如何为嵌套模板类提供推导？ [英] How to provide deduction guide for nested template class?

问题描述

根据[ temp.deduct.guide/3 ]：

（...）推导指南的声明范围应与相应的
相同类模板，对于成员类模板，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. (...)

但是下面的示例似乎未在两个 [gcc] 和 [clang]

But below example doesn't seem to compile in both [gcc] and [clang].

#include <string>

template <class>
struct Foo {
    template <class T>
    struct Bar {
        Bar(T) { }
    };
    Bar(char const*) -> Bar<std::string>;
};

int main() {
    Foo<int>::Bar bar("abc");
    static_cast<void>(bar);
}

嵌套模板类的推导指南的正确语法是什么？也许这是正确的，但是编译器尚不支持？

What is the correct syntax of deduction guide for nested template class? Or maybe this one is correct but it isn't yet supported by the compilers?

---

类似的语法，但没有嵌套类，在gcc中都可以正常编译和clang：

---

Similar syntax but without nested class compiles fine both in gcc and clang:

#include <string>

template <class T>
struct Bar {
    Bar(T) { }
};
Bar(char const*) -> Bar<std::string>;

int main() {
    Bar bar("abc");
    static_cast<void>(bar);
}

推荐答案

[temp.deduct.guide] 包含以下句子：

扣除指南的声明范围应与相应的类模板相同，对于成员类模板，应使用

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.

这表明您的示例应该可以工作-成员类模板明确支持推导指南，只要它们是在相同的作用域和访问权限中声明（这将是类作用域和 public -检查并检查）。

This suggests that your example should work - deduction guides are explicitly supported for member class templates, as long as they're declared in the same scope and access (which would be the class scope and public - check and check).

这是 gcc错误79501 （由Richard Smith提出）。

This is gcc bug 79501 (filed by Richard Smith).

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