Re: C2248: cannot access protected member
"Igor Tandetnik" wrote:
Interesting. So &X::foo and &Y::foo have the exact same value
and the same type (pointer-to-member-of-X), but access check
fails for the former and passes for the latter on purely
syntactical grounds.
I think, there is more to it than just syntax. If `Y' actually
contains `foo' member, then "&Y::foo" yields
"pointer-to-member-of-Y" and compilation rightfully fails. So, it
seems that compiler searches starting from `Y' namespace upward
for the `foo' name. The scope where `foo' is found determines
resulting type.
But that does seem to mean I can call a protected method on an
unrelated class:
class X {
protected:
void foo();
};
class Z : public X {};
class Y : public X {
public:
void bar() {
Z z;
// z.foo(); // doesn't compile
void (X::*pf)() = &Y::foo;
(z.*pf)(); // compiles; calls z.foo()
}
};
Looks to me like a hole in the standard.
Moreover, suppose "void (X::*pf)() = &Y::foo;" statement won't
compile. Then you could legally circumvent it with `static_cast'
(5.2.9/9):
Z z;
// z.foo(); // doesn't compile
void (Y::*pf)() = &Y::foo;
void (X::*pf2)() = static_cast<void (X::*)()>(pf);
(z.*pf2)(); // compiles; calls z.foo()
Alex
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