How to invoke pointer to member function when it's a class data member?

Question

struct B
{
  void (B::*pf)(int, int);  // data member
  B () : pf(&B::foo) {}
  void foo (int i, int j) { cout<<"foo(int, int)\n"; } // target method
};

int main ()
{
  B obj;
  // how to call foo() using obj.pf ?
}

In above test code, pf is a data member of B. What's the grammar rule to invoke it ? It should be straight forward, but I am not getting a proper match. e.g. If I try obj.*pf(0,0); then I get:

error: must use ‘.*’ or ‘->*’ to call pointer-to-member function in ‘pf (...)’, e.g. ‘(... ->* pf) (...)’

Magyar

see : ideone.com/p3a5G – Coaster

possible duplicate of Calling C++ class methods via a function pointer – Kenny

@CiroSantilli六四事件法轮功纳米比亚威视, No both are different. Your linked question explains, "How to call class member using function pointer?", While I am asking "<...same...>, when the function pointer itself is a class member also.". Notice that answers to both the questions are different. – Magyar

I agree that this has one extra step. but I felt it could be easily deduced once you know the function pointer syntax: (obj.*ptr), just that here ptr == (obj.ptr). But well, since you disagree I'll retract ;-) – Kenny

Answer 1

Like this:

(obj.*obj.pf)(0, 1);

Member access (.) has a higher precedence than a pointer to member operator so this is equivalent to:

(obj.*(obj.pf))(0, 1);

Because function call also has higher precedence than a pointer to member operator, you can't do:

obj.*obj.pf(0, 1) /* or */ obj.*(obj.pf)(0, 1)

As that would be equivalent to:

obj.*(obj.pf(0, 1)) // grammar expects obj.pf to be a callable returning a
                    // pointer to member

Zimmerman

+1, thx; it's unique; what would have happened if obj was a member of B ? :) – Magyar

@iammilind: obj has type B so it can't be a member of B as it would imply a recursive class definition. Can you clarify what you mean? – Zimmerman

No. even if there was a member like obj inside B; it won't be a problem until you have (....) enclosing the signature. I am clear with that problem now. So, either it will be a compile error or it will work fine. – Magyar

Answer 2

pf is a method pointer, and you want to invoke the method it points to, so you have to use

(obj.*obj.pf)(1, 2);

It says the object obj you invoke the method pointed by pf

See result here :

http://ideone.com/p3a5G

Coaster

Answer 3

The syntax is quite unnatural but a consequence of C++ precedence rules...

(obj.*obj.pf)(1, 2);

Halfbound