Thanks a lot for your help, few questions follow:
Try D/double(steps).
ok. I'll do this.
Matthias, I seem to recall that the reason we disabled generic multiplication was that it was causing ambiguities with user defined types. Do you think it would be a good idea to enable arithmetic with any built-in arithmetic type?
it would be great.
Try boost::math?
oh, of course! :) const double pi = boost::math::constants::pi<double>();
(***) it seems strange that I need to take .value() here. The result is dimensionless, why can't it directly convert to complex<double> ? There is a risk that .value() would be expressed in millimeters or angstroms or whatever (maybe depending on dd), and then .value() would return a wrongly scaled number?
This is an idiosyncrasy of overload resolution. std::complex::operator+= is a template, so an implicit conversion doesn't work. The template argument can't be deduced.
OK. So I have to use .value(). The question here: should I expect here some scaling problems? For example: a = 1nm b = 1km c = (a/b).value() Will c become 1.0, with units scaling being lost, due to directly taking .value() without taking care of unit scaling. Or would it be correctly 1e-12 ? In my problem I am calculating exponent then multiply and divide by length. In fact the exponent argument is supposed to be dimensionless also! And as you see I multiply wavenumber by length. I assume the conversions to ::one are good here :) Another queestion: complex<double> i(0,1); quantity<wavenumber,complex<double> > k=2*pi/lambda; // [....] a += (dd*exp(i*k*r)/r).value(); do I really need to define wavenumber k as being complex? After all it's multiplied by 'i' so it should get converted automatically?
quantity<length> wavelength ( 632.8 *nano*meters); quantity<length> wavelength = 632.8 *nano*meters; This isn't assignment. It's copy construction and requires an implicit conversion which is forbidden.
why it is forbidden? I think it would be more convenient to allow this. best regards -- Janek Kozicki http://janek.kozicki.pl/ |