Function Pointers and Argurments

Question

I have been trying to understand a post from geeksforgeeks and am not able to understand a code snippet. Here is the following post:

https://www.geeksforgeeks.org/returning-a-function-pointer-from-a-function-in-c-cpp/

In this post they give and example code as the following:

Based on the following code

#include <stdio.h>
 
int add(int a, int b) {
    return a + b;
}
 
int subtract(int a, int b) {
    return a - b;
}
 
int (*operation(char op))(int, int) {
    if (op == '+') {
        return &add;
    } else if (op == '-') {
        return &subtract;
    } else {
        return NULL;
    }
}
 
int main() {
    int (*func_ptr)(int, int) = operation('+');
    printf("%d\n", func_ptr(4, 5));
    return 0;
}

I'm not understanding how a function be declared like this. Based on my knowledge of function pointers, a function pointer can be declared as such:

int func(int x, int y){
     return x*y;
}

int (*func_ptr) (int, int);
func_ptr = &func;

The description is not very clear as well from the post:

"In this example, we have two functions called add() and subtract() that take two integer arguments and return an integer value. We also have a function called operation() that takes a character argument op and returns a function pointer to add() or subtract() depending on the value of op. Finally, in main(), we call operation() with ‘+’ as the argument and store the result in a function pointer called func_ptr. We then call func_ptr() with 4 and 5 as arguments and print the result to the console."

Any help would be appreciated!

Answer 1

Well, it's just how the C syntax works when you want to return a function pointer.

Not sure if this will make it more clear but it tries to explain what the individual parts of the statement means:

int (*operation(char op))(int, int) 
      \-------/
        Name of this function

int (*operation(char op))(int, int) 
               \-------/
            Arguments for this function

int (*operation(char op))(int, int) 
    ^^                  ^^        ^
    This function returns a function pointer

int (*operation(char op))(int, int) 
                         \---------/
                              The argument types taken by the function the
                              function pointer points to.

int (*operation(char op))(int, int) 
\-/
 The return type of the function the function pointer points to.

The syntax can be simplified using a typedef like:

typedef int (*func_pointer)(int, int);

func_pointer operation(char op) {
    ...
}