tail recursion in gcc without return statement

Apparently the following code works in GCC. I tried that code in onlinegdb.

# include <stdio.h>

int calc_gcd (int a, int b) {
    int r = a % b;
    if (r == 0) {
        return b;
    } else {
        calc_gcd (b, r);
    }
}

int main() {
    int a, b, gcd, dividend, divisor;
    printf ("Enter two numbers: ");
    scanf ("%d%d", &a, &b);
    dividend = (a > b) ? a : b;
    divisor = (a < b) ? a : b;
    gcd = calc_gcd (dividend, divisor);
    printf ("GCD = %d\n", gcd);
    return 0;
}

But it fails in clang 13 with following results

tail_recursion_gcd.c:15:1: warning: non-void function does not return a value in all control paths [-Wreturn-type]
}
^
1 warning generated.
Enter two numbers: 15 10
GCD = 127       // garbage

I’m not getting it. Clearly what GCC allows isn’t intuitive, you have to return from a function.

I’ve tried the following but that doesn’t work in gcc

# include <stdio.h>

int useless_func()
{
    3050;
}
int main() {
    printf("result = %d", useless_func());
    return 0;
}

The output is result = 0

>Solution :

Your code is bugged. It is allowed to return from an int function without returning a value only if the value will never be consumed. This is a holdover from K&R C that should no longer be used.

int calc_gcd (int a, int b) {
    int r = a % b;
    if (r == 0) {
        return b;
    } else {
        calc_gcd (b, r);
    }
}

Clearly incorrect. You want.

int calc_gcd (int a, int b) {
    int r = a % b;
    if (r == 0) {
        return b;
    } else {
        return calc_gcd (b, r);
    }
}

In fact this specific code tends to work at -O0 because the return value is left over left over in the register is the return value you want.