Re: Class template specialization with template parameter

From:
Victor Bazarov <v.Abazarov@comAcast.net>
Newsgroups:
comp.lang.c++
Date:
Wed, 14 May 2008 16:06:19 -0400
Message-ID:
<g0fgnr$vdf$1@news.datemas.de>
flopbucket wrote:

Hi,

First, thanks for the reply.

Foo<int> uses the normal template, but Foo<std::map<T1, T2> > for any
types of T1 and T2 uses the specifalization?

Not sure what you're asking here.


What I was trying to say was that if I declare:

Foo<int> myFoo();

The instantiation will be for the normal/not-specialized template.


Yes. Of course, 'myFoo' in that case is a function, but the template
is still instantiated, I believe.

But if I declare:

Foo<std::map<int, std::string> > myFoo2();


Same thing. It's a function. But the 'std::map'-based specialisation
should be used, of course.

it will use the specialization... and that regardless of the types
used for std::map (in this case, int and std::string), any
"Foo<std::map<typeA, typeB> > myFooExample()" will always
instantiate the specialization.


Yes. Or do you still have a doubt about that?

V
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