Re: Is exact 0 "stronger" than inexact 0.0?
Marcin 'Qrczak' Kowalczyk 23 Oct 2005 18:17 UTC
Aubrey Jaffer <xxxxxx@alum.mit.edu> writes:
> (* 0 +inf.0) ==> +nan.0
> ...
> (/ 0 0.0) ==> unspecified
> (/ 0.0 0) ==> +nan.0
> (/ 0.0 0.0) ==> +nan.0
>
> Why is only (/ 0 0.0) out of this set unspecified?
I don't know. In my model and implementation:
(* 0 +inf.0) ==> 0
(/ 0 0.0) ==> 0
(/ 0.0 0) ==> error
(/ 0 0) ==> error
In general my recipe for evaluating such operations is as follows:
1. Replace 0.0 with a very small positive number, +inf.0 with a very
large number, -0.0 and -inf.0 analogously, +nan.0 with any number.
0 stays exact.
2. Perform the operation.
3. Try to perform the inverse substitution. If the answer would depend
on the initial choice of substituted numbers:
a) If the result can't decide between 0.0 and -0.0, answer 0.0
(in most rounding modes) or -0.0 (when rounding towards -inf).
b) If the result can't decide between a number and an error,
answer the number (this happens e.g. for (/ 0 +nan.0)).
c) Otherwise answer +nan.0.
This algorithm doesn't explain two-argument atan near zero, which
is treated specially, as per IEEE and other languages (the algorithm
would answer +nan.0).
Disclaimer: I haven't checked whether atan is the only special case,
I formulated the rules just now.
Note that in my model (- 0 x) is (- x), but (- 0.0 x) is different
(assuming the default rounding mode). Also (+ A (* B +i)) is A+Bi;
this would not be true if complex numbers had both real and imaginary
part of the same exactness, because adding 0.0 is not identity at -0.0.
--
__("< Marcin Kowalczyk
\__/ xxxxxx@knm.org.pl
^^ http://qrnik.knm.org.pl/~qrczak/