将文字常量分配给右值引用时会发生什么? [英] What happens when you assign a literal constant to an rvalue reference?

问题描述

这当然是一个挑剔的问题，主要是出于好奇心.假设我们有以下内容:

This is admittedly a nit-picky question that is mainly driven by curiosity. Suppose we have the following:

int x = 5;
int&& xref = std::move(x);
std::cout << "Before assignment x: " << x << std::endl;
std::cout << "Before assignment xref: " << xref << std::endl;
xref = 10;
std::cout << "After assignment x: " << x << std::endl;
std::cout << "After assignment xref: " << xref << std::endl;

预期的输出是:

// Before assignment x: 5
// Before assignment xref: 5
// After assignment x: 10
// After assignment xref: 10

这很有道理. std::move将x转换为xvalue，并允许我们将其内存位置绑定到xref并相应地修改其内容.现在说我们有以下内容:

This makes sense. std::move converts x to an xvalue and allows us to bind its memory location to xref and modify its contents accordingly. Now lets say we have the following:

int&& xref = 5;
std::cout << "Before assignment xref: " << xref << std::endl;
xref = 10;
std::cout << "After assignment xref: " << xref << std::endl;

int x = 5;
std::cout << "After assignment x: " << x << std::endl;

输出直观:

// Before assignment xref: 5
// After assignment xref: 10
// After assignment x: 5

这具有总体意义.我们希望能够将常量文字5绑定到xref，因为5是prvalue.我们还希望xref是可变的.我们进一步期望常量文字5的值是不可修改的(如上述代码片段的最后两行中有些书呆子所示).

This make overall sense. We expect to be able to bind the constant literal 5 to xref because 5 is a prvalue. We also expect that xref be mutable. We further expect that the value of constant literal 5 isn't modifiable (as shown somewhat pedantically in the last two lines of the above snippet).

所以我的问题是，这到底是怎么回事? C ++如何知道不修改常量文字5的值，又如何为xref保持足够的身份才能知道赋值已将其更改为10.当绑定到常量文字时，是否在分配给xref后创建新变量?在C ++ 03中，这个问题从未出现过，因为只有const引用可以绑定到右值.

So my question is, what exactly is going on here? How does C++ know not to modify the value of the constant literal 5 yet maintain sufficient identity for xref to know that it's been changed to 10 by the assignment. Is a new variable being created upon assignment to xref when its bound to a constant literal? This question never came up in C++03 since only const references could be bound to rvalues.

推荐答案

构造一个临时变量，该临时变量根据文字的值进行初始化，并且其持续时间与引用的时间一样长.您可以使用此对象执行自己喜欢的操作.

A temporary is constructed, initialised from the value of the literal, and it lasts as long as the reference. You can do what you like with this object.

就生命周期而言，这与您编写const int& x = 5一样；只是，在那里，您正在使用自动创建的临时对象这一事实被掩盖了，因为const阻止您通过突变对其进行证明.

In terms of lifetime, this is just the same as if you'd written const int& x = 5; only, there, the fact that you're working with an automatically-created temporary object is masked because the const prevents you from proving it with a mutation.

[C++14: 8.5.3/5]: [..] 如果T1是非类类型，则会创建类型为" cv1 T1"的临时文件，并复制-从初始值设定项表达式初始化(​​8.5).然后将引用绑定到临时目录. [..]

[C++14: 8.5.3/5]: [..] If T1 is a non-class type, a temporary of type "cv1 T1" is created and copy-initialized (8.5) from the initializer expression. The reference is then bound to the temporary. [..]

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