C ++中向量的初始容量
使用默认的constuctor创建的 std :: vector 的 capacity（）是什么？我知道 size（）是零。我们可以声明一个默认构造的向量不会调用堆内存分配？
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?
这样就可以使用单个分配创建一个任意保留的数组，比如 std :: vector< int> iv; iv.reserve（2345）; 。让我们说，由于某种原因，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.
，在Linux上（g ++ 4.4.5，内核2.6.32 amd64）
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;
}
印刷 0,10 。这是一个规则，还是STL供应商的依赖？
printed 0,10. Is it a rule, or is it STL vendor dependent?
标准没有指定初始 capacity 应该是，所以你依赖于实现。一个常见的实现将启动容量为零，但不能保证。另一方面，没有办法更好地你的策略 std :: vector< int> iv; iv.reserve（2345）; 因此坚持下去。
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.