How to return a range::view in a std::expected?

Question

Is there a way to return a range::view in a std::expected construct? For example, in the pseudo code below, what can I write instead of the rangeType?

std::expected<rangeType, error> get_range()
{
    if (some_condition(x))
        return std::unexpected(errorcode);

    return data|ranges::views::take(x);
}

Answer 1

When returning a std::expected<DifficultTypeToSpell, E> from a function, one can avoid spelling the type entirely, by using the monadic interface of std::expected. Start with a std::expected<T, E> which contains either the actual error, or some default constructed T (which could even be void), and transform it with the expression that actually gives you the type that's difficult to spell.

So the code could look something like this:

auto get_range()
{
    return [] -> std::expected<void, error> {   // start with a simple type
        if (some_condition(x)) 
            return std::unexpected(error{});
        return {};
    }().transform([] {    // and then transform to the desired type
            return data | std::ranges::views::take(x);
    });
}