Is it possible to construct a `std::span` from a view in C++20?

Question

This example program does not compile, because the transform_view cannot be converted to a std::span:

class Foo {
private:
    std::vector<std::string> strings = { "a", "b", "c" };

public:
    std::span<const char*> getStrings() {
        return strings | std::views::transform([](const std::string& str) { return str.c_str(); });
    }
};

int main() {
    Foo foo;
    auto strings = foo.getStrings();

    for (auto s : strings)
        std::cout << s << std::endl;
}

I know that it is not possible to construct containers (like std::vector) yet, however I don't quite understand, why it is not possible to construct a std::span from it. I found this answer, that stated, that currently the only container that can be constructed from an arbitrary range is std::span, so I expected the above example to work.

Is there any way to create a span from a range? Or is there any other way to return a generic view from a method, without using auto (which is not allowed for virtual methods)?

Answer 1

Is it possible to construct a std::span from a view in C++20?

It is possible to construct a span from any contiguous range (of appropriate underlying type). The problem here:

std::span<const char*> getStrings() {
    return strings | std::views::transform([](const std::string& str) { return str.c_str(); });
}

is that the adapted range you're producing isn't contiguous, it's only random access. A span<char const*> has to refer to char const*s that are contiguous in memory, and that's definitely not going to be the case here. That's why this doesn't work.

But that doesn't mean that no adapted ranges can be converted to a span. For instance, this would work:

std::span<std::string> getStrings() {
    return strings | std::views::take(5);
}

Since views::take can preserve contiguity (in a way that transform cannot, for reasons that hopefully are clear).

Answer 2

std::span is a pointer and a size. It points into an array (or nowhere). There is no array of const char *.

You can't get a std::span from a std::deque either.