views:

69

answers:

1
struct B1{
  int d;
  void fb(){};
};

struct B2 : B1{
  using B1::d;
  using B1::fb;

  int d;               // why this gives error?
  void fb(){}          // and this does not?
};

int main(){}

Is it because, B1::fb() is treated as B1::fb(B1*) and B2::fb() treated as B2::fb(B2*)? That is, does the implicit parameter, help in distinguishing these?

$13.3.1/4-

For nonconversion functions introduced by a using-declaration into a derived class, the function is considered to be a member of the derived class for the purpose of defining the type of the implicit object parameter.

+7  A: 

The C++ standard (C++03 §7.3.3/12) explains:

When a using-declaration brings names from a base class into a derived class scope, member functions in the derived class override and/or hide member functions with the same name and parameter types in a base class (rather than conflicting).

In your example, B2::fb() hides the B1::fb() introduced by the using declaration.

As for why it is ill-formed to have both using B1::d; and int d; in the definition of B2, the C++ standard (C++03 §7.3.3/10) explains:

Since a using-declaration is a declaration, the restrictions on declarations of the same name in the same declarative region also apply to using-declarations.

So, it is ill-formed for the same reason that the following is ill-formed: it results in two objects with the same name in a single declarative region:

struct S { int d; int d; };
James McNellis
and int d conflicts with the previous declaration of it from the using...
diverscuba23
what is the real impliciation of $13.3.1/4 in that case?
Chubsdad
@chubsdad: Since `B2::fb()` hides `B1::fb()`, `B1::fb()` is not considered as a candidate function during overload resolution, so §13.3.1/4 does not apply.
James McNellis