Re: Class template specialization with template parameter

From:

Greg Herlihy <greghe@mac.com>

Newsgroups:

comp.lang.c++

Date:

Wed, 14 May 2008 13:20:24 -0700 (PDT)

Message-ID:

<760ff606-eac4-4681-9a1c-c0fcff9e84a5@y18g2000pre.googlegroups.com>

On May 14, 12:51 pm, flopbucket <flopbuc...@hotmail.com> 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();

I believe that you mean to declare an object (and not a function):

Foo<int> myFoo;

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

Yes.

But if I declare:

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

Again, I believe that you want:

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

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.

The answer depends on how many template parameters "std::map"
requires. The usual number is two - the two specified by the C++
Standard. An implementation however is allowed to require additional
temmplate type parameters for a std::map. So, assuming that std::map
requires only two type parameters, then the answer is "yes".

Greg