On Wed, Dec 16, 2009 at 20:23, Dave Angel <[email protected]> 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 > > _______________________________________________ Tutor maillist - [email protected] To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
