c++ - C++ 中 (x) 的类型

Question

给定:

decltype(auto) f1()
{
    int x = 0;
    return x;  // decltype(x) is int, so f1 returns int
}
decltype(auto) f2()
{
    int x = 0;
    return (x);  // decltype((x)) is int&, so f2 returns int&
}

(取自 Scott Meyer 的 Effective Modern C++)。

现在，如果我找到了正确的段落，C++ 标准的第 7.1.5.2 节简单类型说明符 [dcl.type.simple] 说:

If e is an id-expression or a class member access (5.2.5 [expr.ref]), decltype(e) is defined as the type of the entity named by e

该部分的示例是:

struct A { double x; }

const A* a = new A();

decltype((a->x)); // type is const double&

现在，我想知道为什么decltype((x))在书中被推导出为int&。

Answer 1

相关标准引用为:

N4140 [dcl.type.simple]/4: For an expression e, the type denoted by decltype(e) is defined as follows:

• 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 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.

由于x是左值，并且表达式是带括号的，所以使用第三条规则，所以decltype((x))是int&.