为什么字符串文字上的decltype不产生数组类型? [英] Why does decltype on a string literal not yield an array type?

问题描述

该标准在§2.13.5/8中将字符串文字的类型定义为:

The standard defines a string literal's type, in §2.13.5/8, as:

普通字符串文字和UTF-8字符串文字也称为窄字符串文字.窄字符串文字的类型为"n const char数组"，其中n是如下定义的字符串的大小，并且具有静态存储持续时间(3.7).

Ordinary string literals and UTF-8 string literals are also referred to as narrow string literals. A narrow string literal has type "array of n const char", where n is the size of the string as defined below, and has static storage duration (3.7).

因此，例如，"sss"应该具有类型char const[4](除非我读错了它).

Therefore, for example, "sss" should have a type char const[4] (unless I'm reading it incorrectly).

但是这个简单的代码段:

But this simple snippet:

std::cout << std::boolalpha << std::is_pointer<decltype("sss")>::value << '\n';
std::cout << std::boolalpha << std::is_array<decltype("sss")>::value;

给予:

false
false

我想念什么?

推荐答案

字符串文字是左值([expr.prim.general]/p1):

String literals are lvalues ([expr.prim.general]/p1):

文字是主要表达式.其类型取决于其形式(2.13).字符串文字是左值；所有其他文字都是prvalue.

A literal is a primary expression. Its type depends on its form (2.13). A string literal is an lvalue; all other literals are prvalues.

当表达式expr是左值表达式([dcl.type.simple]/p4)时，

decltype(expr)返回左值引用:

decltype(expr) returns an lvalue-reference when the expression expr is an lvalue expression ([dcl.type.simple]/p4):

对于表达式e，由decltype(e)表示的类型定义为 如下:

For an expression e, the type denoted by decltype(e) is defined as follows:

• 如果e是未括号化的id表达式或未括号化的类成员访问权限(5.2.5)，则decltype(e)是实体的类型 由e命名.如果没有这样的实体，或者e命名一组 重载的功能，程序格式不正确；
• 否则，如果e是x值，则decltype(e)是T&& ;，其中T是e;的类型.
• 否则，如果e为左值，则decltype(e)为T& ;，其中T为e的类型；
• 否则，decltype(e)是e的类型.

• if e is an unparenthesized id-expression or an unparenthesized class member access (5.2.5), decltype(e) is the type of the entity named by e. If there is no such entity, or if e names a set of overloaded func- tions, the program is ill-formed;
• otherwise, if e is an xvalue, decltype(e) is T&&, where T is the type of e;
• otherwise, if e is an lvalue, decltype(e) is T&, where T is the type of e;
• otherwise, decltype(e) is the type of e.

字符串文字是N const char的数组，但是您遇到的是decltype的作用.您真正拥有的是类型char const(&)[N]，不是 char const[N].

String literals are arrays of N const char, but what you are experiencing is the effect of decltype. What you really have is the type char const(&)[N], not char const[N].

只需删除引用，即可获得所需的行为:

Simply removing the reference should give you the behavior you desire:

std::is_array<std::remove_reference_t<decltype("sss")>>::value;

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