I've been messing about with member-function pointers in relation to a previous question. In the code below I call methods on a class (B) that change a variable (count) in it, but I never make an instance of this class. Why does this work?
#include <iostream>
#include <string>
#include <map>
class A;
typedef int (A::*MEMFUNC)(int, int);
#define HANDLER(aclass, aproc) (MEMFUNC)(&aclass::aproc)
enum
{
    ADD=1,
    MUL,
    SUB,
    DIV
};
class B
{
    int count;
public:
    B() : count(0) {}
    ~B() {}
    int multiply(int x, int y) { count++; return x*y*count; }
    int divide(int x, int y) { count++; if (y!=0) return (x/y)*count; else return 0; }
};
class A
{
    std::map< int, MEMFUNC > funcs;
public:
    A() { AddLocals(); }
    ~A() {}
    int CallLocal(int nID, int x, int y)
    {
        MEMFUNC f = funcs[nID];
        if (f) return (this->*f)(x, y);
        else return 0;
    }
    void AddLocals()
    {
        Add(ADD, HANDLER(A, plus));
        Add(MUL, HANDLER(B, multiply));
        Add(SUB, HANDLER(A, subtract));
        Add(DIV, HANDLER(B, divide));
    }
    void Add(int nID, MEMFUNC f) { funcs[nID] = f; }
    int plus(int x, int y) { return x+y; }
    int subtract(int x, int y) { return x-y; }
};
int main()
{
    A aA;
    int a,b,c,d;
    a = aA.CallLocal(ADD,8,2);
    b = aA.CallLocal(MUL,8,2);
    c = aA.CallLocal(SUB,8,2);
    d = aA.CallLocal(DIV,8,2);
    std::cout << "a = " << a << "\n" 
              << "b = " << b << "\n" 
              << "c = " << c << "\n" 
              << "d = " << d << "\n";
    return 0;
}
(sorry, me again, but this member-function pointers are making me itch)