r值引用返回类型语义?

Does a return type like this represent something meaningful in c++11?

这样的返回类型是否表示c++11中的一些有意义的东西?

template <typename R>
R&& grabStuff();

T instance = grabStuff<T>();

I would hope that grabStuff should throw a compile-time error if R does not have a move constructor, since this would seem to disallow the return type to use a copy constructor

如果R没有move构造函数，我希望grabStuff应该抛出编译时错误，因为这似乎不允许返回类型使用复制构造函数

3 个解决方案

#1

As always, when returning references you must return a reference to something that's still alive after the function returns. How you do that is up to you. Example:

与往常一样，当返回引用时，您必须返回一个引用，该引用在函数返回后仍然存在。如何做到这一点取决于你自己。例子:

T global_thing;

T && get() { return std::move(global_thing); }

struct Foo { Foo(T &&); /* ... */ };

int main()
{
    global_thing.reset();
    Foo a(get());
    global_thing.reset();
    Foo b(get());
}

The more typical example for returning an rvalue reference is std::move itself, though, which returns a reference to the thing you pass into it (and thus it's the caller's responsibility to provide a valid input).

返回一个rvalue引用的更典型的例子是std::move本身，它返回一个对您传入它的东西的引用(因此调用者有责任提供一个有效的输入)。

#2

It may be meaningful depending on what you want to do with it, and how you've implemented the function.

它可能有意义，这取决于您想要对它做什么，以及您如何实现这个函数。

In fact, std::move return type is T&& (rvalue reference) which is meaningful, as it defines the very purpose of the existence of std::move in the C++11 library:

实际上，std::move return类型是T&& (rvalue reference)，这是有意义的，因为它定义了std:::在c++ 11库中移动的目的:

std::move
std::移动

#3

If the return type of a function is an rvalue reference, then the result of the function call is an xvalue; if the return type is non-reference, then the result of the function call is a prvalue.

如果函数的返回类型是一个rvalue引用，那么函数调用的结果就是一个xvalue;如果返回类型是非引用的，那么函数调用的结果就是一个prvalue。

Both xvalue and prvalue are rvalues, there are some minor differences between them, more like differences between reference and non-reference. For example, an xvalue may have an incomplete type, while a prvalue shall usually have a complete type or the void type. When typeid is applied to an xvalue whose type is a polymorphic class type, the result refers to the dynamic type; and for prvalue, the result refers to the static type.

xvalue和prvalue都是rvalues，它们之间有一些细微的差异，更像是reference和non-reference之间的差异。例如，xvalue可能有一个不完整的类型，而prvalue通常有一个完整的类型或void类型。当将typeid应用于类型为多态类类型的xvalue时，结果引用动态类型;对于prvalue，结果引用静态类型。

For your declaration statement T instance = grabStuff<T>();, if T is a class type, I think there is no difference between xvalue and prvalue in this context.

对于您的声明语句T instance = grabStuff ();如果T是类类型，我认为在这个上下文中xvalue和prvalue之间没有区别。

The initializer is an rvalue, so the compiler prefers a move constructor. But if no move constructor is declared, and a copy constructor with const reference parameter is declared, then this copy constructor will be chosen, and there is no error. I don't know why do you want this to be an error. If this is an error, any old code will be incorrect when copy-initialize some object from an rvalue.

初始化器是一个rvalue，所以编译器更喜欢一个move构造函数。但是如果没有声明move构造函数，并且声明了带有const引用参数的复制构造函数，那么将选择这个复制构造函数，并且没有错误。我不知道你为什么要把它当成一个错误。如果这是一个错误，那么当从rvalue复制初始化某个对象时，任何旧代码都将是不正确的。

#1