C puzzle: how will you print something on console?

Question

/*you cannot change anything from here below*/
main()
{
exit(0);
}
/*you cannot change anything from here up*/

This was asked during an interview.

I was told to print something on console.

anybody?

Answer 1

Really surprised that nobody posted this yet:

#include <stdio.h>

#if 0

/*you cannot change anything from here below*/
main()
{
exit(0);
}
/*you cannot change anything from here up*/

#endif

int main()
{
   printf("Hello, World!");
   return 0;
}

Prints at runtime and no undefined behavior whatsoever.

Answer 2

weird question...

int main(void)
{
    printf("hello");
    return 0;
}
#define main int lol
/*you cannot change anything from here below*/
main()
{
exit(0);
}
/*you cannot change anything from here up*/

Answer 3

#include <stdio.h>
#define exit(c) return puts("foobar"),0

over main

Answer 4

One implementation defined way would be to use the pragma directives to print during compilation.

#pragma message "Compiling " __FILE__ "..."

Or, you could do this with some macros and a printf (but not without introducing UB in some aspect or the other) at runtime.

#define exit(x) printf("Hello, world!")
int main() {
 exit(0); 
 return 0; /* if pre-C99 */
}

Answer 5

If you interpreted the question to mean you could not or were not allowed to edit the file by commenting out /* */ or using #ifdef _COMMENT_ME_OUT__ #endif respectively above and below the section you are not allowed to edit and then introducing a new main, then you should give an answer of using another .c file.

Answer 6

#include <stdio.h>

#pragma message("Some foobar")

#error This is an error message

int main()

{

exit(0);

}

I think the interviewer wanted to know if you're aware of the #error directive ... just my 2 cents.

Answer 7

If you cannot find a workaround to edit that file, then use a different c file.

Answer 8

Most answers involve the #define c-preprocessor instruction to change what the program means. Most compilers also support something like

#pragma startup foo()

details depend on the compiler vendor. You can make code run BEFORE main(*) is called that way.

Answer 9

#define exit(x) (printf("Bye"))
int main(int argc,char* argv)
{
    exit(0);
    getchar();
    return 0;
}

Answer 10

Solution 1.

This works without any preprocessor directives in cl and gcc, although I've not tested to make sure I'm not using any extensions:

#include <stdio.h>
static void exit() {
   printf("Hello world");
}
/*you cannot change anything from here below*/
main()
{
exit(0);
}
/*you cannot change anything from here up*/

I think it's valid but I can't remember if masking a standard library function is allowed or not.

Solution 2

As several other answers have specified, you can use preprocessor directives, eg:

• #define main or exit to be something that calls ifdef
• use #if 0 to prevent the existing code being compiled
• using #pragma message or #error to print a message at compile time
• using #pragma startup to use a different function as main or to run start-up code before main.

Solution 3

If your compiler supports any C++ features in addition to C, there are many answers:

• Declare a class with a constructor and a static variable of that type
• Put the existing "main" function into a separate namespace (or class definition) and write a different global main

Solution 4

I also looked for any way of forcing a run-time error (stack overflow, out of memory, null dereference, tc), which would normally cause the program to print something, but couldn't find any way that didn't involve running extra code, in which case the extra code might as well be printf.