Question

This is a functor:

class add_x {

 int x;

public:
  add_x(int x) : x(x) {}
  int operator()(int y) { return x + y; }

};

And from main I can do this:

add_x add10(10); // create my functor 
int i = add10(20); // add 20 and return it 

How can I combine the functor with typedef?

For instance, I came across this:

typedef int (*myfuncType)(float f, char c1,char c2);
myfuncType pt2Function = NULL; 

But what am I defining here exactly? operator ()?

Answer 1

Function pointer is – as its name says – just a pointer to function. You cannot use a pointer to function to point to functor, which is essentially a class with overloaded () operator.

You may be interested in std::function from C++11:

#include <functional>

(...)
std::function<int(int)> fn = add_x(10);
int i = fn(20);

---

Edit: Ok, I see, what you are asking for.

Maybe some code will make things clear.

int fn(float a, char b, char c) { ... }

(...)

typedef int (*myFuncType)(float f, char c1, char c2);

myFuncType ptrToFun = &fn;

(*fn)(1.0f, 'a', 'z');

This typedef creates a “pointer-to-function” type. If you declare a variable of this type, then you’ll be able to assign to it a pointer to function (extracted with regular & operator) and – for example – pass to another function or just store (or call this function).

std::function is a lot more flexible, because you can store in it a pointer to function, a functor or even a lambda and call them normally.