用于定义完整功能模板的语法如何专门化 class 的朋友？

Question

我在 cppreference（“模板友元运算符”部分）看到了以下示例，已简化

template<typename> struct Foo;
template<typename T> void bar(Foo<T>) {}

template<typename T>
struct Foo {
    // friend void bar<T>(Foo f);  // (1)
    friend void bar<>(Foo f);  // (2)
};

int main() {
    bar(Foo<int>{});
}

此处 (1) 或 (2) 都同样有效。我知道在 (1) 中，编译器说 bar 的一个特定实例（即带有某某参数的 bar<T>）是 Foo<T> 的朋友，但我知道不明白 (2).

中发生了什么

(2) 中的 bar 编译器使用什么类型的参数？这通常是如何工作的？该语法是如何调用的，它是否在标准中的某处定义？

Answer 1

参见 C++17 [temp.friend]/1:

A friend of a class or class template can be a function template or class template, a specialization of a function template or class template, or a non-template function or class. For a friend function declaration that is not a template declaration:

if the name of the friend is a qualified or unqualified template-id, the friend declaration refers to a specialization of a function template, otherwise,

if the name of the friend is a qualified-id and a matching non-template function is found in the specified class or namespace, the friend declaration refers to that function, otherwise,

if the name of the friend is a qualified-id and a matching function template is found in the specified class or namespace, the friend declaration refers to the deduced specialization of that function template (17.8.2.6), otherwise,

the name shall be an unqualified-id that declares (or redeclares) a non-template function.

名字bar<T>和bar<>都是template-ids，所以如果你使用任何一个，那么它指的是函数模板 ::bar。如 [temp.arg.explicit] 中指定：

Template arguments can be specified when referring to a function template specialization by qualifying the function template name with the list of template-arguments ... A template argument list may be specified when referring to a specialization of a function template ... in a friend declaration. Trailing template arguments that can be deduced (17.8.2) or obtained from default template-arguments may be omitted from the list of explicit template-arguments. ... If all of the template arguments can be deduced, they may all be omitted; ...

因此，friend 声明中bar<> 的含义是，应进行模板参数推导以确定引用bar 的哪个特化。当然，推导的特化是采用 Foo<T> 类型参数的特化。所以带有 bar<> 和 bar<T> 的两个声明具有相同的含义。

用于定义完整功能模板的语法如何专门化 class 的朋友？

How does that syntax for defining a full function template specialization a friend of a class?

c++

templates

friend-function

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