同时迭代两个或多个容器的最佳方法是什么 [英] What's the best way to iterate over two or more containers simultaneously

问题描述

C++11 提供了多种迭代容器的方法.例如:

C++11 provides multiple ways to iterate over containers. For example:

for(auto c : container) fun(c)

std::for_each

for_each(container.begin(),container.end(),fun)

但是，推荐的方法是迭代两个(或更多)相同大小的容器以完成以下操作:

However what is the recommended way to iterate over two (or more) containers of the same size to accomplish something like:

for(unsigned i = 0; i < containerA.size(); ++i) {
  containerA[i] = containerB[i];
}

推荐答案

派对迟到了.但是:我会遍历索引.但不是使用经典的 for 循环，而是使用基于范围的 for 循环在索引上:

Rather late to the party. But: I would iterate over indices. But not with the classical for loop but instead with a range-based for loop over the indices:

for(unsigned i : indices(containerA)) {
    containerA[i] = containerB[i];
}

indices 是一个简单的包装函数，它返回索引的(惰性求值)范围.由于实现(虽然简单)有点太长，无法在此处发布，您可以在 GitHub 上找到实现.

indices is a simple wrapper function which returns a (lazily evaluated) range for the indices. Since the implementation – though simple – is a bit too long to post it here, you can find an implementation on GitHub.

此代码与使用手动的经典 for 循环一样高效.

This code is as efficient as using a manual, classical for loop.

如果这种模式经常出现在您的数据中，请考虑使用另一种模式，该模式 zip 生成两个序列并生成一系列元组，对应于成对的元素:

If this pattern occurs often in your data, consider using another pattern which zips two sequences and produces a range of tuples, corresponding to the paired elements:

for (auto& [a, b] : zip(containerA, containerB)) {
    a = b;
}

zip 的实现留给读者作为练习，但它很容易从 indices 的实现中得到.

The implementation of zip is left as an exercise for the reader, but it follows easily from the implementation of indices.

(在 C++17 之前，您必须改为编写以下代码:)

(Before C++17 you’d have to write the following instead:)

for (auto items&& : zip(containerA, containerB))
    get<0>(items) = get<1>(items);

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