匹配别名模板作为模板参数 [英] Matching alias template as template argument

问题描述

考虑以下代码:

#include <type_traits>

template<template<class...> class T, class... U>
struct is_specialization_of : std::false_type{};

template<template<class...> class T, class... U>
struct is_specialization_of<T, T<U...>> : std::true_type{};

template<class T, class U = int>
struct test{};

// (1) ok
static_assert(is_specialization_of<test, test<int>>::value, "1");

template<class T>
using alias = test<T>;

// (2) fails
static_assert(is_specialization_of<alias, alias<int>>::value, "2");

int main()
{
}

为什么(2)(即使用别名模板的static_assert)失败?

Why does (2), i.e. static_assert that uses alias template, fail?

(2)中的模板参数推导过程与(1)中的模板推论过程有何不同?

How does the template argument deduction process in (2) differ from the one in (1)?

推荐答案

这是

This is CWG issue 1286. The question is: are alias and test equivalent? There used to be an example in [temp.type] which suggested that y and z have the same type here:

template<template<class> class TT> struct X { };
template<class> struct Y { };
template<class T> using Z = Y<T>;
X<Y> y;
X<Z> z;

该示例已作为 CWG缺陷1244 的一部分得到纠正. a>-正确表示 [temp.alias] 中没有措辞实际上指定别名模板与其别名模板等效.那里唯一的措辞是指别名模板专业化的等效性:

The example was corrected as part of CWG defect 1244 - which indicated correctly that there is no wording in [temp.alias] that actually specifies that alias templates are equivalent to the templates they alias. The only wording there refers to equivalence of alias template specializations:

当 template-id 引用别名模板的专业化时，它等同于通过替换其 template-arguments 表示别名模板的 type-id 中的 template-parameters .

When a template-id refers to the specialization of an alias template, it is equivalent to the associated type obtained by substitution of its template-arguments for the template-parameters in the type-id of the alias template.

在此示例中，意图显然是y和z do 具有相同的类型，这意味着Z和Y实际上是等效的.但是除非并且直到决议中的措词获得通过，否则它们不会被采纳.今天，alias和test与不等价，但alias<int>和test<int> 是.这意味着is_specialization_of<alias, alias<int>>是is_specialization_of<alias, test<int>>，其中alias在test中是唯一的，与您的部分专业知识不匹配，因此是false_type.

The intent is apparently that y and z do have the same type in this example, meaning that Z and Y are actually equivalent. But unless and until the wording in the resolution is adopted, they are not. Today, alias and test are not equivalent but alias<int> and test<int> are. This means that is_specialization_of<alias, alias<int>> is is_specialization_of<alias, test<int>>, where alias is unique from test, which would not match your partial specialization and thus be false_type.

此外，即使采用#1286中的措词，test和alias仍然不等价 ，原因很明显，原因是test采用两个模板参数，而别名采用一个模板参数.解决方案措辞中的示例模仿了您的示例，并在此处阐明了意图:

Moreover, even with the adoption of the wording in #1286, test and alias are still not equivalent for the obvious reason that test takes two template parameters and alias takes one template parameter. The example in the resolution wording mimics your example and clarifies the intent here:

template<typename T, U = T> struct A;

// ...

template<typename V>   
  using D = A<V>;      // not equivalent to A:
                       // different number of parameters

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