是否在移动std :: vector时需要保留容量？ [英] Is the capacity required to be preserved when moving a std::vector?

问题描述

请考虑以下代码：

std::vector vec;
vec.reserve(500);
size_t cap = vec.capacity();

std::vector newVec = std::move(vec);
assert(cap == newVec.capacity());

在几乎所有执行过程中，这都可以工作。我不在乎什么实现。我想知道标准的要求。移动到向量与原始容量相同吗？或者将assert触发器？

In pretty much any implementation you run across, this will work. I don't care about what implementations do. I want to know what the standard requires. Will the moved-to vector have the same capacity as the original? Or will the assert trigger?

推荐答案

看看标准，看起来什么都不需要从move构造函数， amaurea说，如果移动构造函数试图分配或释放内存，它将完全击败移动语义的目的，所以我希望所有实现中的容量保持不变。

Looking at the standard, it appears that nothing is required from the move constructor, however as @amaurea says, it would completely defeat the purpose of move semantics if the move constructor were to try and allocate or deallocate memory, so I would expect the capacity to remain the same in all implementations.

23.2.1一般容器需求

>

X u(a);
X u = a;

声明/注释前/后条件

需要： T 是CopyInsertable为X（见下文）。

post： u == a

Requires: T is CopyInsertable into X (see below).
post: u == a

标准只需要 newVec = = vec 。由于 std :: vector :: operator == ， newVec 不需要考虑容量与 vec 相同的容量。

The standard only requires that newVec == vec. As capacity is not taken into consideration for std::vector::operator==, newVec need not necessarily have the same capacity as vec.

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