How to use the range version of `transform()` with two ranges?

Question

The header <algorithm> contains a version of std::transform() taking a two input sequences, an output sequence, and a binary function as parameters, e.g.:

#include <algorithm>
#include <iostream>
#include <iterator>
#include <vector>

int main()
{
    std::vector<int> v0{1, 2, 3};
    std::vector<int> v1{4, 5, 6};
    std::vector<int> result;
    std::transform(v0.begin(), v0.end(), v1.begin(), std::back_inserter(result),
                   [](auto a, auto b){ return a + b; });
    std::copy(result.begin(), result.end(),
              std::ostream_iterator<int>(std::cout, " "));
    std::cout << '\n';
}

C++20 introduced range algoirthms which does include std::ranges::views::transform(R, F) and its implementation std::ranges::views::transform_view. I can see how to use this transform() with one range, e.g.:

#include <algorithm>
#include <iostream>
#include <iterator>
#include <ranges>
#include <vector>

int main()
{   
    std::vector<int> v0{1, 2, 3}; 
    for (auto x: std::ranges::views::transform(v0, [](auto a){ return a + 3; })) {
        std::cout << x << ' ';
    }   
    std::cout << '\n';
}   

However, there is no version supporting more than one range. So, the question becomes: How to use the range version of transform() with two (or more) ranges? On objective of this approach is to benefit from the lazy evaluation of views and avoid the creation of an intermediate sequence (result in the non-ranges version above). A potential use could look like this (putting the function argument in front to make it easier for a potential solution allowing even more than two ranges):

for (auto v: envisioned::transform([](auto a, auto b){ return a + b; }, v0, v1) {
    std::cout << v << ' ';
}
std::cout << '\n';

Answer 1

您正在寻找的是 range-v3 调用的算法以及我们在P2214zip_with中建议以名称添加到 C++23 的算法。C++20中没有这样的算法。zip_transform

在那之前，range-v3 版本正是您的用例：

for (auto v : zip_with([](auto a, auto b){ return a + b; }, v0, v1)) {
    std::cout << v << ' ';
}

它可以处理任意数量的范围。

请注意，这里没有管道版本，就像常规zip.

Answer 2

The way you would like to use transform, where you take an arbitrary number of input ranges is not possible directly with what's available in <algorithm> as of C++20. You can of course write such an algorithm yourself without too much effort.

The example with 2 input ranges can be implemented in C++20 like this:

std::ranges::transform(v0, v1,
                       std::ostream_iterator<int>(std::cout, " "),
                       std::plus{});    

Here's a demo, and this is specifically the last overload of transform listed here.

不幸的是，没有办法编写这样的等效版本：

for (auto v : std::views::transform(v0, v1, std::plus{})) // no, unfortunately    
  std::cout << v << " ";

但上面的实现做了同样的事情。它当然满足您不必单独存储结果的要求；它们可以在生成时打印。