我保证指针std :: vector元素在向量移动后是有效的吗？ [英] Am I guaranteed that pointers to std::vector elements are valid after the vector is moved?

问题描述

考虑这个例子：

  std :: vector< int> v1 = {1，2，3}; 
 const int * i =& v1 [1]; 
 std :: vector< int> v2（std :: move（v1））; 
 std :: cout<< * i < std :: endl; 
 

即使在许多STL实现中，这可能会工作，我保证标准没有重新分配在移动 std :: vector 时执行，并且内部缓冲区支持 v2 与以前一样 v1 之一？我无法在互联网上或标准本身上找不到此信息。

解决方案

这是 LWG开放问题2321 [着重于我的]

移动容器应（通常）需要保留迭代器

[...]

[by Stephan T. Lavavej]

23.2.1 [container.requirements.general] / 10说除非另有说明，否则没有swap（）函数无效任何引用，指针，
或引用容器的元素的迭代器
交换[注意：end（）迭代器不引用任何元素，因此
可能无效。 - end note]。 但是，移动构造函数和
移动赋值运算符不会给出类似的无效
保证。保证需要几个异常，所以我不
相信毯子语言/ 11除非另有规定
（显式地或通过其他
函数定义函数），调用容器成员函数或将
容器作为参数传递给库函数，不会使
迭代器无效，或更改该容器中的对象的值。
适用。

[2014-02-13 Issaquah]

STL以提供更新的文字。 移至开启。

$ b

> [...]

没有一个容器的移动构造函数[...]（ array 除外）使引用源容器的元素的任何引用，指针或迭代器无效。 [注意： end（）迭代器不会引用任何元素，因此可能会失效。 ]

所以，这是一个开放的问题，指针不能通过移动而失效）。但是，它不是正式接受（但？）为缺陷。据我所知，所有主要的实现并不使指针在移动构造时无效，并且它似乎是一个通常（隐式）提供的保证。

Considering this example:

std::vector<int> v1 = { 1, 2, 3 };
const int* i = &v1[1];
std::vector<int> v2(std::move(v1));
std::cout << *i << std::endl;

Even though in many STL implementations this will probably work, am I guaranteed by the standard that no reallocations are performed when a std::vector is moved, and the internal buffer backing v2 is the same that used to be the one of v1? I wasn't able to find this information neither on the Internet, nor on the standard itself.

解决方案

This is LWG open issue 2321 [emphasis mine]

Moving containers should (usually) be required to preserve iterators

[...]

[by Stephan T. Lavavej]
23.2.1 [container.requirements.general]/10 says that unless otherwise specified, "no swap() function invalidates any references, pointers, or iterators referring to the elements of the containers being swapped. [Note: The end() iterator does not refer to any element, so it may be invalidated. — end note]". However, move constructors and move assignment operators aren't given similar invalidation guarantees. The guarantees need several exceptions, so I do not believe that blanket language like /11 "Unless otherwise specified (either explicitly or by defining a function in terms of other functions), invoking a container member function or passing a container as an argument to a library function shall not invalidate iterators to, or change the values of, objects within that container." is applicable.

[2014-02-13 Issaquah]

General agreeement on intent, several wording nits and additional paragraphs to hit.

STL to provide updated wording. Move to Open.

Proposed resolution:

[...]

no move constructor [...] of a container (except for array) invalidates any references, pointers, or iterators referring to the elements of the source container. [Note: The end() iterator does not refer to any element, so it may be invalidated. — end note]

So, this is an open issue, with general agreement on its basic solution (pointer shall not be invalidated by moving). However, it isn't officially accepted (yet?) as a defect. As far as I know, all major implementations do not invalidate pointers when move-constructing, and it seems to be a generally (implicitly) provided guarantee.

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