From: Steven Watanabe (steven_at_[hidden])
Date: 2007-02-13 17:30:08
AMDG
Noah Roberts <roberts.noah <at> gmail.com> writes:
>
> "Converting the second to match the first" doesn't make sense to me. I
> guess you might be talking about operands of '+'? Yeah, not necessary.
>
> The answer is pretty simple. All arithmetic operators work in the base
> system. The result is converted on assignment if necessary.
Ok. What happens here:
std::cout << (x + y) << std::endl;
Is the unit printed
a. the unit of x
b. the unit of y
c. the base unit
>
> >
> > Both of these require storing twice as much information
> > as the current implementation.
>
> There is one condition in which that is necessary. That is if you are
> using a value that is entered by the user and the calculation loop keeps
> requesting it from whatever object it is provided by. Then, in order
> that the conversion isn't repeatedly done you would want to store two
> values...the user entered value and the converted value. The reason you
> wouldn't just store the converted value is double creep...the number
> might change on the user. I imagine that a developer who needs to can
> work with this so that it does not happen...so that the calculation loop
> is not asking for the same conversion on the same value repeatedly.
>
> Any other time it is sufficient that you keep only the user entered
> value and the unit (ie the conversion factor) to translate it into the
> base system...OR...the converted value in the base system without a
> conversion factor.
>
The former is exactly what I said. It stores
two doubles instead of one. Further,
temp = a + b;
c = temp * d;
is not equivalent to
c = d * (a + b);
Every operation incurs an extra multiplication
per operand. This is a far cry from zero overhead.
The latter does not have any significant
benefit over the current implementation.
In Christ,
Steven Watanabe