Why the printf statement inside the fun(int i) is not executed?

Question

This is a code that has a recursive loop. I got the output as 199, but the printf statement marked as --A is not executed. Please tell me why??

#include <stdio.h>

int fun(int); // function defined

int main()
{
    printf(" %d ", fun(200)); // function called
    return 0;
}

int fun(int i) // entire body of the function
{
    static int d = 0;
    d++;

    if (i % 2)
        return (i++);
    else 
        return fun(fun(i - 1));

    printf("%d ", d);// ------A

} // function fun(int) ends here.

Answer 1

From :

if ( i%2 ) return (i++);
else return fun(fun( i - 1 ));

Both parts are returning if the function converges. So printf is unreachable code.

Answer 2

因为return终止当前功能。

Answer 3

return returns control of the program to the calling function. So nothing after return is called inside a function is executed. If you want the printf to be called, put it before return .