C++中向量的初始容量
使用默认构造函数创建的 std::vector 的 capacity() 是多少?我知道 size() 为零.我们可以声明默认构造的向量不会调用堆内存分配吗?
What is the capacity() of an std::vector which is created using the default constuctor? I know that the size() is zero. Can we state that a default constructed vector does not call heap memory allocation?
通过这种方式，可以使用单个分配创建具有任意保留的数组，例如 std::vector.假设出于某种原因，我不想在 2345 上启动 size().
This way it would be possible to create an array with an arbitrary reserve using a single allocation, like std::vector<int> iv; iv.reserve(2345);. Let's say that for some reason, I do not want to start the size() on 2345.
例如，在 Linux(g++ 4.4.5，内核 2.6.32 amd64)上
For example, on Linux (g++ 4.4.5, kernel 2.6.32 amd64)
#include <iostream>
#include <vector>
int main()
{
using namespace std;
cout << vector<int>().capacity() << "," << vector<int>(10).capacity() << endl;
return 0;
}
打印0,10.这是规则，还是取决于 STL 供应商?
printed 0,10. Is it a rule, or is it STL vendor dependent?
推荐答案
该标准没有指定容器的初始 capacity 应该是多少，因此您依赖于实现.一个常见的实现将从零开始容量，但不能保证.另一方面，没有办法改善 std::vector 所以坚持下去.
The standard doesn't specify what the initial capacity of a container should be, so you're relying on the implementation. A common implementation will start the capacity at zero, but there's no guarantee. On the other hand there's no way to better your strategy of std::vector<int> iv; iv.reserve(2345); so stick with it.
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