我可以通常/总是使用std :: forward而不是std :: move？ [英] Can I typically/always use std::forward instead of std::move?

问题描述

我一直在看Scott Meyers的在C ++和Beyond 2012会议上讨论通用引用，到目前为止，一切都有意义。然而，观众成员在大约50分钟时问一个问题，我也想知道。 Meyers说他不关心答案，因为它是非惯用的，会愚蠢他的心，但我仍然感兴趣。

I've been watching Scott Meyers' talk on Universal References from the C++ and Beyond 2012 conference, and everything makes sense so far. However, an audience member asks a question at around 50 minutes in that I was also wondering about. Meyers says that he does not care about the answer because it is non-idiomatic and would silly his mind, but I'm still interested.

代码如下：

// Typical function bodies with overloading:
void doWork(const Widget& param)   // copy
{
  // ops and exprs using param
}
void doWork(Widget&& param)        // move
{
  // ops and exprs using std::move(param)
}

// Typical function implementations with universal reference:
template <typename T>
void doWork(T&& param)             // forward => copy and move
{
  // ops and exprs using std::forward<T>(param)
}

注意，当我们采用右值引用时，我们知道我们有一个右值，所以我们应该 std :: move 它保留这是一个右值的事实。当我们采用通用引用（ T&&& ，其中 T 是推导类型）时， c $ c> std :: forward 以保留它可能是一个左值或右值的事实。

The point being that when we take an rvalue reference, we know we have an rvalue, so we should std::move it to preserve the fact that it's an rvalue. When we take a universal reference (T&&, where T is a deduced type), we want std::forward to preserve the fact that it may have been an lvalue or an rvalue.

所以问题是：因为 std :: forward 保存传递给函数的值是否为左值或右值， std :: move 简单地将其参数转换为右值，我们可以使用 std :: forward everywhere吗？在所有使用的情况下， std :: forward 的行为如 std :: move std :: move ，或者在Meyers的概括中漏掉了一些重要的行为差异？

So the question is: since std::forward preserves whether the value passed into the function was either an lvalue or an rvalue, and std::move simply casts its argument to an rvalue, could we just use std::forward everywhere? Would std::forward behave like std::move in all cases where we would use std::move, or are there some important differences in behaviour that are missed out by Meyers' generalisation?

我不建议任何人都应该这样做，因为正如Meyers正确地说，它完全是非惯用的，但下面也是 std :: move 的有效使用：

I'm not suggesting that anybody should do it because, as Meyers correctly says, it's completely non-idiomatic, but is the following also a valid use of std::move:

void doWork(Widget&& param)         // move
{
  // ops and exprs using std::forward<Widget>(param)
}

推荐答案

两个是非常不同的和互补工具。

The two are very different and complementary tools.

• std :: move 推断参数，并无条件创建一个右值表达式。这适用于实际的物件或变数。

• std::move deduces the argument and unconditionally creates an rvalue expression. This makes sense to apply to an actual object or variable.

std :: forward 模板参数（你必须指定这个！），并神奇地创建一个左值引用或右值表达式，取决于类型是什么（通过添加&& 规则）。这只适用于推导的模板函数参数。

std::forward takes a mandatory template argument (you must specify this!) and magically creates an lvalue reference or an rvalue expression depending on what the type was (by virtue of adding && and the collapsing rules). This only makes sense to apply to a deduced, templated function argument.

也许下面的例子说明了一点：

Maybe the following examples illustrate this a bit better:

#include <utility>
#include <memory>
#include <vector>
#include "foo.hpp"

std::vector<std::unique_ptr<Foo>> v;

template <typename T, typename ...Args>
std::unique_ptr<T> make_unique(Args &&... args)
{
    return std::unique_ptr<T>(new T(std::forward<Args>(args)...));  // #1
}

int main()
{
    {
        std::unique_ptr<Foo> p(new Foo('a', true, Bar(1,2,3)));
        v.push_back(std::move(p));                                  // #2
    }

    {
        v.push_back(make_unique<Foo>('b', false, Bar(5,6,7)));      // #3
    }

    {
        Bar b(4,5,6);
        char c = 'x';
        v.push_back(make_unique<Foo>(c, b.ready(), b));             // #4
    }
}

一个现有的，具体的对象 p ，我们想从它，无条件。只有 std :: move 才有意义。没有什么可以转发在这里。我们有一个命名的变量，我们想从它移动。

In situation #2, we have an existing, concrete object p, and we want to move from it, unconditionally. Only std::move makes sense. There's nothing to "forward" here. We have a named variable and we want to move from it.

另一方面，情况＃1接受任何类型的参数的列表，每个参数需要被转发为与原始呼叫中相同的值类别。例如，在＃3中，参数是临时表达式，因此它们将作为右值转发。但我们也可以在构造函数调用中混合命名对象，如情况＃4，然后我们需要转发为lvalues。

On the other hand, situation #1 accepts a list of any sort of arguments, and each argument needs to be forwarded as the same value category as it was in the original call. For example, in #3 the arguments are temporary expressions, and thus they will be forwarded as rvalues. But we could also have mixed in named objects in the constructor call, as in situation #4, and then we need forwarding as lvalues.

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