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
struct X::type*>
{
};

It's because B::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 from X 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

---

However, if you change the specialization to this:

template
struct X >
{
};

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