STL算法的可组合性 [英] Composability of STL algorithms

问题描述

STL算法在C ++中是一个非常有用的东西。

The STL algorithms are a pretty useful thing in C++. But one thing that kind of irks me is that they seem to lack composability.

例如，我有一个向量< pair< int ，int>> 并希望将其转换为仅包含第二的向量< int> $ c>成员的对。这很简单：

For example, let's say I have a vector<pair<int, int>> and want to transform that to a vector<int> containing only the second member of the pair. That's simple enough:

std::vector<std::pair<int, int>> values = GetValues();
std::vector<int> result;

std::transform(values.begin(), values.end(), std::back_inserter(result),
    [] (std::pair<int, int> p) { return p.second; });

或者也许我想过滤向量只有那些第一成员是偶数的对。也很简单：

Or maybe I want to filter the vector for only those pairs whose first member is even. Also pretty simple:

std::vector<std::pair<int, int>> values = GetValues();
std::vector<std::pair<int, int>> result;

std::copy_if(values.begin(), values.end(), std::back_inserter(result),
    [] (std::pair<int, int> p) { return (p.first % 2) == 0; });

但是如果我想同时做两个呢？没有 transform_if 算法，并使用 transform 和 copy_if 似乎需要分配临时向量来保存中间结果：

But what if I want to do both? There is no transform_if algorithm, and using both transform and copy_if seems to require allocating a temporary vector to hold the intermediate result:

std::vector<std::pair<int, int>> values = GetValues();
std::vector<std::pair<int, int>> temp;
std::vector<int> result;

std::copy_if(values.begin(), values.end(), std::back_inserter(temp),
    [] (std::pair<int, int> p) { return (p.first % 2) == 0; });

std::transform(values.begin(), values.end(), std::back_inserter(result),
    [] (std::pair<int, int> p) { return p.second; });

这对我来说似乎比较浪费。我可以想到避免临时向量的唯一方法是放弃 transform 和 copy_if ，然后使用 for_each （或常规for循环，无论哪种适合您的想法）：

This seems rather wasteful to me. The only way I can think of to avoid the temporary vector is to abandon transform and copy_if and simply use for_each (or a regular for loop, whichever suits your fancy):

std::vector<std::pair<int, int>> values = GetValues();
std::vector<int> result;

std::for_each(values.begin(), values.end(),
    [&result] (std::pair<int, int> p) 
        { if( (p.first % 2) == 0 ) result.push_back(p.second); });

我在这里缺少什么？是否有一个很好的方法来将两个现有的STL算法组合成一个新的STL算法而不需要临时存储？

Am I missing something here? Is there a good way to compose two existing STL algorithms into a new one without needing temporary storage?

推荐答案

您可以使用 Boost.Range适配器以实现组合。

You're right. You can use Boost.Range adaptors to achieve composition.

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