Typedef function pointer?

Question

I'm learning how to dynamically load DLL's but what I don't understand is this line typedef void (*FunctionFunc)();I have a few questions. If someone is able answer them I would be grateful.1. Why is `typedef` used?2. The syntax looks odd; after `void` should there not be a function name or something? It looks like an anonymous function.3. Is a function pointer created to store the memory address of a function?So I'm confused at the moment; can you clarify things for me?

Answer 1

typedef is a way to define your own type definition (type alias) to make declarations easier to read. So, in your case, the typedef is creating a type alias called "FunctionFunc", which is a "pointer to a function that takes no arguments and returns nothing (void)":

#include <iostream>

void f() {

  std::cout << "Function f called\n";

}

int main(int argc, const char * argv[]) {

   

  typedef void (*FunctionFunc)();

   

  FunctionFunc funcPtr = f;

   

  funcPtr();

   

  return 0;

}

In the example above, I defined a variable (funcPtr) and initialized it to the address of the function (f). You may ask "why do this? Why not just call f directly?". The answer is that funcPtr is a pointer and therefore can be reassigned to point to another function of the same function prototype :)

I hope this helps!