Understanding snippet of code

Question

Im reading a book and there is no online code reference so I don't understand couple of things. Complete code:

#include <stdio.h>
/* The API to implement */
struct greet_api    {
  int (*say_hello)(char *name);
  int (*say_goodbye)(void);
};
/* Our implementation of the hello function */
int say_hello_fn(char *name)    {
  printf("Hello %s\n", name);
  return 0;
}
/* Our implementation of the goodbye function */
int say_goodbye_fn(void)    {
  printf("Goodbye\n");
  return 0;
}
/* A struct implementing the API */
struct greet_api greet_api =     {
  .say_hello = say_hello_fn,
  .say_goodbye = say_goodbye_fn
};
/* main() doesn't need to know anything about how the
* say_hello/goodbye works, it just knows that it does */
int main(int argc, char *argv[])    {
  greet_api.say_hello(argv[1]);
  greet_api.say_goodbye();
  printf("%p, %p, %p\n", greet_api.say_hello, say_hello_fn, &say_hello_fn);
  exit(0);
}

I have never seen before:

 int (*say_hello)(char *name);
 int (*say_goodbye)(void);

and:

struct greet_api greet_api =    {
  .say_hello = say_hello_fn,
  .say_goodbye = say_goodbye_fn
};

I kind a understand what is happening there, but as I stated before I have never seen that before.

Those things are double brackets in first snippet. Equal sign, dots and function name w/o brackets in second snippet.

If someone could post some good articles or titles of what is that so I can look online it, I would really appreciate it.

Answer 1

These are respectively function pointers and designated initializers. The function names without brackets are just the way you take a function's address to store it in a function pointer (the & is facultative in this case).

Answer 2

The first two are function pointers (How do function pointers work) and then, the author initializes these two pointers via the designated initializers (check this answer).