什么是 T&&(双和号)在 C++11 中是什么意思? [英] What does T&& (double ampersand) mean in C++11?

问题描述

我一直在研究 C++11 的一些新特性，我注意到一个是声明变量时的双 & 号，比如 T&&var.

I've been looking into some of the new features of C++11 and one I've noticed is the double ampersand in declaring variables, like T&& var.

首先，这只野兽叫什么?我希望谷歌允许我们搜索这样的标点符号.

For a start, what is this beast called? I wish Google would allow us to search for punctuation like this.

它究竟是什么意思?

乍一看，它似乎是一个双引用(就像 C 风格的双指针 T** var)，但我很难想到它的用例.

At first glance, it appears to be a double reference (like the C-style double pointers T** var), but I'm having a hard time thinking of a use case for that.

推荐答案

它声明了一个 右值参考(标准提案文档).

It declares an rvalue reference (standards proposal doc).

这是对右值的介绍参考.

这是微软标准库之一对右值引用的精彩深入研究 开发人员.

Here's a fantastic in-depth look at rvalue references by one of Microsoft's standard library developers.

注意: MSDN 上的链接文章(Rvalue 参考:VC10 中的 C++0x 功能，第 2 部分")是对 Rvalue 参考的非常清晰的介绍，但做出了陈述关于在 C++11 标准草案中曾经是正确的 Rvalue 引用，但在最后一个标准中却不是这样的！具体来说，它在不同的地方说右值引用可以绑定到左值，这曾经是真的，但后来被改变了.(例如 int x; int &&rrx = x; 不再在 GCC 中编译)——drawbarbs 2014 年 7 月 13 日16:12

CAUTION: the linked article on MSDN ("Rvalue References: C++0x Features in VC10, Part 2") is a very clear introduction to Rvalue references, but makes statements about Rvalue references that were once true in the draft C++11 standard, but are not true for the final one! Specifically, it says at various points that rvalue references can bind to lvalues, which was once true, but was changed.(e.g. int x; int &&rrx = x; no longer compiles in GCC) – drewbarbs Jul 13 '14 at 16:12

C++03 引用(现在在 C++11 中称为左值引用)之间的最大区别在于，它可以像临时对象一样绑定到右值，而不必是 const.因此，这种语法现在是合法的:

The biggest difference between a C++03 reference (now called an lvalue reference in C++11) is that it can bind to an rvalue like a temporary without having to be const. Thus, this syntax is now legal:

T&& r = T();

右值引用主要提供以下内容:

rvalue references primarily provide for the following:

移动语义.现在可以定义移动构造函数和移动赋值运算符，它们采用右值引用而不是通常的 const-lvalue 引用.移动的功能类似于副本，只是它不必保持源不变；事实上，它通常会修改源，使其不再拥有移动的资源.这对于消除无关副本非常有用，尤其是在标准库实现中.

Move semantics. A move constructor and move assignment operator can now be defined that takes an rvalue reference instead of the usual const-lvalue reference. A move functions like a copy, except it is not obliged to keep the source unchanged; in fact, it usually modifies the source such that it no longer owns the moved resources. This is great for eliminating extraneous copies, especially in standard library implementations.

例如，复制构造函数可能如下所示:

For example, a copy constructor might look like this:

foo(foo const& other)
{
    this->length = other.length;
    this->ptr = new int[other.length];
    copy(other.ptr, other.ptr + other.length, this->ptr);
}

如果这个构造函数传递了一个临时对象，那么副本就没有必要了，因为我们知道临时对象会被销毁；为什么不利用已经分配的临时资源呢?在 C++03 中，没有办法阻止复制，因为我们无法确定我们是否被传递了一个临时的.在 C++11 中，我们可以重载移动构造函数:

If this constructor were passed a temporary, the copy would be unnecessary because we know the temporary will just be destroyed; why not make use of the resources the temporary already allocated? In C++03, there's no way to prevent the copy as we cannot determine whether we were passed a temporary. In C++11, we can overload a move constructor:

foo(foo&& other)
{
   this->length = other.length;
   this->ptr = other.ptr;
   other.length = 0;
   other.ptr = nullptr;
}

注意这里的巨大差异:移动构造函数实际上修改了它的参数.这将有效地移动"临时放入正在构造的对象中，从而消除不必要的复制.

Notice the big difference here: the move constructor actually modifies its argument. This would effectively "move" the temporary into the object being constructed, thereby eliminating the unnecessary copy.

