Re: C++0x, confusion with using-declaration in multiple base classes.

From:
"A. McKenney" <alan_mckenney1@yahoo.com>
Newsgroups:
comp.lang.c++.moderated
Date:
Mon, 25 Apr 2011 01:04:33 CST
Message-ID:
<d12ee770-3734-48f6-a3a5-4f25784bb4d0@p23g2000vbl.googlegroups.com>
On Apr 22, 11:03 pm, Johannes Schaub <schaub.johan...@googlemail.com>
wrote:

... the following is fine:

 struct A {
   void f();
 };

 struct B : A {
   using A::f;
 };

 struct C : A {
   using A::f;
 };

 struct D : B, C {

 };

 void g() { &D::f; }

In other words, member name lookup as described by 10.2 only finds A::f.


I don't understand why this is not ambiguous.
A call to B::f() is not the same as C::f(),
since B::f() is operating on the first
A subobject (from B), while C::f() is operating
on the second A subobject.

To make it more obvious, consider a more realistic example:
---------------------------------
#include <iostream>

struct A { int x; A(int i = 0) : x(i) {} int f() { return x; } };
struct B : public A { using A::f; B() : A(1) {} };
struct C : public A { using A::f; C() : A(2) {} };
struct D : public B, C {};

int main() {
A a; std::cout << a.f() << std::endl; // prints out 0
B b; std::cout << b.f() << std::endl; // prints out 1
C c; std::cout << c.f() << std::endl; // prints out 2
// D d; std::cout << d.f() << std::endl; // would print out what?
}
-----------------------------------

The last line doesn't compile under g++, but if it did,
what would you want it to print out?

The case of virtual inheritance is different, since
B::f and C::f are in this case the same code acting
on the same subobject.

But I ask: what does allowing this very special case
add to the language? And is it enough to justify
making the language more complicated? Especially
given that there is a well-known, easy-to-understand,
and quite general way to deal with the ambiguity?

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