std::vector 向下调整大小 [英] std::vector resize downward

问题描述

C++ 标准似乎没有声明任何一方对容量的副作用resize(n)，用 n 或 clear().

The C++ standard seems to make no statement regarding side-effects on capacity by either resize(n), with n < size(), or clear().

它确实声明了 push_back 和 pop_back - O(1)

It does make a statement about amortized cost of push_back and pop_back - O(1)

我可以设想一个执行通常类型的容量更改的实现ala CLRS 算法(例如，放大时加倍，将 size 减小到 < capacity()/4 时减半).(Cormen Lieserson Rivest Stein)

I can envision an implementation that does the usual sort of capacity changes ala CLRS Algorithms (e.g. double when enlarging, halve when decreasing size to < capacity()/4). (Cormen Lieserson Rivest Stein)

有人有任何实施限制的参考吗?

Does anyone have a reference for any implementation restrictions?

推荐答案

以较小的尺寸调用 resize() 对 vector 的容量没有影响.它不会释放内存.

Calling resize() with a smaller size has no effect on the capacity of a vector. It will not free memory.

从 vector 释放内存的标准习惯用法是用一个空的临时 vector swap() 它: std::vector().swap(vec);.如果要向下调整大小，则需要从原始向量复制到新的局部临时向量，然后将结果向量与原始向量交换.

The standard idiom for freeing memory from a vector is to swap() it with an empty temporary vector: std::vector<T>().swap(vec);. If you want to resize downwards you'd need to copy from your original vector into a new local temporary vector and then swap the resulting vector with your original.

更新: C++11 添加了一个成员函数 shr​​ink_to_fit() 出于此目的，这是一个将 capacity() 减少到 size() 的非绑定请求.

Updated: C++11 added a member function shrink_to_fit() for this purpose, it's a non-binding request to reduce capacity() to size().

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