C89 mode, it prints C89 and when it is compiled and run in C99 mode, it prints C99. But I am not able to figure out how this program works.

Can you explain how the second argument of printf works in the above program?

回答1:

C99 allows //-style comments, C89 does not. So, to translate:

C99:

 printf("C%d\n",(int)(90-(-4.5     /*Some  comment stuff*/
                         -4.5)));
// Outputs: 99

C89:

printf("C%d\n",(int)(90-(-4.5/      
                         -4.5)));
/* so  we get 90-1 or 89 */

回答2:

the line comment // is introduced since C99. Therefore your code is equal to this in C89

#include <stdio.h>

int main(){

    printf("C%d\n",(int)(90-(-4.5/
-4.5)));

    return 0;
}
/* 90 - (-4.5 / -4.5) = 89 */

and equal to this in C99

#include <stdio.h>

int main(){

    printf("C%d\n",(int)(90-(-4.5
-4.5)));

    return 0;
}
/* 90 - (-4.5 - 4.5) = 99*/

回答3:

Because // comments only exist in C99 and later standards, the code is equivalent to the following:

#include <stdio.h>

int main (void)
{
  int vers;

  #if   __STDC_VERSION__ >= 201112L
    vers = 99; // oops
  #elif __STDC_VERSION__ >= 199901L
    vers = 99;
  #else
    vers = 90;
  #endif

  printf("C%d", vers);

  return 0;
}

Correct code would be:

#include <stdio.h>

int main (void)
{
  int vers;

  #if   __STDC_VERSION__ >= 201112L
    vers = 11;
  #elif __STDC_VERSION__ >= 199901L
    vers = 99;
  #else
    vers = 90;
  #endif

  printf("C%d", vers);

  return 0;
}

来源：https://stackoverflow.com/questions/31115453/how-does-the-below-program-output-c89-when-compiled-in-c89-mode-and-c99-when