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Re: [fpu] '/' operator




--- Rudolf Usselmann <rudi@asics.ws> wrote:
> on 9/3/00 19:19, Jamil Khatib at
> jamilkhatib75@yahoo.com wrote:
> > 
> ...
> > but what about this exponent equations:
> > 
> > E1 = e1 - 127
> > E2 = e2 - 127
> > E1 - E2 = (e1 - 127 )- (e2 - 127)
> > E1 - E2 = e1 - 127 - e2 + 127
> > = e1 - e2 
> > Please correct me if I am wrong
> 
> No, no, no, the default equation for the exponent
> is:
> exp_result = expa - expb + 127;
> 
Why? can you explain more?

> However, there are a lot of exceptions. Read a good
> book.
> 
> >> 2) Why do you add 3 bits at the beginning of the
> >> exponents ?
> >> 3) Why do you add 8 bits at the end of both
> >> fractions ?
> > 
> > I am doing the conversion to the extended format
> which
> > is 32 bit for fraction and 11 bit for exp
> 
> That is wrong. There is no need to convert to
> extended format.
> 
I thought that the standard mentioned that, and we
discussed it on the mailing list and no one said that
I should not do it!!!

> >> 4) The reason you think that the '/' operator
> does
> >> not work
> >> is because you do not understand how to recover
> >> precision. That
> >> is done by shifting opa 23 bits to the left.
> > 
> > Could you explaine more because I thought when I
> added
> > the 8 bit to the right I am doing shift to the
> left!!
> 
> Yes, I understand that, but it's still wrong. You
> want to
> add 23 bit to opa, and none to opb.
> 
why? 

> >> 5) You do not handle de-normalized numbers at
> all.
> >> 
> > 
> > How did you get this conclusion?
> 
> Looking at your code.
> 
> rudi
> 
Anyhow is there is real bugs they could be fixed from
the begining I posted my code lot of times before but
I did not get any comment? where is the cooperation?


Jamil Khatib

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