can template alias be used for partial specialization?

Question

Given a template alias

template<unsigned U>
using uint_ = integral_constant<unsigned,U>;

The partial specialization of

template<class T,class P>
struct size{};

as

template <class T,unsigned U>
struct size<T,uint_<U>>{};

generates a warning astemplate parameter can not be deduced for clang 3.1 while no warning is generated with gcc 4.7

So, is it malformed code?

I see no reason why that should not work. Alias templates are substituted immediately. — Johannes Schaub - litb, Aug 28 '12 at 10:59
Not only it gives warning in clang 3.1 but it ignores the specialization if I try to instantiate it. However using integral_constant directly in place of alias works as usual. I guess it is clang bug. — abir, Aug 28 '12 at 11:15
@abir: That means template alias isn't implemented correctly in Clang. — Nawaz, Aug 28 '12 at 12:02

score 5 · Accepted Answer · answered Aug 28 '12 at 10:58

5

The code is perfectly fine in C++11. The Clang's warning can be ignored.

answered Aug 28 '12 at 10:58

Nawaz

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score 3 · Answer 2 · answered Aug 28 '12 at 20:29

Another guy said that this is a Clang bug. You can work it around if you change the using declaration like this

template<unsigned T, unsigned U = T>
using uint_ = integral_constant<unsigned,U>;

As an educated guess, apparently Clang does not correctly update the identity of the template parameter that appears in the type-id. So it thinks in your example that the resulting type uint_<U> refers to the first parameter of the partial specialization (because within uint_ that is the case, but not in the point of use). Alternatively you can exchange the order at the point of use

template <unsigned U,class T>
struct size<T,uint_<U>>{};

can template alias be used for partial specialization?

2 Answers2