Get projected value from std::ranges algorithms

Question

I am using algorithms from std::ranges (max and max_element) with a projection. Is it possible for the result to also be the projected value? Currently I have to call the projection function again on the returned value.

Example: Here I want the size of the longest string, but the algorithms return only the string or an iterator to it.

int main()
{
    const std::vector<std::string> vec = {
        "foo",
        "hello",
        "this is a long string",
        "bar"
    };

    //r1 is a string. r2 is an iterator
    const auto r1 = std::ranges::max(vec, {}, &std::string::size);
    const auto r2 = std::ranges::max_element(vec, {}, &std::string::size);
    
    //I have to call size() again
    std::cout << r1 << '\n' << *r2 << '\n';
    std::cout << r1.size() << '\n' << r2->size() << std::endl;
}

Compiler Explorer

Answer 1

You're using an algorithm (max/max_element) on the original range, which can't do anything but give you an element/iterator into the range.

If you want just the projected values, do the projection (via a views::transform) to get the lengths first, and then find the maximum of that

auto const lens = std::views::transform(vec, &std::string::size);

const auto r1 = std::ranges::max(lens);
const auto r2 = std::ranges::max_element(lens);

std::cout << r1 << '\n' << *r2 << '\n';  // prints 21 21

Here's a demo.

---

As mentioned in this answer, taking the address of std::string::size is not permitted, so you should use a lambda instead. In general though, taking a projection based on a member function works just fine, so long as it's not a std function.