[boost] Re: Re: rational with finite precision

From: Jonathan Turkanis (technews_at_[hidden])
Date: 2004-12-12 14:35:29

Andras Erdei wrote:
> Mickey Moore wrote:

>> ... What
>> can you expect it to do in such case?
>
> return the (unique) best approximation, in this
> case 3197/1540, or (an easier way out) to state
> in the docs that boost::rational<> only works
> with unlimited precision types

This is essentially what the docs currently state: "The rational number class is
designed for use in conjunction with an unlimited precision integer class." It
would be easy to prevent rational from being instantiated with built-in integral
types, but this would just make the library less useful. Perhaps the disclaimer
should be featured more prominently.

>> If someone wants the best approximation rather than the exact
>> answer, they would probably just be using floating point.  rational
>> is used when people want an exact answer, not just an approximation.
>
> if i understand you correctly, what you say is that
> rational<> will never be instantiated with finite
> precision integers anyway (no exact answer then), so
> there is no reason not to change the docs
>
> afaik rationals (and interval arithmetic and numerous
> other workarounds) are used in applications where
> floating point arithmetic gives you unusable results
> (CAD, electronic circuit design and so on), because
> finite rational arithmetic has much nicer properties
> than finite floating-point

It sounds like you are describing a numeric type which could coexist with
boost::rational. Perhaps you should implement it and post it.

Jonathan