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';