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Subject: Re: [boost] [qvm] scalar_cast
From: Nikolay Mladenov (nikolay.mladenov_at_[hidden])
Date: 20121105 22:38:03
On Mon, Nov 5, 2012 at 4:38 PM, Emil Dotchevski <emildotchevski_at_[hidden]>wrote:
> On Sat, Nov 3, 2012 at 4:00 PM, Nikolay Mladenov
> <nikolay.mladenov_at_[hidden]> wrote:
> > I am trying qvm in a small project that uses complex numbers and have
> > expressions of the kind :
> > cv3 = c1*rv3_1 + c2*rv3_2
> > (the v3's are vectors, the c's are complex and the r's are real).
> >
> > I wrote it as :
> >
> > cv3
> > = scalar_cast<complex>(rv3_1)*c1
> > + scalar_cast<complex>(rv3_2)*c2;
> >
> > but this seems to invoke 6 complex constructors (unnecessarily)
> > I have not tested to see if the constructors will be optimized away,
> > but it seems to me that if :
> > v_traits<vector_scalar_cast_>::scalar_type == complex , bit
> > v_traits<vector_scalar_cast_>::r/ir return real
> > the expression will compile and no complex constructors will be invoked.
>
> If I understand correctly, you mean that if a complex can be
> implicitly constructed from a float, then r/ir can return float and
> it'll all work. I don't think that it is a safe assumption to make in
> general, that any scalar type would define a suitable implicit
> constructor.
>
No. the reason it works is because (i'll use [i] for indexing)
rv1[0]*c1+rv2[0]*c2
is well defined linear combination of two complex numbers with real
coefficients
the cast is only needed to force rv3_1*c1 's result to be complex ( the
default operator tries to build result with the same scalar type as the
rv3_1)
no implicit conversions or constructions are needed to calculate the result.
it is almost as if operator * and / should have return type from the kind
of deduce_v2<vec1, scalar2>
>
> That said, it might be a good idea to specify in the documentation
> that the return value from r/ir must be implicitly convertible to
> ::scalar_type, which would allow a partial specialization for your
> complex type of the vector_scalar_cast_ and the other related
> scalar_cast type templates. Are you willing to give this a try?
>
> (As for scalar multiplication requiring the scalar on the right,
> you're right it should work from the left too.)
>
I think I understood the reason for the right only scalar multiplication:
it is probably because
scalar*vec%X
will do multiple times the work of
vec%X * scalar
> Thanks,
> Emil Dotchevski
> Reverge Studios, Inc.
> http://www.revergestudios.com/reblog/index.php?n=ReCode
>
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