Passing std::ranges::views as parameters in C++20

Question

I have a method that prints a list of integers (my actual method is a bit more complicated but it is also read-only):

void printElements(const std::vector<int> &integersList)
{
    std::for_each(integersList.begin(), integersList.end(), [](const auto& e){
        std::cout << e << "\n";
    });
}

Now suppose I have the following vector:

std::vector<int> vec{1,2,3,4,5,6,7,8,9,10};

Then I want to print even numbers only. To do this, I thought of employing the new std::ranges feature in C++20. I know that you can do this as follows:

auto evenList = vec | std::views::filter([](auto i){ return i % 2 == 0; });

Now I would like to call printElements(evenList), however this obviously won't compile. What would be the solution to this? Also can I write a single function that will both print a std::vector and an object of the same type as my evenList? Or do I need to write two separate functions?

Ted Lyngmo · Accepted Answer · 2022-08-22 13:52:32Z

21

You can make printElements take any object by making it a function template. This will instantiate it for views as well as vectors.

#include <algorithm>
#include <iostream>
#include <ranges>

void printElements(std::ranges::input_range auto&& range) {
    std::ranges::for_each(range, [](const auto& e) { std::cout << e << '\n'; });
}

Demo

or:

// ...
#include <iterator>

void printElements(std::ranges::input_range auto&& range) {
    using T = std::ranges::range_value_t<decltype(range)>;
    std::ranges::copy(range, std::ostream_iterator<T>(std::cout, "\n"));
}

Demo

edited Aug 22, 2022 at 13:52

answered Aug 22, 2022 at 11:10

Ted Lyngmo

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3 Comments

Mr. Clear Over a year ago

How can I define what type the values in the range shall have? What does the "auto" do there? This syntax looks weird?

Ted Lyngmo Over a year ago

@Mr.Clear If you want to constrain it further than it already is, you could use require or make a concept that combines input_range and the set of types you want to approve it for. auto is there because it's required for this kind of usage of type-constraints. Weird syntax? C++20 gave us some news for sure but I don't think it's that weird.

user202729 Jul 28 at 2:42

One possible reason why there isn't a specific input_range<T> is explained in stackoverflow.com/a/64228354/5267751

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