STL向量：移动向量的所有元素 [英] STL vector: Moving all elements of a vector

问题描述

我有两个STL向量 A 和 B ，我想清除<$ c的所有元素$ c> A 并将 B 的所有元素移动到 A code> B 。简单地说，我想这样做：

I have two STL vectors A and B and I'd like to clear all elements of A and move all elements of B to A and then clear out B. Simply put, I want to do this:

std::vector<MyClass> A;
std::vector<MyClass> B;
....
A = B;
B.clear();

由于 B 可能很长， k * O（N）执行此操作，其中 k 是常数， N 是 max（size_of（A），size_of（B））。我想知道是否可以有一个更有效的方式这样做。我可以想到的一件事是定义 A 和 B 作为指针，然后复制指针在常量时间和清除out B 。

Since B could be pretty long, it takes k*O(N) to do this operation, where k is a constant, and N is max(size_of(A), size_of(B)). I was wondering if there could be a more efficient way to do so. One thing that I could think of is to define A and B as pointers and then copy pointers in constant time and clear out B.

推荐答案

使用C ++ 11 as：

Using C++11, it's as simple as:

A = std::move(B);

现在 A 由 B 持有， B 现在为空。这避免复制：内部表示简单地从 B 移动到 A ，所以这是一个 O（1）解决方案。

Now A contains the elements that were previously held by B, and B is now empty. This avoids copying: the internal representation is simply moved from B to A, so this is an O(1) solution.

对于C ++ 03，如Prætorian所说，可以交换向量。有一个专门的 std :: swap 函数，它以 std :: vector s为参数。这有效地交换内部表示，所以你最终避免创建他们持有的元素的副本。此函数在 O（1）复杂性中也有效。

As for C++03, as Prætorian states, you could swap the vectors. There is a specialization of the std::swap function, which takes std::vectors as its arguments. This effectively swaps the internal representation, so you end up avoiding creating copies of the elements held by them. This function works in O(1) complexity as well.

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