So how does exit() work?

If I use exit(), GCC doesn't give a warning:

int main()
{
    exit(EXIT_SUCCESS);
}

If we use any other function, we will definitely meet such a warning:

warning: control reaches end of non-void function
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How does exit() make the parent function get its return value without using return(), which the compiler makes?

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On GNU libc, exit is declared with __attribute__((__noreturn__)), which tells gcc that the function does not return.

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From the docs:

The status argument is returned to the host environment.

And

Issuing a return statement from the main function is equivalent to calling the exit function with the return value as its argument.

This is implemented (in this case) via a function declaration attribute (__noreturn__) which tells the compiler that it can be treated as a return (or, more correctly, that the function will not return, so control will never reach the end of main).

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instead of

 exit(EXIT_SUCCESS);

have

 return(EXIT_SUCCESS);

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The problem is that you failed to include stdlib.h, and that you're not compiling with -std=c99. In C99, main implicitly returns 0 if you run off the end of it.