调整大小以减少计数时，在 vector::resize() 中重新分配空间? [英] Space reallocated in vector::resize() when resized to lower count?

问题描述

根据Stroustrup:C++ 编程语言 :-

当一个 vector 被调整大小以容纳更多(或更少)元素时，它的所有元素都可能是搬到了新的地点."

"When a vector is resized to accommodate more (or fewer) elements, all of its elements may be moved to new locations."

这是否成立，即使向量被重新调整为更小的尺寸?

Is this holds true, even if the vector is re-sized to smaller size ?

推荐答案

情况 1: 如果请求的新大小大于当前 std::vector::capacity() 那么所有元素都将被重新定位.
情况 2: 如果请求的新大小小于当前的 std::vector::capacity()，则不会重新定位元素.

Case 1: If the new size being requested is greater than the current std::vector::capacity() then all elements will be relocated.
Case 2: If the new size being requested is lesser than the current std::vector::capacity() then there will be no relocation of elements.

标准证据:

标准将vector::resize()的效果定义为:

C++11 标准 23.3.6.3/12 向量容量:

void resize(size_type sz, const T& c);

效果:

if (sz > size())
     insert(end(), sz-size(), c);
else if (sz < size())
    erase(begin()+sz, end());
else
; // do nothing

正如 @DavidRodríguez-dribeas 正确指出的那样，std:: 的迭代器失效规则:vector::insert() 操作是:

As @DavidRodríguez-dribeas correctly points out, Iterator invalidation rules for std::vector::insert() operation are:

23.3.6.5 矢量修饰符

1 [insert,push_back,emplace,emplace_back]

1 [insert,push_back,emplace,emplace_back]

备注:如果新容量大于旧容量，则导致重新分配.如果没有发生重新分配，插入点之前的所有迭代器和引用仍然有效.

Remarks: Causes reallocation if the new size is greater than the old capacity. If no reallocation happens, all the iterators and references before the insertion point remain valid.

本质上这意味着:插入点之前的所有迭代器和引用都不会受到影响，除非新容器的大小大于以前的容量，因为在这种情况下，所有元素都可能移动到新位置，从而使指向原始位置的指针/迭代器无效.自 resize() 仅擦除/插入容器末尾的元素^{[注 1]}.控制因素归结为根据当前容量请求的大小.

Essentially this means: All iterators and references before the point of insertion will be unaffected, unless the new container size is greater than the previous capacity because in such a scenario all elements might be moved to new locations thus invalidating pointers/iterators to original location.Since resize() only erases/inserts elements at the end of the container^{[Note 1]}.The governing factor boils down to size being requested as against current capacity.

因此情况 1 结果.

在情况2中std::vector::erase()会被调用，此时的失效规则是:

In Case 2 std::vector::erase() will be called and the invalidation rule in this case is:

23.3.6.5 矢量修饰符

迭代器擦除(const_iterator position);

iterator erase(const_iterator position);

3 效果:在擦除点或之后使迭代器和引用无效.

3 Effects: Invalidates iterators and references at or after the point of the erase.

由于^[注1]，元素只会在最后被移除，不需要所有元素都重新定位.

Since ^{[Note 1]}, elements will be only removed at end and there is no need of relocation of all elements.

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