`std::vector<primitive>::clear()` 是一个常数时间操作吗? [英] Is `std::vector<primitive>::clear()` a constant time operation?

问题描述

对向量调用 clear() 将调用存储在向量中的任何内容的析构函数，这是一个线性时间操作.但是当向量包含诸如 int 或 double 之类的原始类型时，情况是否如此?

Calling clear() on a vector will call the destructors of whatever is stored in the vector, which is a linear time operation. But is this the case when the vector contains primitive types like int or double?

推荐答案

我相信答案取决于实现.它需要最多个线性时间，但某些实现可能会选择对此进行优化.

I believe the answer is implementation dependent. It takes at most linear time, but some implementations may choose to optimize this.

Per '清除矢量会影响其容量吗?'，即使在调用 .clear 时，MSVC 和 G++ 都不会降低其向量的容量.查看 G++ 头文件，很明显 .clear 是具有默认分配器的常量时间，只要元素是标量(原始算术类型或指针).

Per 'Does clearing a vector affect its capacity?', neither MSVC nor G++ decrease the capacity of their vectors, even when .clear is called. Looking at the G++ headers, it is evident that .clear is constant-time with the default allocator, as long as the elements are scalar (primitive arithmetic types or pointers).

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