是一个xvalue的生命延长，当它被绑定到一个const lvalue引用？ [英] Is an xvalue's lifetime extended when it is bound to a const lvalue reference?

问题描述

如果我写下面的代码：

#include <iostream>

using namespace std;

int main()
{
  cout << &(int &&)123 << endl;
  return 0;
}

然后 g ++ ：

foo.cc: In function ‘int main()’:
foo.cc:7:20: error: taking address of xvalue (rvalue reference)

好，感谢

Ok, thanks to What are rvalues, lvalues, xvalues, glvalues, and prvalues? I get that an xvalue means that it's about to "expire", which makes sense. But now if I do this:

#include <iostream>

using namespace std;

int main()
{
  const int &x = (int &&)123;
  cout << &x << endl;
  return 0;
}

这个工作很好，会打印一个地址。所以，我有几个问题：

This "works" just fine and will print an address. So, I have a few questions:

1. 如果值即将到期，为什么我可以参考呢？参考不会保留原始对象的活动（右？）。


2. 这样的引用是否导致未定义的行为？例如。因为我们引用的对象可能已被破坏？

一般来说，有一种方法可以知道右值引用的生命周期？

In general is there a way to know the lifetime of an rvalue reference?

推荐答案

第12.2节第4-5段说，生命周期示例

Clause 12.2, paragraphs 4-5, says that the lifetime is extended in the second example

有两种上下文，其中临时表在与完整表达式结束不同的点处被销毁。 ...

There are two contexts in which temporaries are destroyed at a different point than the end of the full-expression. ...

第二个上下文是当引用绑定到临时时。引用绑定到的临时或临时（引用绑定的子对象的完整对象）在引用的生命周期内仍然存在，除了：

（这里没有任何例外适用） / p>

The second context is when a reference is bound to a temporary. The temporary to which the reference is bound or the temporary that is the complete object of a subobject to which the reference is bound persists for the lifetime of the reference except:
(none of the exceptions apply here)

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