In the example below, why is B::f() called even though it is private?
I know this fact that : Access is checked at the call point using the type of the expression used to denote the object for which the member function is called.
#include <iostream>
class A {
public:
virtual void f() { std::cout << "virtual_function"; }
};
class B : public A {
private:
void f() { std::cout << "private_function"; }
};
void C(A &g) { g.f(); }
int main() {
B b;
C(b);
}
Because the standard says so:
[C++11: 11.5/1]:The access rules (Clause 11) for a virtual function are determined by its declaration and are not affected by the rules for a function that later overrides it. [ Example:class B { public: virtual int f(); }; class D : public B { private: int f(); }; void f() { D d; B* pb = &d; D* pd = &d; pb->f(); // OK: B::f() is public, // D::f() is invoked pd->f(); // error: D::f() is private }—end example ]
The example is the same as yours, lol.
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