Re: A non-dependent name becomes a dependent name?

From:

Alberto Ganesh Barbati <AlbertoBarbati@libero.it>

Newsgroups:

comp.lang.c++.moderated

Date:

2 Nov 2006 18:22:21 -0500

Message-ID:

<nMu2h.26602$uv5.193528@twister1.libero.it>

pongba@gmail.com ha scritto:

template<typename T>
struct B
{
//void f();
};

template<typename T>
struct D:B<T>
{
using B<T>::f;
D()
{
f();
}
};

int main()
{}

According to the standard, f() in D::D() is clearly a non-dependent
name, therefore looked up and bound at the time it's used.
But, since "using B<T>::f" introduced a name f into D, the name look up
will find it, thus bind the f() call to B<T>::f which is clearly a
dependent name.
Does this make f() a dependent name again? Or it's just a case where a
non-dependent name bounded to a dependent name?

?14.6.2/1 couldn't be clearer:

---
[...] In an expression of the form:

   postfix-expression ( expression-list_opt )

where the postfix-expression is an identifier, the identifier denotes a
dependent name if and only if any of the expressions in the
expression-list is a type-dependent expression (14.6.2.2). [...]
---

As there's no expression-list in your case, f is never dependent. The
presence of the using declaration is irrelevant. Whether this is good or
bad, I can't tell. Certainly it's not the first not-so-intuitive name
lookup rule in the realm of templates. Fortunately, there's always the
"this->f()" notation if you want to be sure the name is dependent.

HTH,

Ganesh

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