4 Apr
2019
4 Apr
'19
9:51 p.m.
Every time you touch a variant object, you must consider the possibility of it being valueless. No I don't. It just works and I explained why. Could you please point out specifically where I made a mistake in my reasoning in the mail before where I discussed the behavior of all member functions? And again: I'm not arguing that one is better than the other. I'm saying it doesn't matter, but I'd much perfer a design that is in line with std::variant instead of having yet another slightly incompatible version of variant that doesn't offer any significant benefit (would be a different story if it would always offer the strong exception guarantee).