C++ Function Pointers to an Object

Question

I'm not sure if this is possible in C++. I know you can pass a pointer to a function or static member function as a parameter. I want a function pointer for a specific object, so that when the function is executed, it is done on the object.

class MyClass
{
   public:
      MyClass(int id){mId = id;}
      void execute(){cout<<mId<<endl;}
   private:
      int mId;
};


MyClass obj1(1);
MyClass obj2(2);

typedef (Executor)();
Executor ex1 = &obj1::execute();
Executor ex2 = &obj2::execute();

So when ex1 is executed, "1" should be printed and if ex2 is execute, "2" is printed. Is this possible?

Answer 1

The facility that handles this is the function template bind:

auto ex1 = std::bind(&MyClass::execute, obj1);

You can store a bind in a function object:

std::function<void()> ex1 = std::bind(&MyClass::execute, obj1);

Note that by default bind will store obj by value; you can store a reference with ref:

std::function<void()> ex1 = std::bind(&MyClass::execute, std::ref(obj1));

A related facility is mem_fn, which wraps a member function pointer:

void (MyClass::*ex1)() = &MyClass::execute;  // raw member function pointer
ex1(obj1);

auto ex1 = std::mem_fn(&MyClass::execute);   // mem_fn wrapper
ex1(obj)

However, because mem_fn doesn't bind an instance, you have to supply the instance each time you call it.

---

In order to avoid writing the class name when binding a member function, you can use a macro:

#define BIND_MEM_FN(o,m) \
    std::bind(&std::remove_reference<decltype(o)>::type::m, (o))

A macro is necessary because you can only form a member function pointer from its type and name, and you cannot pass a name (an unqualified-id) to a function.

Answer 2

It's possible, but not the way you describe it. You can do it the way proposed by the above comments but more elegant and C++ way would be to use functors. Functor is basically an object which has operator () overloaded. In your case it can be something like this:

class MyClass
{
   public:
       MyClass(int id){mId = id;}
       void operator()(){cout<<mId<<endl;}
   private:
   int mId;
};

MyClass obj1(1);
MyClass obj2(2);

obj1();
obj2();

This way your object actually mimics function behavior. Here you can read more: http://en.wikipedia.org/wiki/Function_object#In_C_and_C.2B.2B

Answer 3

Member function pointers:

typedef void (MyClass::*Executor)();
Executor ex1 = &MyClass::execute;
Executor ex2 = &MyClass::execute;

obj1.*ex1();
obj2.*ex2();

EDIT: Corrected the code as mentioned in the comments.

I made a mistake because I rarely use (member) function pointers indeed. But not because I use std::bind, but because I use polymorphism :)

Actually I don't remember the last time I used a (member) function pointer except in interfacing with OS APIs. Member function pointers are better than non-member function pointers, and std::bind/std::function might be better than member function pointers, but polymorphism is better than all of the above.