控制如何到达这个非void function的结尾？

Question

int test_ex(int num) {
  
  if (num >= 1000) {
    printf("+");
  } else {
    return num;
  }
  
}

int main(void) {  
   test_ex(123812);
}

I'm confused here as est_ex returns an int, but both if/else clauses return ints, right? So how would it be possible for there to be a non-int return here, I've been stuck on this for a long time. I tried adding a redundant return 0 but that messes up the output I want.

Any advice?

Answer 1

If the function "reaches" the end brace it returns. If it is supposed to return a value, then the behavior is undefined.

Update: it is undefined behavior only if the return value of the function is used:

from C99 6.9.1 Function definitions:

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

Since the snippet of code in the example doesn't use the return value, there is no undefined behavior.

Answer 2

You are missing that you do not return any int if num >= 1000 so,

int test_ex(int num) {
  // No matter you should return
  if (num >= 1000) {
    printf("+");
    return num;
  } else {
    return num;
  }

}