自动&&变量不是右值引用 [英] auto&& variable's are not rvalue reference

问题描述

为什么要自动&&不是右值引用?

Why auto&& is not rvalue reference?

Widget&& var1 = Widget(); // rvalue reference
auto&& var2 = var1; //var2 not rvalue reference

下面是右值参考示例

void f(Widget&& param); // rvalue reference
Widget&& var1 = Widget(); // rvalue reference

为什么var2不是右值引用，而f和var2是右值引用?

Why var2 is not rvalue reference but f and var2 are rvalue references?

推荐答案

确定了初始化程序的类型后，编译器将使用函数中模板参数推导的规则来确定将替换关键字auto的类型.调用(有关详细信息，请参见模板参数deduction#Other上下文).关键字auto可能附带修饰符，例如const或&，它们将参与类型推导.

Once the type of the initializer has been determined, the compiler determines the type that will replace the keyword auto using the rules for template argument deduction from a function call (see template argument deduction#Other contexts for details). The keyword auto may be accompanied by modifiers, such as const or &, which will participate in the type deduction.

例如，给定

const auto& i = expr;

i的类型恰好是虚构中的参数u的类型

The type of i is exactly the type of the argument u in an imaginary

template template<class U> 
void f(const U& u)

如果函数调用f(expr)已编译.

If the function call f(expr) was compiled.

通常，可以这样认为.

 template template<class U> 
    void f(paramtype u)

因此，auto&&可以根据初始化程序推导为左值引用或右值引用.

Therefore, auto&& may be deduced either as an lvalue reference or rvalue reference according to the initializer.

在您的情况下，假想模板看起来像

In your case , imaginary template would look like

 template template<class U> 
        void f(U&& var2){}
f(var1) 

在这里，var1被命名为rvalue，它被视为lvalue，因此var2将被推导为lvalue.

Here ,var1 is named rvalue which is being treated as lvalue, so var2 will be deduced as lvalue .

请考虑以下示例:

auto&& var2 = widget() ; //var2 is rvalue reference here .
int x=10;
const int cx=10;
auto&& uref1 = x; // x is int and lvalue, so uref1's type is int&
auto&& uref2 = cx; // cx is const int and lvalue,  so uref2's type is const int&
auto&& uref3 = 27; // 27 is int and rvalue,  so uref3's type is int&&

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