Re: typename and sizeof

From:

Ulrich Eckhardt <eckhardt@satorlaser.com>

Newsgroups:

microsoft.public.vc.language

Date:

Wed, 05 Dec 2007 12:07:31 +0100

Message-ID:

<kcdi25-l9e.ln1@satorlaser.homedns.org>

Mycroft Holmes wrote:

template <class T>
struct A
{ typedef char type;};

template <class T>
struct B
{
static const int N = sizeof(A<T>::type); // note 'typename' missing here
};

int main()
{
return B<int>::N;
}

Indeed, I would expect a 'typename' to be necessary. Just wondering, does it
hurt to add one?

while it obviously gives error if we try to declare a member in B, whose
type is A<T>::type (without 'typename'). it still compiles if I add a
partial specialization where 'type' is not a type

template <>
struct A<int>
{ static const int type = -99; };

Yes, and that is perfectly fine. The point is that 'sizeof' has two
syntaxes:

1. sizeof reference
2. sizeof (type)

In the first case, brackets are not necessary but also don't hurt (the
expression '(reference)' just yields 'reference'). In the second case, the
brackets around the type are of course necessary.

So, the sizeof-expression doesn't need 'A<T>::type' to be a type to work
correctly and therefore the typename isn't necessary. If the standard
allows that is a totally different thing though, if you want a proper
answer I would rather ask in comp.lang.c++.moderated.

cheers

Uli