嵌套模板类型的用户定义推论指南 [英] User defined-deduction guide for nested template-types

问题描述

在这里的示例中，GCC和Clang不会产生相同的行为:

In the example here, GCC and Clang does not produce the same behavior :

• https://godbolt.org/z/8W6hznEjo

#include <tuple>

template <typename T>
struct some_type;
template <template <typename...> typename T, typename ... Ts>
struct some_type<T<Ts...>>
{
    template <typename U>
    class nested
    {
        U member;
    public:
        nested(U &&){}
    };

    // non-namespace scope user-deduction-guide : OK with Clang, fix the deduction issue
    template <typename U>
    nested(U&&) -> nested<U>;
};

void func()
{
    using pack_type = std::tuple<int, char>;
    some_type<pack_type>::nested{
        [](auto &&){}
    };
}

简而言之，我们有一个模板参数类型，一个嵌套类型本身就是模板参数，并且模板参数之间没有任何关系.

In short, we have a template-parametered type, with a nested type which is itself template-parametered, and template parameters have no relationship between each others.

template <typename T>
struct some_type;
template <template <typename...> typename T, typename ... Ts>
struct some_type<T<Ts...>>
{
    template <typename U>
    class nested // <- nested type, where `U` as no relationship with `T<Ts...>`
    {
        U member;
    public:
        nested(U &&);
    };
};

该标准指定: http://eel.is/c++ draft/temp.deduct.guide#3

[...]推导指南应在与相应类模板相同的范围内声明，并且对于成员类模板，应具有相同的访问权限.[...]

[...] A deduction-guide shall be declared in the same scope as the corresponding class template and, for a member class template, with the same access. [...]

Q2 :如果对Q1的回答否，当命名空间类型和命名空间类型之间没有关系时，为嵌套类型创建用户定义的推导指南的语法是什么?嵌套类型的模板参数?

我希望语法接近:

Q2 : If no to Q1, what is the syntax to create a user-defined deduction guide for nested type, when there is no relationship between namespace-type and nested-type template-parameters ?

I'd expect a syntax close to :

template <template <typename...> typename T, typename ... Ts>
template <typename U>
some_type<T<Ts...>>::nested<U>::nested(U&&) -> nested<U>;

但是，嵌套< U> 是错误，因为它需要推导的类型...来推导它.

However, nested<U> is wrong, as it required a deduced type ... to deduce it.

此外，这被解释为尾随返回类型为 void 的函数.

Also, this is interpreted as a function with trailing return type void.

template <template <typename...> typename T, typename ... Ts>
template <typename U>
typename some_type<T<Ts...>>::template nested<U>::nested(U&&) -> nested<U>;

谢谢您的时间.

推荐答案

快速修复

我发现的唯一解决方法是仅为Clang添加用户定义的推导指南，
就可维护性而言，这是次优的方式.

godbolt上的实时示例，
或查看下面的资源.

• GCC 不允许在非命名空间上下文中使用推导， Clang 允许.
在这种情况下，
• requires语需要一个演绎指南， GCC 不需要

• GCC does not allow deduction guide in non-namespace context, Clang does.
• Clang requires a deduction guide in this context, GCC does not

NB:张贴此消息时，clang干线是11.0.1，gcc干线是10.2

#include <tuple>

template <typename T>
struct type
{
    template <typename U>
    struct nested
    {
        template <typename ... nested_Ts>
        nested(U &&, std::tuple<nested_Ts...> &&)
        {}

    };
    #if __clang__
    // here, user-defined deduction guide only for Clang
    template <typename U, typename ... Ts>
    nested(U&&, std::tuple<Ts...>&&) -> nested<U>;
    #endif
};

void instanciate_symbols()
{
    using type = type<int>;
    [[maybe_unused]] auto value = type::nested{'a', std::tuple{42, .42f}};
}

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