Why does call to exit function create a new basic block

I have a function defined as:

int f_2() {
  rand();
  return 10;
}

clang breaks it up into 3 basic blocks. This is understandable.

however when I replace the call to rand() by exit(0), then it breaks it into 4 basic blocks. Wikipedia (http://en.wikipedia.org/wiki/Basic_block) says that functions that cannot return can be at the end of a basic block.

How does clang know that exit() function does not return? I am compiling my code with clang -c.

Solution

clang is a C compiler. It's permitted to take advantage of guarantees made by the language standard for standard library functions.

There may also be something in the particular implementation of <stdlib.h> that marks exit() as a function that doesn't return, perhaps using a language extension or the _Noreturn keyword added by the 2011 ISO C standard.

Another example of this: the call sin(0.0), with -O1 or higher, compiles to a literal 0.0, because the compiler knows about the sin function. (Which means that a program that calls sin(0.0) needs to be linked with -lm only if you don't optimize it.)

This is all permitted because a program that defines its own (non-static) function with the same name as a standard library function has undefined behavior; the compiler needn't consider the possibility that a call to exit or sin does anything other than what the standard specifies for those functions.