自动类型推导中的 Const

Question

我正在阅读 Scott Meyers 的 Effective modern C++。项目 1 包含以下示例：

template<typename T>
void f(T& param);       // param is a reference
int x = 27;             // x is an int
const int cx = x;       // cx is a const int
f(cx);                  // T is const int,
                        // param's type is const int&

项目 3 中出现以下示例：

Widget w;
const Widget& cw = w;
auto myWidget1 = cw;             // auto type deduction:
                                 // myWidget1's type is Widget

根据第 1 项，我预计 myWidget1 的类型为 const Widget。我错过了什么吗？

Answer 1

在大多数情况下auto遵循模板参数推导规则：

§ 7.1.6.4 [dcl.spec.auto]/p6:

Once the type of a declarator-id has been determined according to 8.3, the type of the declared variable using the declarator-id is determined from the type of its initializer using the rules for template argument deduction. Let T be the type that has been determined for a variable identifier d. Obtain P from T by replacing the occurrences of auto with either a new invented type template parameter U or, if the initializer is a braced-init-list (8.5.4), with std::initializer_list<U>. The type deduced for the variable d is then the deduced A determined using the rules of template argument deduction from a function call (14.8.2.1).

§ 14.8.2.1 [temp.deduct.call]/p2:

If P is not a reference type:

• [...]

• If A is a cv-qualified type, the top level cv-qualifiers of A's type are ignored for type deduction.

如果你想让myWidget1是const Widget&类型，应该声明为引用类型，例如：

auto& myWidget1 = cw;
//  ^

DEMO

Answer 2

auto myWidget1 = cw;遵循Meyers书中模板参数类型推导的第三条规则，即按值传递。 当您按值传递时，cv 限定符和引用将被忽略，因为您正在获取该对象的新副本，因此您并不真正关心您从中复制的旧对象是 const 还是引用。