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Re: Explicit alias template specialization

From: Arthur O'Dwyer <arthur.j.odwyer_at_[hidden]>
Date: Fri, 19 Nov 2021 10:16:04 -0500
On Fri, Nov 19, 2021 at 10:07 AM Nicolas Weidmann via Std-Proposals <
std-proposals_at_[hidden]> wrote:

> Thank you for your replies.
> But the problem you are raising with your example already exists now:
> template<typename> struct no_default_alias3;
> template<> struct no_default_alias3<int> {
> using type = std::vector<int>;
> };
> template<typename T> using alias3 = typename no_default_alias3<T>::type;
> template<typename T> void bar(alias3<T>) { }

Here the signature of `bar` (looking through the "transparent" aliases) is

    template<typename T> void bar(no_default_alias3<T>::type) { }

The template parameter `T` cannot be deduced, because it's "firewalled"
behind a member typedef of a dependent type. This is the primary use-case
for C++20 `std::identity` — as a type-deduction firewall:

    template<class T> void baz1(T a, T b);
    template<class T> void baz2(T a, typename std::identity<T>::type b);
    int main() {
        baz1(sizeof(int), 0); // deduction fails: `a` says T=size_t but
`b` says T=int
        baz2(sizeof(int), 0); // deduction succeeds: `a` says T=size_t,
and `b` does not contribute to deduction

We also define
    template<class T> identity_t = typename std::identity<T>::type;
as an alias template so that we don't have to write out the whole long
thing every time. This is "free" because the compiler will just look
straight through aliases: `identity_t<T>` is exactly equivalent to
`std::identity<T>::type` in every place it's written.


Received on 2021-11-19 09:16:18