Re: Constraints => Traits

From:
IR <no_email@use.net.invalid>
Newsgroups:
comp.lang.c++.moderated
Date:
11 Dec 2006 16:31:46 -0500
Message-ID:
<Xns9896D2F84ED0Zjx3HdoP0Qwt@194.177.96.26>
Greg Herlihy wrote:

Michael Aaron Safyan wrote:

I presently have types CopyableCONSTRAINT<T> and
CloneableCONSTRAINT<T> defined to require these respective
functions and to generate informative compiler errors in their
absence.

I would like very much to define CopyableTrait<T> and
CloneableTrait<T> using these constraints. My current definition
for CopyableTrait<T> looks as follows (and CloneableTrait follows
the same pattern):

template<typename T>
class CopyableTrait
      : public Trait<false>{};

template<typename T>
class CopyableTrait< typename CopyableCONSTRAINT<T>::copyabletype

      : public Trait<true>{};

Unfortunately, the definition above does not seem to work


With some template metaprogramming, it is possible to determine
whether a class has a clone() or copy() method with the desired
signature. (Though I am curious what is the difference between a
clone and a copy?).

Here's one technique that I found it on the web:

     #include <iostream>
     #include <ios>

     template<class T, T (T::*)() const>
     struct mf_bind
     {
         typedef T type;
     };

     template<class T1, class T2>
     struct has_copy_member
     {
         static const bool value = false;
     };

     template<class T>
     struct has_copy_member<T,
                           typename mf_bind<T, &T::copy>::type>
     {
         static const bool value = true;
     };

     // A and B test classes
     class A
     {
     public:
         A copy() const;
     };

     class B {};

     using std::cout;

     int main()
     {
         cout << std::boolalpha;

         cout << "A has copy member? ";
         cout << has_copy_member<A, A>::value << "\n";

         cout << "B has copy member? ";
         cout << has_copy_member<B, B>::value << "\n";
     }

      Program Output:

      A has copy member? true
      B has copy member? false

To use these templates for your purposes, just declare
CopyableTrait as shown below (no specialization is needed):

      template<typename T>
      class CopyableTrait
           : public Trait< has_copy_member<T, T>::value>
      {};

Greg


I have been struggling with this issue too. SFINAE works well when
you know the exact signature of the function, but...

CloneableTrait should (IMHO) also be true for polymorphic types:

struct A
{
   virtual A* Clone() const;
};

struct B: public A
{
   A* Clone() const;
};

But one could also write B using a covariant return type.
So the constraint on CloneableTrait<T> is not to have a
T* Clone() const function, but rather S* Clone() const, where S
is either T or one of it's base classes.

And to some extent, CopyableTrait<T> should (still IMHO) be true for
any S Copy() const where S is publicly convertible to T (if this
leads to a split, my understanding is that the implementor of T is
the one who should be blamed).

Such compile-time constraints are kinda easy to assert:

template<typename T>
class CopyableConstraint
{
   template<typename Any>
   static void Helper(Any (T::*)() const)
   {
     // here, some static assert to ensure that
     // Any is convertible to T
   }
   static void Enforce()
   {
     Helper(&T::Copy);
   }
public:
   CopyableConstraint()
   {
     void (*enforce)() = &Enforce;
   }
};

However, no matter how hard I tortured that kind of code to try to
make it fit the SFINAE traits design, I couldn't manage it as of
today.

Managing to push SFINAE beyond the limit of having to know the exact
function signature would definitely allow some decent compile-time
reflection, but all the search I done on so far the subject yielded
no interesting results. :(

The only solution I found so far to have a trait class reflecting
this kind of complex constraints, it to specialize the trait.

Cheers,
--
IR

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