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.