C ++中矢量的初始容量(Initial capacity of vector in C++)

C ++中矢量的初始容量(Initial capacity of vector in C++)

使用默认构架创建的std::vector的capacity()是多少？ 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？

这样，可以使用单个分配来创建一个具有任意保留的数组，如std::vector<int> iv; iv.reserve(2345); std::vector<int> iv; iv.reserve(2345); 。 让我们说，由于某种原因，我不想在2345年启动size() 。

例如，在Linux上（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 。 这是一个规则，还是STL厂商依赖？

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?

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.

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; }

printed 0,10. Is it a rule, or is it STL vendor dependent?

最满意答案

该标准没有规定容器的初始capacity应该是什么，因此您依赖于实现。 一个共同的实施将开始在零的能力，但不能保证。 另一方面，没有办法改善你的策略std::vector<int> iv; iv.reserve(2345); std::vector<int> iv; iv.reserve(2345); 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.