std :: vector.pop_back（）改变向量的容量？ [英] Does std::vector.pop_back() change vector's capacity?

问题描述

如果我使用 resize（）和 reserve（）分配一个std :: vector到一定的大小和容量在我的程序的开始，是否可能 pop_back（）可能打破保留容量并导致重新分配？

If I allocated an std::vector to a certain size and capacity using resize() and reserve() at the beginning of my program, is it possible that pop_back() may "break" the reserved capacity and cause reallocations?

推荐答案

否。缩小向量容量的唯一方法是交换技巧

No. The only way to shrink a vector's capacity is the swap trick

template< typename T, class Allocator >
void shrink_capacity(std::vector<T,Allocator>& v)
{
   std::vector<T,Allocator>(v.begin(),v.end()).swap(v);
}

，即使不能保证按照标准工作。 （虽然很难想象一个实现它不能工作。）

and even that isn't guaranteed to work according to the standard. (Although it's hard to imagine an implementation where it wouldn't work.)

据我所知，下一版本的C ++标准（以前是C ++ 0x，但现在成为C ++ 1x）将有 std :: vector<> :: shrink_to_fit（）。

As far as I know, the next version of the C++ standard (what used to be C++0x, but now became C++1x) will have std::vector<>::shrink_to_fit().

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