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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