Lack of return statement in recursive function, still works

Question

I'm enrolled in a CS class and we are learning about recursion. The teacher showed us some code and I'm wondering why it works. buscar is a recursive search function in an array v[]. It returns the position of x in v[]

int buscar(int x, int v[], int n) {
    if(n == 0) return -1;
    if(x == v[n-1]) return n-1;
    buscar(x, v, n-1);
}

int main() {
    int v[] = {5, 3, 9, 12};
    printf("%d\n", buscar(5, v, 4));
}

The problem here is

buscar(x, v, n-1);

Shouldn't it return 'nothing' and terminate with some error? When I run this program, the result is correct as expected. Why does it work the same as

return buscar(x, v, n-1);

This is not how C works: you can generally never prove that a program is correct by running it. The only way to be sure that a program is correct is to prove that it follows the rules. — Kerrek SB, Mar 26 '19 at 14:16
You are right. there should have been `return buscar(x, v, n-1);` instead. In C#, in example, this won't compile and the compiler says : *"not all code paths return a value"* — Cid, Mar 26 '19 at 14:20

score 0 · Accepted Answer · answered Mar 26 '19 at 14:18

As others commented, your code shows undefined behaviour.

The fact that you "accidentally" (or coidentially) get the results you expect probably is a result of the different levels of functions using the same processor register for returning function results.

Lack of return statement in recursive function, still works

1 Answers1