What is the meaning in C when a function is in another function's name (not in the arguments)?

Question

Why is "void (*parse(char *op))(int, int)" written like this? (main added to give use case), it gets called from a pointer "fun" with argv[2] without the ints (line 18)... and then again as "fun" with the ints (line 21)?

void    (*parse(char *op))(int, int)
{
    if (!strcmp(op, "+"))
        return(other stuff...); 
    else
        return (0);
}
int     main(int argc, char *argv[]){
    int a;
    int b;
    void    (*fun)(int, int);

    if (argc == 4){
        a = atoi(argv[1]);
        fun = parse(argv[2]);
        b = atoi(argv[3]);
        if (fun)
            fun(a, b);
        else
            return(0);
    }
    return(0);
}

How does it technically work, and it is just a showoff with a simpler way to write it or is this the only correct grammar?

Answer 1

parse() is a function that returns a function pointer. You can use typedef with the type of the returned function to make it look a lot nicer, though:

#include <string.h>
#include <stdlib.h>

// parser_func describes a function that takes two ints and returns nothing
typedef void (*parser_func)(int, int);

// Like this one.
void somefunc(int a, int b) {}

// And parse() is a function that takes a char* and returns a
// pointer to a function that matches parser_func.
parser_func parse(char *op) {
  if (!strcmp(op, "+"))
    return somefunc;
  else
    return NULL;
}

int main(int argc, char *argv[]) {
  int a;
  int b;
  parser_func fun; // Function pointer

  if (argc == 4) {
    a = atoi(argv[1]);
    fun = parse(argv[2]); // Assign a function to fun
    b = atoi(argv[3]);
    if (fun) {
      fun(a, b); // And call it if it's not null
    }
  }
  return 0;
}