Re: [boost] fixed pt binary lib

From: Matt Gruenke (mgruenke_at_[hidden])
Date: 2008-08-01 13:52:43

John C. Femiani wrote:
> Matt Gruenke wrote:
>> John C. Femiani wrote:
>>> Matt Gruenke wrote:
>>>> Speaking of division, while I'd expect /= to return the same type 
>>>> as the LHS operand, the way I've previously implemented fixed point 
>>>> division is to return:
>>>>    result_int_bits = (numer_int_bits + denom_frac_bits + 1)
>>>>    result_frac_bits = (numer_frac_bits + denom_int_bits - 1)
>>> So then what happens to the value of X after I say  Z = X /=Y?
>>> Do you mean '/', instead of '/='?
>>
>> Yes, I was describing the behavior of my operator/().  I hadn't 
>> written an operator/=().
>>
> I think perhaps the issue with division is that rounding policy should 
> be rethought. Also:
>>  base_type tmp = (base_type)val / (base_type)x.val;
>>     val = tmp << (frac_bits);
> should be something like
>
>    static const int DENOMINATOR=  1 << frac_bits;
>    base_type tmp =  rounding_policy::divide(val * DENOMINATOR,   x.val);
>
> I think the old way lost a lot of precision because it divided before 
> multiplying.e.g.

By replying to my message, it seems like you might be confusing two 
implementations.  The one I was describing is my own private code, which 
worked the way you're suggesting.

> Then for multiplying:
>
>>     base_type tmp = (base_type)val * (base_type)x.val;
>>     val = tmp >> (frac_bits);
> could possibly be:
>
>    val = rounding_policy::divide( val * x.val,    DENOMINATOR);

The approach I took for operator*() was to return a larger type that 
lost no precision or range from the multiplication.  Then the rounding 
happens when you assign the result to a smaller type.  This both gives 
the user a natural way to control how much precision is preserved and 
avoids overhead from any unnecessary rounding (i.e. rounding more 
frequently than the user might want or need).

Matt