“int main (vooid)”? How does that work?

问题

I recently had to type in a small C test program and, in the process, I made a spelling mistake in the main function by accidentally using vooid instead of void.

And yet it still worked.

Reducing it down to its smallest complete version, I ended up with:

int main (vooid) {
    return 42;
}

This does indeed compile (gcc -Wall -o myprog myprog.c) and, when run, it returns 42.

How exactly is this valid code?

Here's a transcript cut and pasted from my bash shell to show what I'm doing:

pax$ cat qq.c
int main (vooid) {
    return 42;
}

pax$ rm qq ; gcc -Wall -o qq qq.c ; ./qq

pax$ echo $?
42

回答1:

It's simply using the "old-style" function-declaration syntax; you're implicitly declaring an int parameter called vooid.

回答2:

It's valid code, because myprog.c contains:

int main (vooid) // vooid is of type int, allowed, and an alias for argc
{     
  return 42; // The answer to the Ultimate Question
}

vooid contains one plus the number of arguments passed (i.e., argc). So, in effect all you've done is to rename argc to vooid.

回答3:

In C, the default type for a function argument is int. So, your program is treating the word vooid as int main(int vooid), which is perfectly valid code.

回答4:

It is only gcc -std=c89 -Wall -o qq qq.c and gcc -std=gnu89 -Wall -o qq qq.c don't emit a warning. All the other standards emit a warning about implicit type int for vooid.

int main(chart) behaves the same way as does int main (vooid).

return vooid; returns the number of command line arguments.

I tested with gcc 4.4.5 on Debian testing system.

来源：https://stackoverflow.com/questions/4987415/int-main-vooid-how-does-that-work

标签

syntax

main