Message du 09/05/11 23:40 De : "Stewart, Robert" A : "'boost@lists.boost.org'" Copie à : Objet : Re: [boost] [convert] no-throw and fallback feature
Vicente BOTET wrote:
De : "Stewart, Robert"
Vicente BOTET wrote:
If we see optional as a kind of conditional tuple and we imagine something like a conditional tie (let me call it 'ctie'), we could be able to write
ctie(i,b) = try_convert_to(s);
ctie will left i unchanged if b is true.
if b is false.
ctie won't change i if the conversion succeeds?
I presume that no assignment would be done to i if b is false.
Surely that is the right interpretation.
Yes.
That actually means writing this:
int i; bool b; ctie(i, b) = try_converting_to(s); if (!b) { i = fallback; std::cerr << "using fallback\n"; }
Compare that to this:
int i(fallback); if (!try_converting_to(s, i)) { std::cerr << "using fallback\n"; }
The fist is more on the line of functional paradigm. The second on the procedural one. I understand that the procedural one can be more efficient, but the functional one separates clearly what are the in a,d the out parameters. Are both alternatives exclusive?
They aren't exclusive, but they'd need different names and, again, I suspect we'd be getting into the too-many-choices conundrum.
Couldn't overload be used here?
Besides, the same thing is done with the following which is simpler and less tricky:
int i; if (try_converting_to(s, i)) ...
If conversion to bool is implicit the bool parameter will be redundant and the preceding could be expressed as
int i; if ((opt_tie(i) = try_convert_to(*int*)(s))) ...
Ah, right. That's a curious approach to it. You're suggesting, if I have all the parts aligned right, that try_convert_to() return optional and that opt_tie() assign to its argument iff the optional is set. That means there's an optional-based API and the following is also possible:
int i(fallback); opt_tie(i) = try_convert_to(s);
Yes this something I could use. Best, Vicente