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Subject: Re: [boost] [convert] Version in the Vault ready for review.
From: Stewart, Robert (Robert.Stewart_at_[hidden])
Date: 2009-04-09 11:44:07
On Thursday, April 09, 2009 11:05 AM
Gottlob Frege wrote:
> On Wed, Apr 8, 2009 at 4:59 PM, Stewart, Robert
> <Robert.Stewart_at_[hidden]> wrote:
> > On Monday, April 06, 2009 1:23 PM
> > Gottlob Frege wrote:
> >>
> >> A)
> >> int i = convert<int>::from(str); // throws if necessary
> >>
> >> B)
> >> int i = convert<int>::from(str, 0); // default => no throw
>
> Let's split case 'C' into 'C' and 'D':
>
> >> C)
> >> convert<int>::result res;
> >> int i = convert<int>::from(str, res); // no throw - error/status
>
> D)
>
> >> result returned in 'res'
> >> int i = convert<int>::from(str, 0, res); // default + status
>
> Case 'C' is problematic - let's get rid of it:
> i = from(str, res);
> is difficult when the type is NOT default constructible. So let's get
> rid of it.
> If you want a status result, supply a default (case D):
> i = from(str, 0, res);
>
> *Alternatively*, since we know the type, we could probably use some
> type-inference tricks to only allow case C to compile when the type is
> default constructible:
>
> C1:
> int i = convert<int>::from(str, res); // OK
>
> - compiles as int is default constructible.
> - does NOT throw as you get status back in res.
>
> C2:
> class C { public: C(int, int); };
> C c = convert<int>::from(str, res); // does NOT compile
> C c = convert<int>::from(str, C(1, 2), res); // default required
I'm in favor of keeping C) for symmetry. Since from() is a member function of a class template, it will not be instantiated unless used and, if used, it will fail to compile. Thus, employing Boost.ConceptCheck can produce a helpful error message to explain the problem.
_____
Rob Stewart robert.stewart_at_[hidden]
Software Engineer, Core Software using std::disclaimer;
Susquehanna International Group, LLP http://www.sig.com
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