std::vector.pop_back() 会改变向量的容量吗?

Does std::vector.pop_back() change vector#39;s capacity?(std::vector.pop_back() 会改变向量的容量吗?)

问题描述

如果我在程序开始时使用 resize() 和 reserve() 将 std::vector 分配到特定大小和容量，是否有可能pop_back() 可能会破坏"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

没有.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);
}

即使这样也不能保证按照标准工作.(虽然很难想象它不会工作的实现.)

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，C++ 标准的下一个版本(以前是 C++0x，但现在变成了 C++1x)将具有 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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