Fw: [Cst-1b] numerical analysis 98/3/10

[Kevin Chan]

----- Original Message ----- 
From: Kevin Chan <kkfc2@cam.ac.uk>
To: <spj23@cam.ac.uk>
Sent: Tuesday, May 30, 2000 11:18 AM
Subject: Re: [Cst-1b] numerical analysis 98/3/10


If you do it this way then the answer overall will be 0.125 x 2^-31

This is because although the "hidden bit" is excluded from the
representation of the significand it is not ignored when multiplied by the
exponent
 
Not sure if that makes much sense to you but according to your
representation we will get

1 x 2^-34  =  .125^-31 which is not the answer we are looking for.

Does that make any sense to you??
 
 
 Kev
 
 
> 
> ----- Original Message -----
> From: <spj23@cam.ac.uk>
> To: CST 1b Revision List <cst-1b@srcf.ucam.org>
> Sent: Tuesday, May 30, 2000 11:27 AM
> Subject: Re: [Cst-1b] numerical analysis 98/3/10
> 
> 
> > hi
> >
> > i was just wondering about the answer to the first one - this is how i
> > thought you did it...
> >
> > it's a normalised number as it's 1.x so we don't worry about
> > the 1 as it's taken care of in the hidden bit.
> >
> > 0.125 is an eighth, ie 2^-3, so in total we get 2^(-3-31) = 2^(-34)
> > we store exponents as e + eMAX, so store the exponent as -34+127 = 93
> > so _I_ THINK the answer is
> > 0 for the sign bit
> > 93 in binary for the exponent
> > 0 for the significand
> >
> > ??? thanks
> >
> >
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>