Re: specialization on function functors

From:
Yechezkel Mett <ymett.on.usenet@gmail.com>
Newsgroups:
comp.lang.c++.moderated
Date:
Tue, 17 Feb 2009 10:07:18 CST
Message-ID:
<cf94d4cf-204b-4b76-91e0-d067e582af5f@x9g2000yqk.googlegroups.com>
On Feb 17, 12:31 am, p7ere...@gmail.com wrote:

Hi,

I am building a template that takes an argument class F which is a
functor template argument. Normally this does equality testing with
the == sign so that a global scope template function Equals can do the
job. Moreover, the F argument uses a T argument that is also part of
the template class signature. My question is this, is there a way to
specialize when F argument is the global function Equals and not a
true functor?

template <class T, class F>
class MyClass {
     F f;
     T t1, t2;

     bool Compare ()
     {
          return F(t1, t2);
     }

};

This would be the unspecialized version.
Here's the global scope function:

template <class T>
bool Equals (T t1, T t2)
{
     return t1 == t2;

}

typedef MyClass<int, Equals> cls;


I think you'll find that won't compile -- Equals is not a type.

How to specialize when F==Equals?


If Equals were a type (like so:

struct Equals
{
     template <class T>
     bool operator()(T t1, T t2)
     {
         return t1 == t2;
     }
};

), you could specialise like this:

template <class T>
class MyClass<T, Equals> {
     // specialised members here
};

Why do yu want to specialise it?

Yechezkel Mett

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