移动构造函数将用于临时和非常量左值引用，这些引用使用 std::move 函数显式转换为右值引用(它只是执行转换).以下代码都为 f1 和 f2 调用了移动构造函数:

The move constructor would be used for temporaries and for non-const lvalue references that are explicitly converted to rvalue references using the std::move function (it just performs the conversion). The following code both invoke the move constructor for f1 and f2:

foo f1((foo())); // Move a temporary into f1; temporary becomes "empty"
foo f2 = std::move(f1); // Move f1 into f2; f1 is now "empty"

完美转发.右值引用允许我们正确地转发模板函数的参数.以这个工厂函数为例:

Perfect forwarding. rvalue references allow us to properly forward arguments for templated functions. Take for example this factory function:

template <typename T, typename A1>
std::unique_ptr<T> factory(A1& a1)
{
    return std::unique_ptr<T>(new T(a1));
}

如果我们调用factory(5)，参数将被推导出为int&，它不会绑定到文字5，即使foo 的构造函数接受一个 int.好吧，我们可以改用 A1 const&，但是如果 foo 通过非常量引用来获取构造函数参数怎么办?为了创建一个真正通用的工厂函数，我们必须在 A1& 和 A1 const& 上重载工厂.如果工厂采用 1 个参数类型，那可能没问题，但每个额外的参数类型都会将必要的重载集乘以 2.这很快就无法维护.

If we called factory<foo>(5), the argument will be deduced to be int&, which will not bind to a literal 5, even if foo's constructor takes an int. Well, we could instead use A1 const&, but what if foo takes the constructor argument by non-const reference? To make a truly generic factory function, we would have to overload factory on A1& and on A1 const&. That might be fine if factory takes 1 parameter type, but each additional parameter type would multiply the necessary overload set by 2. That's very quickly unmaintainable.

rvalue 引用通过允许标准库定义一个 std::forward 函数来解决这个问题，该函数可以正确地转发左值/右值引用.有关 std::forward 工作原理的更多信息，请参阅这个极好的答案.

rvalue references fix this problem by allowing the standard library to define a std::forward function that can properly forward lvalue/rvalue references. For more information about how std::forward works, see this excellent answer.

这使我们能够像这样定义工厂函数:

This enables us to define the factory function like this:

template <typename T, typename A1>
std::unique_ptr<T> factory(A1&& a1)
{
    return std::unique_ptr<T>(new T(std::forward<A1>(a1)));
}

现在参数的右值/左值在传递给 T 的构造函数时被保留.这意味着如果用右值调用工厂，则用右值调用 T 的构造函数.如果用左值调用 factory，则用左值调用 T 的构造函数.改进后的工厂函数由于一条特殊规则而起作用:

Now the argument's rvalue/lvalue-ness is preserved when passed to T's constructor. That means that if factory is called with an rvalue, T's constructor is called with an rvalue. If factory is called with an lvalue, T's constructor is called with an lvalue. The improved factory function works because of one special rule:

当函数参数类型为表单 T&& 其中 T 是模板参数和函数参数是 A 类型的左值，A& 类型是用于模板参数推导.

When the function parameter type is of the form T&& where T is a template parameter, and the function argument is an lvalue of type A, the type A& is used for template argument deduction.

因此，我们可以像这样使用工厂:

Thus, we can use factory like so:

auto p1 = factory<foo>(foo()); // calls foo(foo&&)
auto p2 = factory<foo>(*p1);   // calls foo(foo const&)

重要的右值引用属性:

• 对于重载解析，左值更喜欢绑定到左值引用，而右值更喜欢绑定到右值引用.因此，为什么临时对象更喜欢调用移动构造函数/移动赋值运算符而不是复制构造函数/赋值运算符.
• 右值引用将隐式绑定到右值和作为隐式转换结果的临时变量.即 float f = 0f;国际&&i = f; 格式良好，因为 float 可以隐式转换为 int;引用将指向作为转换结果的临时对象.
• 命名的右值引用是左值.未命名的右值引用是右值. 这对于理解为什么 std::move 调用是必要的很重要:foo&&r = foo();foo f = std::move(r);

• For overload resolution, lvalues prefer binding to lvalue references and rvalues prefer binding to rvalue references. Hence why temporaries prefer invoking a move constructor / move assignment operator over a copy constructor / assignment operator.
• rvalue references will implicitly bind to rvalues and to temporaries that are the result of an implicit conversion. i.e. float f = 0f; int&& i = f; is well formed because float is implicitly convertible to int; the reference would be to a temporary that is the result of the conversion.
• Named rvalue references are lvalues. Unnamed rvalue references are rvalues. This is important to understand why the std::move call is necessary in: foo&& r = foo(); foo f = std::move(r);

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