getting function pointer to protected member function of base class

Question

I've got a class which exposes a protected member function of a base class. Is there a way to get a function pointer to the exposed function?

class B
{
protected:
  void foo() {}
};

class D : protected B
{
public:
  using B::foo;
};


void(D::*test)() = &D::foo; // error C2248: 'B::foo' : cannot access protected member declared in class 'D'

Vaughn Cato · Accepted Answer · 2014-04-08 13:56:19Z

2

It's a bit awkward, but if you can't change the original classes, you could make a derived class to give you access:

struct E : D {
  static void (D::*fooPtr())() { return &D::foo; }
};

void(D::*test)() = E::fooPtr();

answered Apr 8, 2014 at 13:56

Vaughn Cato

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Comments

sp2danny · Accepted Answer · 2014-04-08 13:52:23Z

0

in a way there is;

void foo_exposed() { foo(); } // in 'D'

but it gets a new name..

answered Apr 8, 2014 at 13:52

sp2danny

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1 Comment

Daan Posthuma Over a year ago

Doesn't have to: void foo() { B::foo(); }. It is indeed a solution but I would prefer not changing my original code.

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