"typename"何时可以与明确引用类型的标识符一起使用? [英] When can `typename` be used with identifiers that unambiguously refer to a type?

问题描述

通常，typename用于在标识符可能指向类型或其他类型的情况之间进行歧义消除:

Normally, typename is used to disambiguate between cases where an identifier may refer to a type, or may refer to something else:

template<class T>
void foo(typename T::type value) {
   // ...
}

当标识符已经是类型时，什么时候可以使用typename?

1..如果已经有一个使用此名称的课程，可以使用它吗?

When can typename be used when an identifier is already a type?

1. Can it be used if there's already a class with this name?

class MyClass{};

void foo(typename MyClass value) {}

2..它可以与声明为类型的模板参数一起使用吗?

2. Can it be used with a template parameter that's declared as a type?

template<class T>
void foo(typename T value) {}

3.可以与明确类型的内部类一起使用吗?

3. Can it be used with inner classes that are unambiguously types?

class A {
   public:
    class B {};
};

// Compiles; no typename necessary
void foo(A::B value) {} 
// This compiles too on gcc and msvc
void bar(typename A::B value) {}

编译器解释

情况1: MSVC认为这可以； gcc和clang抛出错误

Compiler interpretation

Case 1: MSVC considers this OK; gcc and clang throw an error

情况2: MSVC认为这样做可以； gcc和clang抛出错误

Case 2: MSVC considers this OK; gcc and clang throw an error

情况3: A::B无疑是一种类型，但是gcc和clang现在允许使用typename.

Case 3: A::B is unambiguously a type, but gcc and clang now permit the use of typename.

推荐答案

关键字typename仅由C ++允许语法以引入模板类型参数或在限定名称前，即包含::令牌的内容.

The keyword typename is only permitted by the C++ syntax to introduce a template type parameter, or before a qualified name, i.e. something containing the :: token.

因此，您的#1和#2格式错误，因为MyClass和T是不合格的名称，不包含任何::.

So your #1 and #2 are ill-formed because MyClass and T are unqualified names, not containing any ::.

在限定名称之前，typename令牌为:

Before a qualified name, the typename token is:

• 基类之前的语法不允许类定义开头的名称，或从属并且是未知专业化的成员
• 否则是可选的，无论是否在模板声明中

• not allowed by the grammar before a base class name in the head of a class definition, or in combination with the class, struct, or union keywords; a qualified name is always treated as a type in these contexts
• otherwise required, if the qualified name is dependent and a member of an unknown specialization
• otherwise optional, whether within a template declaration or not

C ++ 17 [temp.res]/3,5,6:

C++17 [temp.res]/3,5,6:

当 qualified-id 旨在引用不是当前实例成员的类型，而其 nested-name-specifier 引用从属类型时，它必须以关键字typename为前缀，形成一个 typename-specifier . ...

When a qualified-id is intended to refer to a type that is not a member of the current instantiation and its nested-name-specifier refers to a dependent type, it shall be prefixed by the keyword typename, forming a typename-specifier. ...

用作 class-or-decltype (条款[class.derived])或 elaborated-type-specifier 的名称的合格名称被隐式假定为为类型命名，而不使用typename关键字.在嵌套名称说明符中，它立即包含依赖于模板参数的嵌套名称说明符，其中 identifier 或假定使用simple-template-id 命名类型，而不使用typename关键字. [注意:这些构造的语法不允许使用typename关键字. -尾注]

A qualified name used as the name in a class-or-decltype (Clause [class.derived]) or an elaborated-type-specifier is implicitly assumed to name a type, without the use of the typename keyword. In a nested-name-specifier that immediately contains a nested-name-specifier that depends on a template parameter, the identifier or simple-template-id is implicitly assumed to name a type, without the use of the typename keyword. [Note: The typename keyword is not permitted by the syntax of these constructs. - end note]

如果对于给定的一组模板参数，实例化的模板专门化是指表示类型或类模板的 qualified-id ，而 qualified- id 指的是未知专业的成员， qualified-id 应该以typename为前缀，或在其隐式命名上述类型的上下文中使用

If, for a given set of template arguments, a specialization of a template is instantiated that refers to a qualified-id that denotes a type or a class template, and the qualified-id refers to a member of an unknown specialization, the qualified-id shall either be prefixed by typename or shall be used in a context in which it implicitly names a type as described above.

#3格式正确，即使名称不依赖模板参数，也不在模板声明中.

So your #3 is well-formed, even though the name does not depend on a template parameter and isn't even in a template declaration.

请注意，C ++ 20将添加更多上下文，其中typename即使是从属名称，也是可选的，因为可以从上下文中明确确定名称只能表示类型.

Note C++20 will add many more contexts where typename is optional even with a dependent name, because it's possible to unambiguously determine from the contexts that the name can only mean a type.

这篇关于"typename"何时可以与明确引用类型的标识符一起使用?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！