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如何使用 std::views C++20 功能对向量进行排序?

[英]How to sort a vector using std::views C++20 feature?

I want to loop through a vector in a sorted way without modifying the underlying vector .我想在不修改基础 vector 的情况下以排序的方式循环遍历 vector

Can std::views and/or std::range be used for this purpose? std::views和/或std::range可以用于此目的吗?

I've successfully implemented filtering using views , but I don't know if it is possible to sort using a predicate.我已经使用views成功实现了过滤,但我不知道是否可以使用谓词进行排序。

You can find an example to complete here: https://godbolt.org/z/cKer8frvq您可以在此处找到要完成的示例: https://godbolt.org/z/cKer8frvq

#include <iostream>
#include <ranges>
#include <vector>
#include <chrono>

struct Data{
    int a;
};

int main() {

    std::vector<Data> vec = {{1}, {2}, {3}, {10}, {5}, {6}};
  
    auto sortedView = // <= can we use std::views here ?
                           
    for (const auto &sortedData: sortedView) std::cout << std::to_string(sortedData.a) << std::endl; // 1 2 3 5 6 10

    for (const auto &data: vec) std::cout << std::to_string(data.a) << std::endl; // 1 2 3 10 5 6
}

You have to modify something to use std::ranges::sort (or std::sort ), but it doesn't have to be your actual data.您必须修改某些内容才能使用std::ranges::sort (或std::sort ),但它不一定是您的实际数据。

#include <iostream>
#include <ranges>
#include <numeric>
#include <algorithm>
#include <vector>
#include <chrono>

struct Data{
    int a;
    friend auto operator<=> (const Data &, const Data &) = default;
};

int main() {

    std::vector<Data> vec = {{1}, {2}, {3}, {10}, {5}, {6}};

    std::vector<std::size_t> indexes(vec.size());
    std::iota(indexes.begin(), indexes.end(), std::size_t{ 0 }); // 0z in C++23

    auto proj = [&vec](std::size_t i) -> Data & { return vec[i]; };

    std::ranges::sort(indexes, std::less<>{}, proj);
  
    auto sortedView = std::ranges::views::transform(indexes, proj);
                           
    for (const auto &sortedData: sortedView) std::cout << sortedData.a << std::endl; // 1 2 3 5 6 10

    for (const auto &data: vec) std::cout << data.a << std::endl; // 1 2 3 10 5 6
}

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