On 5/5/2011 6:48 AM, Stewart, Robert wrote:
Your use case is the following, right?
optional<int> o(try_convert_cast<int>(str)); int i; if (!o) { std::cout<< "using fallback\n"; i = fallback; } else { i = o.get(); }
That's clearly inconvenient and 5 doesn't apply because the optional won't be set if the conversion fails. To address such inconvenience, you offered convert<T>::result and pair<optional<T>,bool> has been suggested. Isn't optional redundant in the latter?
What happened to: int i = fallback; if (!try_convert_to<int>(str, i)) { cout << "using fallback"; } A good postcondition is that try_convert_to does not modify i if the conversion fails. This achieves the same thing and is honestly more convenient and transparent than the convert<T>::result approach. I suppose you could argue that the preliminary assignment is inefficient, but I can't see a reasonable objection to it. If it really must be avoided, then of course for defaultable types it can look like yours (assigning the fallback in the if), but for non-defaultable types there's no benefit (a default_value<T> invocation is as bad as the preliminary assignment). -Matt