OpenEuphoria

Re: bug in remainder ?

• Posted by Robert Craig <rds at ATTCANADA.NET> Feb 03, 2000
• 414 views

Daniel Johnson writes:
> ? remainder(1, .1)
> 0.1
> ? remainder(1, .01)
> 0.01
> ? remainder(10, 1)
> 0 (as he corrected himself later)

The last case is obviously correct.
The remainder of 10 divided by 1 is 0.

The first two look wrong, but that's because we
humans can represent numbers like .1 exactly in our brains,
whereas Intel 64-bit floating-point has no bit pattern that
corresponds exactly to .1 (or .01), so it uses something
that's "very close to .1". The machine must figure that
10 is 9 times "very close to .1", plus "very close to .1".
So the remainder must be "very close to .1", which looks like
0.1 when you print it, but isn't exactly 0.1.

In the floating point case I simply call a C library routine to
compute the result, so blame it all on C and Intel.  

The moral is: don't expect perfect accuracy
in floating-point calculations.

if 1 = .1+.1+.1+.1+.1+.1+.1+.1+.1+.1 then
    puts(1, "perfect floating-point\n")
else
    puts(1, "fuzzy floating-point\n")
end if

Regards,
     Rob Craig
     Rapid Deployment Software
     http://www.RapidEuphoria.com