On Mon, Nov 5, 2012 at 4:38 PM, Emil Dotchevski <emildotchevski@gmail.com>wrote:
On Sat, Nov 3, 2012 at 4:00 PM, Nikolay Mladenov <nikolay.mladenov@gmail.com> 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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