alias of function pointer

Question

when we have :

           int functon1(int a, int b);
           int function2(int a, int b);
           .....

           typedef int (*callback) (int,int);

           struct gate{
                  char name[50];
                  callback fptr;
                  };

Where did we give and alias with typedef and what the callback fptr is? Is (int ,int ) an alias to (*callback)?

Answer 1

Here in this statement

typedef int (*callback) (int,int);

/* now callback can be used as a type */

You are typedefing a function pointer. Here callback is a function pointer which can point to any function, that functions has to take two argument of int type and returns int

Next time when you do like

callback fptr;

that means fptr is function pointer and that can point to any function who satisfies callback declaration properties.

Next when you initializes fptr like this

struct gate var;

var.fptr = sum; /* here sum is the function which adds 2 int and returns int */

Answer 2

typedef is used for defining a new type. What this typedef int (*callback) (int,int); means is that you are defining a new type named callback. You can use this type to define variables. A variable having the type callback is actually a pointer to a function which takes two ints and returns an int.

This explanation applies to your struct gate inside which you define fptr to be of type callback.

Here is a simple example:

#include <stdio.h>

int function1(int a, int b);
int function2(int a, int b);

typedef int (*callback) (int,int);

struct gate{
     char name[50];
     callback fptr;
};


int main(void) {
    struct gate g;
    g.fptr = function1;
    printf("The sum is: %d\n", g.fptr(4, 5));
    return 0;
}

int function1(int a, int b)
{
    return a + b;
}

Running the above program shows: The sum is: 9