Subject: Re: [boost] [units] - learning to use
From: Steven Watanabe (watanabesj_at_[hidden])
Date: 2011-11-25 09:35:21
On 11/25/2011 05:11 AM, Janek Kozicki wrote:
>>> 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 ?
It will be 1.0. in this case. To be safe
use static_cast, or boost::implicit_cast.
> 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 question:
> 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?
It's the same problem as above with int/double.
>>> 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.
In general, any conversion that causes the
underlying value to change is explicit.
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