From the nimble fingers of Phil Yates (phil.yates@...)
(5/11/2000 12:12 PM) came...

> I'm mucking around with encryption. Can anyone tell me what's the largest
> integer you can work with in FB3 / PPC ? Every digit has to be significant,
> and there can't be any rounding.
> 
> TIA,
> 
> Phil.

This revives shades of the 6^6^6 question that hovered on so many lips some
time ago, does it not?

If you're headed toward 128 bit 'strong' (whatever that might mean when it's
at home...) encryption, the number is 128÷4 = 32 hex digits = 16 integers =
8 long integers. FB^3 won't do the math, of course. If you're headed toward
anything over a single long, you're in for the long haul of inventing your
own maths unit.

I'll bet some of the Wise Ones out there will be able to tell us how to
extract a 'carry' bit out of the micro, which would simplify matters a great
deal. for example:

dim a as unsigned long
dim b as unsigned long
dim longval$

a = &Hffffffff
b = 2

print a,b

print a * b

The '*' operation will overflow the long integer, but does it anyway, with
out any error shown. Why?

My old method for 500 (or 50,000) digit math does the whole business as
strings - pictures of numbers, in effect - but it isn't at all cool. On the
other hand, we do all sorts of math in assembly on little 4 and 8 bit
micros, and do it the long way; access to carry and half carry bits is taken
for granted, and n digit math is fairly straightforward.

Maybe what you need is a $1.50, 8 bit coprocessor...
-- 
Ted Spencer; ted@...
-- 
Who, after wine, talks of war's hardships or of poverty? [Horace]