compile-time cost of alias templates

[Jonathan Wakely]

On 18 December 2011 16:00, Jonathan Wakely wrote:
> I don't know the G++ internals well enough to find this out for
> myself, so I hope Dodji or Jason can answer the question below.
>
> Currently in <functional> we have this helper class to constrain the
> std::bind function template via SFINAE:
>
>  template<typename _Tp>
>    class __is_socketlike
>    {
>      typedef typename decay<_Tp>::type _Tp2;
>    public:
>      static const bool value =
>        is_integral<_Tp2>::value || is_enum<_Tp2>::value;
>    };
>
> I was going to replace the static member with:
>      static const bool value = __or_<is_integral<_Tp2>, is_enum<_Tp2>>::value;
>
> But now that we have alias templates I'm wondering if it would be
> better to write it as:
>
>  template<typename _Tp>
>    using __is_socketlike
>      = __or_<is_integral<typename decay<_Tp>::type>,
>              is_enum<typename decay<_Tp>::type>>;

Or slightly shorter:

  template<typename _Tp, typename _Tp2 = typename decay<_Tp>::type>
    using __is_socketlike = __or_<is_integral<_Tp2>, is_enum<_Tp2>>;

> Does the implementation in G++ mean that using an alias template
> avoids instantiating the separate __is_socketlike class template, or
> is there just as much cost in instantiating an alias template as a
> class template?

I found that when __is_socketlike is an alias instead of a class
template it avoids a call to instantiate_class_template_1 in cp/pt.c,
but I don't know if there's something else being done for the alias
which is equally (or more?) expensive.