問題描述

可能的重復:
在 C++11 range-based 中查找元素的位置for 循環?

我有一個 vector 并且我想迭代它，同時可以訪問每個單獨元素的索引(我需??要將元素及其索引傳遞給功能).我考慮了以下兩種解決方案:

I have a vector and I would like to iterate it and, at the same time, have access to the indexes for each individual element (I need to pass both the element and its index to a function). I have considered the following two solutions:

std::vector<int> v = { 10, 20, 30 };

// Solution 1
for (std::vector<int>::size_type idx = 0; idx < v.size(); ++idx)
    foo(v[idx], idx);

// Solution 2
for (auto it = v.begin(); it != v.end(); ++it)
    foo(*it, it - v.begin());

我想知道是否有更緊湊的解決方案.類似于 Python 的 enumerate.這是我使用 C++11 范圍循環得到的最接近的結果，但必須在私有范圍內定義循環外的索引似乎比 1 或 2 更糟糕:

I was wondering whether there might be a more compact solution. Something similar to Python's enumerate. This is the closest that I got using a C++11 range-loop, but having to define the index outside of the loop in a private scope definitely seems to be like a worse solution than either 1 or 2:

{
    int idx = 0;
    for (auto& elem : v)
        foo(elem, idx++);
}

有沒有辦法(也許使用 Boost)來簡化最新的例子，使索引自包含到循環中?

Is there any way (perhaps using Boost) to simplify the latest example in such a way that the index gets self-contained into the loop?

推薦答案

正如@Kos 所說，這是一件如此簡單的事情，我真的不認為有必要進一步簡化它，個人只會堅持傳統的使用索引循環，除了我會放棄 std::vector::size_type 并簡單地使用 std::size_t:

As @Kos says, this is such a simple thing that I don't really see the need to simplify it further and would personally just stick to the traditional for loop with indices, except that I'd ditch std::vector<T>::size_type and simply use std::size_t:

for(std::size_t i = 0; i < v.size(); ++i)
    foo(v[i], i);

我不太熱衷于解決方案 2.它需要(有點隱藏)隨機訪問迭代器，這不允許您輕松交換容器，這是迭代器的強項之一.如果您想使用迭代器并使其通用(并且當迭代器不是隨機訪問時可能會導致性能下降)，我建議使用std::distance:

I'm not too keen on solution 2. It requires (kinda hidden) random access iterators which wouldn't allow you to easily swap the container, which is one of the strong points of iterators. If you want to use iterators and make it generic (and possibly incur a performance hit when the iterators are not random access), I'd recommend using std::distance:

for(auto it(v.begin()); it != v.end(); ++it)
    foo(*it, std::distance(it, v.begin());

這篇關于C++ 中類似 Python 的循環枚舉的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網！