Thread: What does .* and ->* do?

Im wonderin what do those operators do? My book is lazy and wont tell me.

Originally posted by vVv
Probably because there is no such operator in C++.

Actually, they are.

They actually do consider .* and ->* single operators.

All they do are derefernce and "attach" a member function pointer to an object directly or through a pointer to an object (respectively). I personally always just think of .* as 2 operators: the . operator and the * operator (same for ->*), but they actually are single operators of their own.

Example usage:

Code:

#include<iostream>

class An
{
public:
    void SomeFunction() const;
};

void An::SomeFunction() const
{
    std::cout << "SomeFunction\n";
}

int main()
{
    void (An::*MemFunc)() const = An::SomeFunction;
    An Object,
      *ObjectPointer = &Object;
    (Object.*MemFunc)();
    (ObjectPointer->*MemFunc)();
    return 0;
}

Originally posted by vVv
OK, ISO/IEC 14882-1998 agrees with ``them''.

Ha, yeah -- it's so much easier to just say "them."

Yeah thanx Polymorphic dude