Re: [Boost-users] [Units] Un-scaling dimensionless value

Subject: Re: [Boost-users] [Units] Un-scaling dimensionless value
From: Lindley French (lindleyf_at_[hidden])
Date: 2014-02-17 13:05:12

Solved it this way:

si::dimensionless unscaled;
double f = conversion_factor(u,unscaled);
if (f < 1)
{
     int64_t intf = static_cast<int64_t>(1.0/f + 0.5);
     return val.value()/intf;
}
else
{
     int64_t intf = static_cast<int64_t>(f + 0.5);
     return val.value()*intf;
}

I can understand why double was chosen for scaling, but it doesn't "feel"
right. Oh well, this will do for now.

On Mon, Feb 17, 2014 at 11:17 AM, Steven Watanabe <watanabesj_at_[hidden]>wrote:

> AMDG
>
> On 02/17/2014 07:40 AM, Lindley French wrote:
> > For the life of me I can't figure out where the actual multiplication of
> > the value occurs. Boost Units is very clever template magic, but I doubt
> > anyone could maintain the thing without weeks of studying the design....
> >
>
> detail/conversion_impl.hpp:345
>
> return(destination_type::from_value(static_cast<T2>(source.value() *
> conversion_factor(u1, u2))));
>
> This won't really help you because
> the result of conversion_factor
> returns a double.
>
> The scale is evaluated at detail/conversion_impl.hpp:448
>         return(conversion1::value() * (scale::value() /
> conversion2::value()));
>
> "scale" here is the result of get_scale_list/eval_scale_list.
> get_scale_list collects the scaling factors, eval_scale_list
> multiplies their ::value() together.
>
> Again, scale::value() returns a double, because the
> individual scale factors use double.  (See scale.hpp)
>
> >
> > On Mon, Feb 17, 2014 at 9:48 AM, Lindley French <lindleyf_at_[hidden]>
> wrote:
> >
> >> What's the best way to access the current scaling of a dimensionless
> >> value? Maybe I can back it out myself rather than relying on type
> >> conversion to do it.
> >>
>
> There's a metafunction called get_scale_list,
> which returns an MPL sequence whose members
> contain
>
> static const long base = ...;
> typedef static_rational<...> exponent;
>
> It's not documented, so use it at your own risk.
> It's unlikely to change, but I make no guarantees.
>
> In Christ,
> Steven Watanabe
>
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