Why is return 0 optional?

Why, if I write

int main() 
{ 
    //... 
}

do I not need to write return 0; at the end of the main function? Does the compiler do it for me?

I use GCC / C99.

The most recent C (currently that's C99 with a few amendments) returns 0 from main by default if there is no explicit return statement at the end of the function, and control flows off the function's end (see 5.1.2.2.3 in C99 TC3). This is because most often one would write such a form of return anyway.

In C89 you need to return something there - it has no such implicit return. But the compiler is by no means required to diagnose such a mistake (see 3.6.6.4 in the C89 draft and 6.9.1/12 in C99 TC3).

C99 and C++ special case the main function to return 0 if control reaches the end without an explicit return. This only applies to the main function.

The relevant bit of the C99 spec is 5.1.2.2.3 for the main special case

5.1.2.2.3 Program termination

If the return type of the main function is a type compatible with int, a return from the initial call to the main function is equivalent to calling the exit function with the value
returned by the main function as its argument; reaching the } that terminates the main function returns a value of 0.

6.9.1/12

If the } that terminates a function is reached, and the value of the function call is used by the caller, the behavior is undefined.

You can test this out with gcc:

int foo ( void ) { }
int main( void ) { }

C89 mode ( errors for both functions ):

sandiego:$ gcc src/no_return.c -std=c89 -Wall 
src/no_return.c: In function ‘main’:
src/no_return.c:2: warning: control reaches end of non-void function
src/no_return.c: In function ‘foo’:
src/no_return.c:1: warning: control reaches end of non-void function

C99 mode ( main is a special case ) :

sandiego:$ gcc src/no_return.c -std=c99 -Wall
src/no_return.c: In function ‘foo’:
src/no_return.c:1: warning: control reaches end of non-void function