C++11 的 is_array 函数无法判断字符串字面量

Question

int main()
{
    std::cout << typeid("asd").name() << '\n';
    std::cout << std::is_array_v<decltype("asd")> << '\n';
    std::cout << std::is_pointer_v<decltype("asd")> << '\n';
}

在 Visual Studio 2019 输出：

char const [4]
0
0

GCC 也是一样。字符串字面量难道既不是数组也不是指针？

Answer 1

decl.type.simple/4:

For an expressione, the type denoted bydecltype(e)is defined as follows:

• if e is an unparenthesized id-expression naming a structured binding ([dcl.struct.bind]), decltype(e) is the referenced type as given in the specification of the structured binding declaration;
• otherwise, if e is an unparenthesized id-expression or an unparenthesized class member access, decltype(e) is the type of the entity named bye. If there is no such entity, or ifenames a set of overloaded functions, 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.

expr.prim.literal/1

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

lex.string/8

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.

只有对 id-expression 以及 prvalue ，decltype(e) 才会直接返回 e 的类型。

否则，对 xvalue 返回 T&& ，对 lvalue 返回 T&。

"abs" 的类型是 const char[4]，lvalue
它既不是 id-expression 也不是 xvalue，所以 decltype("abc") 会得到 const char(&)[4]

Answer 2

更新：

decltype("asd") 得到的类型其实是 const char (&)[4]，故

int main()
{
    std::cout << typeid("asd").name() << '\n';
    std::cout << std::is_array_v<std::remove_reference_t<decltype("asd")>> << '\n';
    std::cout << std::is_pointer_v<std::remove_reference_t<decltype("asd")>> << '\n';
}

输出

char const [4]
1
0