Question

I recently wrote a post:
Weird Error in C++ Program: Removing Printout Breaks Program

…in which I was trying to solve a seemingly baffling problem, in which removing a cout statement would break my program.

As it turned out, my problem was that I forgot to return my true/false success flag that I was later using for logic.

But apparently SOMETHING was being returned and that something was always true if I left that cout in, but would seemingly “magically” become false when I took it out.

My question for you all is:
What determines what a c++ function return when no return command is executed within the function? Is there any logic to it?

Obviously forgetting your return type is a bad idea. In this case, though, it was largely due to the nature of my program — a quick hack job. I later decided that it wasn’t worth the effort to include implement an algorithm to determine the success/failure of the function call — but accidentally left behind the code dependent on the return.

Bafflingly g++ gave me no warnings or errors when compiling the executable like so:

g++ main.cc -g -o it_util

My version is:
g++ (GCC) 4.1.2 20080704 (Red Hat 4.1.2-44)

Again, to save others future frustration in case they make the same silly mistake and are met with the same seemingly erratic behavior, can anyone cast light on where a function without a return gets its return value from??

Thanks!!

Answer 1

On x86 calling conventions, the return value for integers and pointers is on the EAX register. The following is an example of that:

int func() {
    if(0) return 5; // otherwise error C4716: 'func' : must return a value
}
int main() {
    int a;
    a = func();
}

Compiling with cl.exe /Zi, MSVC++10:

push    ebp
mov     ebp, esp
push    ecx
call    j_?func@@YAHXZ  ; func(void)
mov     [ebp+a], eax ; assumes eax contains the return value
xor     eax, eax
mov     esp, ebp
pop     ebp
retn

Of course, this is all undefined behavior.