Question

I found this problem in a book.

Problem:

What is the output of the following program ?

#include <stdio.h>
int fun(int,int);
typedef int(*pf) (int,int);
int proc(pf,int,int);

int main()
{
    printf("%d\n",proc(fun,6,6));
    return 0;
}

int fun(int a,int b){
    return (a==b);
}

int proc(pf p,int a,int b){
    return ((*p)(a,b));
}

This code, when run, prints out 1.

I tried understanding it but no it is of no use. What is going in this program and why does it output 1?

Thanks in advance.

Answer 1

proc is indirectly calling fun via a function pointer. The arguments that fun receives are again 6 and 6, and the equality operator evaluates to an int with the value 1 because they are equal. If they were not equal, the == operator would yield 0.