Template partial specialization

Question

Would any one knows according to what rules code below doesn\'t compile?

template 
struct B
{
    typedef T type;
};

template

        

Answer 1

You need to use typename keyword as,

template<class T>
struct X<typename B<T>::type*>
{
};

It's because B<T>::type is a dependent name. So typename is required!

--

EDIT:

Even after putting typename, it isn't compiling. I think it's because deduction of type T in B<T> from X<U> is difficult, or possibly impossible, for the compiler. So I believe its non-deduced context.

See a similar example here and the discussion:

Template parameters in non-deduced contexts in partial specializations

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However, if you change the specialization to this:

template<class T>
struct X<B<T> >
{
};

Then it becomes the deducible context, and so would compile.

Answer 2

How do you think that will work? The compiler will look to see if there is a class T somewhere that has a typedef "type" to your class?

It just won't. Even though it's a pointer.

Remember that presumably your B template is presumably specialised in places so that type is not always T*, but it can't deduce it with reverse engineering.

For those who did not understand my answer fully, what you are asking the compiler to do is find a class U such that B::type is the class you pass in as a parameter.

class Foo;
class Bar;

template<> struct B<Foo>
{
  typedef int type;
};

template<> struct B<Bar>
{
  typedef int type;
};

X<int*> // ambiguous, T is Foo or Bar?

It is difficult to know exactly why you are trying to do what you are. You can do a partial specialization on all pointers and then a total specialization on specific pointers, which could be implement in terms of another template.

Answer 3

Assuming you already added typename as suggested by Nawaz.

The problem is exactly explained in the error message you encounter: "template parameter is not deducible in partial specialization B<T>::type*. The problem is that B<T>::type and T is exactly the same for all types T. Consider the following example:

class MyClass1 {};
typedef typename B<MyClass>::type MyClass2; //(*)

X<MyClass*> obj1;
X<MyClass2*> obj2;

The result of line (*) is a type MyClass2 which is essentially MyClass1. So, obj1 and obj2 should be objects of the same class. Now, which version of template X should they use?

If you would expect the specialised version of X, tell me if the answer should be the same if line (*) is removed (and obviously obj2 as well). Still obj1 should be the specialised version of X as line (*) has nothing to do with it.

But now we expect the compiler to detect that some type can be potentially declared as B<T>::type although we never do this. We expect the compiler to verify all possible template instantiations to check if there is no strange typedef in one of them.

I hope this clarifies why such specialisation cannot be handled by the compiler.

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An alternative that might help

I believe your problem could be attacked by creating a trait class for explicitly marking types that should be handled in a special way. Something like this:

template <bool v>
struct boolean_value {
  static const bool value=v;
};

template <typename T>
struct is_my_interesting_type : public boolean_value<false> {};

class MyClass {
  ...
};

template <>
struct is_my_interesting_type<MyClass> : public boolean_value<true> {};

template <typename T, bool special>
class  InternalX {
  ... //generic version of your template X
};

template <typename T>
class InternalX<T,true> {
  ... //special version of your template X
};

template <typename T>
class X : public InternalX<T,is_my_interesting_type<T>::value> {};

Also, you might be interesting how it is done in boost library, in particular Boost.Type_Traits