Reputation: 3573
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.
Upvotes: 0
Views: 1442
Reputation: 67211
int fun(int,int);
function takes 2 int arguments and returns an int
typedef int(*pf) (int,int);
pf is a function pointer that store the address of address of a function which takes two ints as its agrs and returns an int
int proc(pf,int,int);
proc is a function which takes 3 args first is a function pointer to a function like above and two integer args.
proc(fun,6,6);
above statement calls fun with two args 6 and 6 and returns true if they are equal which is how the result is 1
Upvotes: 0
Reputation: 223187
In main the first line
printf("%d\n",proc(fun,6,6));
is calling proc which is taking argument a function pointer and two integer values. Function pointer pf is defined as typedef int(*pf) (int,int);
This line printf("%d\n",proc(fun,6,6));
will call the function defined as:
int proc(pf p,int a,int b){
return ((*p)(a,b));
}
Now in this function pf holds the pointer to function fun. This will cause the function fun to be called which is returning whether the values of a and b are true or not. Since you have passed 6,6 as the arguments the result will be true and that is why you are getting as 1 as an Answer.
Upvotes: 1
Reputation: 43498
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
.
Upvotes: 2