[Tutor] Need a better name for this function

[Richard D. Moores]

On Wed, Dec 16, 2009 at 20:23, Dave Angel <davea at ieee.org> wrote:
> Richard D. Moores wrote:

> There are conceivably better ways to get at the mantissa of the fp number,
> but you can simply parse the hex digits as I did manually, and add one and
> subtract one from the given mantissa  (the part between the decimal point
> and the 'p').  Then it just remains to figure out which two of those three
> values actually span the desired value.
>
> Using the numbers and strings you supply, the three values would be:
>
> 0x1.bd70a3d70a3d6p-2
> 0x1.bd70a3d70a3d7p-2
> 0x1.bd70a3d70a3d8p-2
>
>
> and the second one is somewhat less than .435, while the 3rd is more.
>
> Now, while this is good enough to do by hand, you have to realize there are
> some places it may not work, and another where it won't.

Dave,

I was hoping to find a way to NOT do it by hand, for the simple cases
such as 0x1.bd70a3d70a3d7p-2 .  I'm weak on hex arithmetic. For these
simple cases, is there a general way to "add" something to the last
digit of a hex value to bump it up and down by 1? After I can do that,
I'll try to deal with the cases you mention below.

Dick

> I'm not sure whether trailing zeroes are always provided in the hex()
> method.  So you may have to pad it out before adjusting the last digit.  I'm
> also not positive how it normalizes the hex value printed out.  As the
> example in the help indicates, there are more than one hex string that can
> be converted to the same float - if you denormalize, it'll still convert it.
>  I just don't know if hex() ever produces a non-normalized value.
>
> More drastically, if you're trying to be complete, is the infinities, the
> NAN values, and the numbers, *very* close to zero, where gradual underflow
> takes effect.  The notion here is that when the exponent gets as negative as
> it can safely store, the standard requires that the number be stored
> denormalized, rather than going immediately to zero.  I don't know how those
> values are represented by the hex() method, but they have fewer significant
> digits, so the adjustment you would need would be in a different column.
>
>
> If I had to write a function to do what you ask, and if I couldn't get
> better specs as to what Python is actually doing with the hex() method, I'd
> work out an algorithm where tweak what seems to be the bottom digit, then
> see if the float has a new value.  If it does not, I'm working on the wrong
> column.
>
> DaveA
>
>