Re: Totally omit default parameter brackets from template?

From:

tony@donotspamme.ar

Newsgroups:

microsoft.public.vc.language

Date:

24 Aug 2006 18:08:31 GMT

Message-ID:

<44edeb1f$0$47973$4fafbaef@reader3.news.tin.it>

"Thomas Tutone" <thomas8675309@yahoo.com> wrote:

I have this template class with a default type (please do not give too
much importance to the actual code, it's just an example, it's not a
useful implementation of anything nor a good example of use of void*):

template <typename Type = void>
class MyClass {
   int Var;
   Type* Get() {
      return((Type*)Var);
   }
   void Set(Type* Address) {
      Var=(int)Address;
   }
};

I can declare instances of this class like:

MyClass<int> Instance1;
MyClass<short> Instance2;
MyClass<void> Instance3;

etc..

the latter can be written also as:
MyClass<> Instance3;

and here's finally the question: I'd like to use also this default form:
MyClass Instance3;

which of course will be perfectly equivalent to MyClass<> and MyClass<void>

How do I do it? I tried some namespace trick, but not successfully (which
doesn't mean it's not the right way to go, but just that I'm too lame to
successfully bring it to life).

namespace detail {
template<typename T = void>
class Foo {};
}

typedef detail::Foo<> Foo;

Best regards,

Tom

Thanks Tom, but this way I cannot use Foo<type> anymore, but just Foo !

At this point I could even stop using templates altogether.

It seems that the typedef doesn't allow any use of Foo with <...>, while
I just want to alias Foo to Foo<> or Foo<void>, but not lose the other
Foo<short> Foo<int> etc..

Is there any workaround?

Thanks,
Tony