Function returns value without return statement

Question

Why does following code has a correct output? int GGT has no return statement, but the code does work anyway? There are no global variables set.

#include <stdio.h> #include <stdlib.h>  int GGT(int, int);  void main() {     int x1, x2;     printf("Bitte geben Sie zwei Zahlen ein: \n");     scanf("%d", &x1);     scanf("%d", &x2);     printf("GGT ist: %d\n", GGT(x1, x2));     system("Pause"); }  int GGT(int x1, int x2) {     while(x1 != x2) {         if(x1 > x2) {             /*return*/ x1 = x1 - x2;         }         else {             /*return*/ x2 = x2 - x1;         }     } } 

Answer 1

For x86 at least, the return value of this function should be in eax register. Anything that was there will be considered to be the return value by the caller.

Because eax is used as return register, it is often used as "scratch" register by callee, because it does not need to be preserved. This means that it's very possible that it will be used as any of local variables. Because both of them are equal at the end, it's more probable that the correct value will be left in eax.