How can I pass arguments to the function pointer in following code?

Question

I tried those code but there is some issue with scope of parameters a and b. please someone help me out here.

#include<conio.h>
#include<stdio.h>
int add(int x, int y) {
    return (x + y);
}
void passptr(int( * fp)(int a, int b)) {
    int result = ( * fp)(a, b);
    printf("%d", result);
}
int main() {
    add(3, 5);
    passptr( & add);
    getch();
    return 0;
}

Answer 1

This kind of call is easier to understand if you typedef your function pointer:

#include<conio.h>
#include<stdio.h>

// addFuncPtr_t is a pointer to a function that:
// - returns int
// - takes two int arguments
typedef int ( *addFuncPtr_t )( int, int );

int add(int x, int y) {
    return (x + y);
}
void passptr(addFuncPtr_t fp, int a, int b) {
    int result = fp(a, b);
    printf("%d", result);
}
int main() {
    add(3, 5);

    // note that the function is passed separately from
    // the arguments - add(3,5) would *call* the function
    // instead of passing the address of the function
    passptr( add, 3, 5 );
    getch();
    return 0;
}