范围v3展平序列 [英] range v3 flattening a sequence

问题描述

所以我最近观看了有关c ++的演讲: https://www.youtube.com/watch?v=mFUXNMfaciE

So I recently watched this talk on c++: https://www.youtube.com/watch?v=mFUXNMfaciE

我对尝试它非常感兴趣.因此，在执行一些玩具程序之后，我将继续学习如何将向量的向量正确展平为向量.根据此处的文档: https://ericniebler.github.io/range-v3/使用ranges::view::for_each可以做到这一点.但是我似乎无法使它正常工作.这是一些最少的代码.

And I was very interested on trying it out. So after some toy program I am stuck on how to properly flatten a vector of vectors into a vector. According to the documentation here: https://ericniebler.github.io/range-v3/ This is possible using ranges::view::for_each. However I just can't seem to get it to work. Here is some minimal code.

#include <range/v3/all.hpp>
#include <iostream>
#include <vector>

int main()
{
    auto nums = std::vector<std::vector<int>>{
        {0, 1, 2, 3},
        {5, 6, 7, 8},
        {10, 20},
        {30},
        {55}
    };

    auto filtered = nums
        | ranges::view::for_each([](std::vector<int> num) { return ranges::yield_from(num); })
        | ranges::view::remove_if([](int i) { return i % 2 == 1; })
        | ranges::view::transform([](int i) { return std::to_string(i); });

    for (const auto i : filtered)
    {
        std::cout << i << std::endl;
    }
}

推荐答案

range-v3错误消息往往非常可怕，以至于这个错误消息实际上比大多数情况都要好:

range-v3 error messages tend to be pretty horrible, so much so that this one is actually better than most:

prog.cc: In lambda function:
prog.cc:16:90: error: no match for call to '(const ranges::v3::yield_from_fn) (std::vector<int>&)'
         | ranges::view::for_each([](std::vector<int> num) { return ranges::yield_from(num); })
                                                                                          ^
In file included from /opt/wandbox/range-v3/include/range/v3/view.hpp:38:0,
                 from /opt/wandbox/range-v3/include/range/v3/all.hpp:21,
                 from prog.cc:1:
/opt/wandbox/range-v3/include/range/v3/view/for_each.hpp:133:17: note: candidate: template<class Rng, int _concept_requires_132, typename std::enable_if<((_concept_requires_132 == 43) || ranges::v3::concepts::models<ranges::v3::concepts::View, T>()), int>::type <anonymous> > Rng ranges::v3::yield_from_fn::operator()(Rng) const
             Rng operator()(Rng rng) const
                 ^~~~~~~~

对于一些对range-v3的概念仿真层有一定了解的人，

清楚地指出，对yield_from的调用失败了，因为传递给它的参数的类型-std::vector<int>-不满足View概念.

to someone with a bit of knowledge of range-v3's concepts emulation layer, this "clearly" states that the call to yield_from failed because the type of the argument you passed to it - std::vector<int> - does not satisfy the View concept.

View概念表示不拥有其元素的范围子集的特征，因此具有所有操作-移动/复制构造/分配，开始，结束和默认构造-可在O(1)中计算. range-v3中的范围组成算法只在视图上起作用，以避免必须处理元素寿命并提供可预测的性能.

The View concept characterizes a subset of ranges that do not own their elements, and therefore have all operations - move/copy construction/assignment, begin, end, and default construction - computable in O(1). The range composition algrebra in range-v3 works only on views to avoid having to deal with element lifetimes and to provide predictable performance.

yield_from拒绝您尝试传递的std::vector，因为它们不是视图，但是您可以通过(1)将向量作为左值代替for_each中的值来轻松地提供视图，以及(2 )产生这些左值的view::all [ DEMO ]:

yield_from rejects the std::vectors you are trying to pass since they are not views, but you could easily provide views by (1) taking the vectors as lvalues instead of by value in for_each, and (2) yielding view::all of those lvalues [DEMO]:

auto filtered = nums
    | ranges::view::for_each([](std::vector<int>& num) {
        return ranges::yield_from(ranges::view::all(num)); })
    | ranges::view::remove_if([](int i) { return i % 2 == 1; })
    | ranges::view::transform([](int i) { return std::to_string(i); });

但是在这种简单情况下，将范围的元素范围展平为一系列的元素已经在range-v3中具有特定用途的视图:view::join.您也可以使用[ DEMO ]:

But in this simple case, flattening a range of ranges of elements into a range of elements already has a purpose-specific view in range-v3: view::join. You may as well use that [DEMO]:

auto filtered = nums
    | ranges::view::join
    | ranges::view::remove_if([](int i) { return i % 2 == 1; })
    | ranges::view::transform([](int i) { return std::to_string(i); });

这篇关于范围v3展平序列的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！