C++ Pointer to virtual function

Question

If you have a struct like this one

struct A {
    void func();
};

and a reference like this one

A& a;

you can get a pointer to its func method like this:

someMethod(&A::func);

Now what if that method is virtual and you don't know what it is at run-time? Why can't you get a pointer like this?

someMethod(&a.func);

Is it possible to get a pointer to that method?

Ulrikeulster

Answer 1

Pointers to members take into account the virtuality of the functions they point at. For example:

#include <iostream>
struct Base
{
    virtual void f() { std::cout << "Base::f()" << std::endl; }
};

struct Derived:Base
{
    virtual void f() { std::cout << "Derived::f()" << std::endl; }
};


void SomeMethod(Base& object, void (Base::*ptr)())
{
    (object.*ptr)();    
}


int main()
{
    Base b;
    Derived d;
    Base* p = &b;
    SomeMethod(*p, &Base::f); //calls Base::f()
    p = &d;
    SomeMethod(*p, &Base::f); //calls Derived::f()    
}

Outputs:

Base::f()
Derived::f()

Renner

Thanks! I didn't realize a member function pointer would take that into account. – Ulrikeulster

@Chris: Member function pointers are smart. That's why they can be larger than usual function pointers. They even take into account the offset of a base-class subobject in case of multiple inheritance – Renner

Answer 2

The way to invoke a function pointer is to also provide its object's instance pointer. This will take care of all virtuality issues:

struct A { void func(); };

int main()
{
  typedef void (A::*mf)();

  A x; // irrelevant if A is derived or if func is virtual

  mf f = &A::func;   // pointer-to-member-function
  A* p = &x;         // pointer-to-instance

  (p->*f)();           // invoke via pointer
  (x.*f)();            // invoke directly
}

---

OK, interesting syntax challenge question: Suppose I have this.

struct Basil { virtual void foo(); virtual ~Basil(); };
struct Derrek : public Basil { virtual void foo(); };

Now if I have Derrek * p or a Basil * p, I can invoke the Basil member via p->Basil::foo(). How could I do the same if I was given a void(Derrek::*q)() = &Derrek::foo?

Answer: It cannot be done. The PMF q alone does not know whether it points to a virtual function, let alone which one, and it cannot be used to look up a base class function at runtime. [Thanks to Steve and Luc!]

Propitious

I think you mean p->*f(). (but now say that func is virtual and could either be A::func or B::func - you don't know what it is. How can you get a pointer to the right one?) – Ulrikeulster

Chris, I edited the typo. In you original question, you do not need to worry, you will always call the correct polymorphic function. – Propitious

@Chris, at the low level, the member function pointer, for virtual functions, contains a vtable index, not an actual function pointer. When the function pointer is invoked the compiler will look up the index in the vtable and call the function in question. – Cribbage

Exactly. Calling via PTMF invokes the exactly same machinery that a straight call to x.func() would cause if x is polymorphic, namely a lookup in the class's vtable of the most derived location of func. – Propitious

The braces around (p->*f) are not optional. Without them the precedence rules will stop it from compiling. – Luddite

@Martin: Thanks, I had forgotten about the parentheses :-) – Propitious

"syntax challenge" so given a pointer to an unknown derived class member function, you want to call the base class member function of the same name? I don't think you can, although I might be wrong. The point about p->Basil::foo() is that you're using a fully-qualified name to refer to the method of Basil - given just a pointer you don't know the name of the function and hence you can't use that same symbol in a fully-qualified name. – Timelag

@Steve: Just discussing that with Luc, apparently it can't be done at all, you're right! The pointer alone does not contain enough information. I'll edit this! – Propitious

Answer 3

You can get a pointer to it, like so:

struct A {
    virtual void foo() = 0;
};

typedef void (A::*LPFOO)();

LPFOO pFoo = &A::foo;

Cribbage