circumventing recursive definition wrt template specialization

[SainTiss]

Hi,

If you've got a template class with lots of methods,
and then you've got a type which works with the template,
except for one method...

What you need to do there is specialize the template for your
type. However, this requires you to copy the whole template, and
change the method, which leads to code duplication...
If there's only 1 template parameter, one can specialize the method
instead of the whole class, but with more template parameters,
this is impossible, because partial specialization of a
method isn't allowed...

Therefore, it would be handy to be able to do this:
template<class A, class T> myClass<A,T*> : public myClass<A,T*>

However, the compiler will complain that you're having a
recursive definition, which is illegal...
Obviously, the compiler thinks that myClass<A,T*> is being
defined here, and therefore no other definition exists...
But in this case, myClass<A,T*> is a template specialization,
and therefore there IS another definition, namely the original
template...

So, is there a reason why this isn't possible, or is this just
something that doesn't happen too often, and therefore not worth
extending the standard?

Thanks,

Hans

[tom_usenet]

On Tue, 04 Nov 2003 14:29:25 GMT, SainTiss <st***@gmx.ne t> wrote:

Hi,

If you've got a template class with lots of methods,
and then you've got a type which works with the template,
except for one method...

What you need to do there is specialize the template for your
type. However, this requires you to copy the whole template, and
change the method, which leads to code duplication...
If there's only 1 template parameter, one can specialize the method
instead of the whole class, but with more template parameters,
this is impossible, because partial specialization of a
method isn't allowed...

Therefore, it would be handy to be able to do this:
template<cla ss A, class T> myClass<A,T*> : public myClass<A,T*>

However, the compiler will complain that you're having a
recursive definition, which is illegal...
Obviously, the compiler thinks that myClass<A,T*> is being
defined here, and therefore no other definition exists...
But in this case, myClass<A,T*> is a template specialization,
and therefore there IS another definition, namely the original
template...
The problem is that specialization is couched in such terms as there
*isn't* an original template. Often the original template wouldn't
make sense or even compile for the type that you are specializing for.
Also, how you you specify that you want to refer to a particular less
specialized version? e.g.

template <class U, class V, class W>
class Foo
{
};

template <class U, class V>
class Foo<U, V, int>
{
};

template <class U, class V>
class Foo<U, int, V>
{
};

template <class U>
class Foo<U, int, int>: public Foo<U, int, int>//which specialization?
{
};

I think it was considered a bit illogical, since a specialization is
by definition the version of the template for the particular
parameters - there is no "original" template, rather a "fall back"
definition. E.g. often this is done:

template <class T>
class Foo; //never defined

//specialisations are defined.
So, is there a reason why this isn't possible, or is this just
something that doesn't happen too often, and therefore not worth
extending the standard?

There's a simple workaround - put the common stuff in a
(implementation only, not interface) base class. To accomodate this
feature, rather than extending the standard, large sections would have
to be rewritten, since it contradicts what is said elsewhere. The
definition as written is reasonably compact and logical, whereas what
this extension would mean in syntax and consistency terms isn't.

Tom

[Rob Williscroft]

SainTiss wrote in news:9J******** **********@phob os.telenet-ops.be:

Hi,

If you've got a template class with lots of methods,
and then you've got a type which works with the template,
except for one method...

What you need to do there is specialize the template for your
type. However, this requires you to copy the whole template, and
change the method, which leads to code duplication...
Here's another common way of doing it:

#include <iostream>

namespace detail
{
template < typename U, typename V >
struct basic_method_he lper;

}

template < typename U, typename V >
struct basic
{
typename
detail::basic_m ethod_helper< U, V >::return_typ e
method( V & v )
{
return detail::basic_m ethod_helper< U, V >::apply( this, v );
}
V multiply;
basic( V f ): multiply( f ) {}
};

namespace detail
{
template < typename U, typename V >
struct basic_method_he lper
{
// non-specialized versions
typedef U return_type;
static return_type apply( basic< U, V > *that, V & v )
{
return U(v * that->multiply);
}
};

template < typename U >
struct basic_method_he lper< U, U >
{
// partialy specialized version
typedef U return_type;
static return_type apply( basic< U, U > *that, U & v )
{
return v * that->multiply;
}
};
}

int main()
{
double d = 2.5;

basic< int, double > bid( 3.5 );
basic< double, double > bdd ( 7 );

std::cerr << bid.method( d ) << "\n";
std::cerr << bdd.method( d ) << "\n";
}
Its a fair bit of typing, but more maintainable than copying a
whole class.
If there's only 1 template parameter, one can specialize the method
instead of the whole class, but with more template parameters,
this is impossible, because partial specialization of a
method isn't allowed...

Therefore, it would be handy to be able to do this:
template<class A, class T> myClass<A,T*> : public myClass<A,T*>

Well why not just allow partial specialization of a function's,
member and non-member. I've read talk that a future standard
might allow this, but wether this would extend to member function's
I don't know (but I can't see why not).

If you want to talk about the why's of the current standard and
it's future direction's comp.std.c++ maybe better than here.

[snip]

Rob.
--
http://www.victim-prime.dsl.pipex.com/