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Subject: Re: [boost] [convert] no-throw and fallback feature
From: Stewart, Robert (Robert.Stewart_at_[hidden])
Date: 2011-05-11 07:08:07

Vicente BOTET wrote:
> If opt_tie is implicitly convertible to a bool as it is
> optional
> > // 2
> > int i(fallback);
> > bool b;
> > opt_tie(i, b) = convert_to<optional<int>>(s);
> > if (!b)
> > {
> > std::cerr << "using fallback\n";
> > }
> can be written as
> // 2 b
> int i(fallback);
> if (!( opt_tie(i) = convert_to<optional<int>>(s)) )
> {
> std::cerr << "using fallback\n";
> }
> I agree however that //2 could be easier to read. Do you think
> that opt_tie must be implicitly convertible to bool or not?

I think, while possible, that's much too "magical" and complicated when compared to this:

int i(fallback);
if (!try_convert_to(s, i))
   std::cerr << "using fallback\n";

Also, assignment within a conditional is error prone, so often rejected. An idiom that relies on it, even though not a normal assignment, will likely also be rejected.

Rob Stewart robert.stewart_at_[hidden]
Software Engineer using std::disclaimer;
Dev Tools & Components
Susquehanna International Group, LLP

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