r值引用返回类型的语义？ [英] r-value reference return type semantics?

问题描述

这样的返回类型在c ++ 11中表示有意义吗？

  template< typename R> 
 R&& grabStuff（）; 
 
 T实例= grabStuff< T>（）; 
  

我希望 grabStuff 应该扔一个如果 R 没有移动构造函数，则会发生编译时错误，因为这似乎不允许返回类型使用复制构造函数

  T global_thing; 
 
 T&& get（）{return std :: move（global_thing）; } 
 
结构Foo {Foo（T&）; / * ... * /}; 
 
 int main（）
 {
 global_thing.reset（）; 
 Foo a（get（））; 
 global_thing.reset（）; 
 Foo b（get（））; 
} 
  

返回右值引用的更典型示例是但是，std :: move 本身会返回对您将传递给它的东西的引用（因此，提供有效输入的内容是调用者的责任）。

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

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

解决方案

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:

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).

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