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Re: division by 0
From: |
Bernard Urban |
Subject: |
Re: division by 0 |
Date: |
29 Mar 2004 11:09:20 +0200 |
User-agent: |
Gnus/5.09 (Gnus v5.9.0) Emacs/21.2 |
Marius Vollmer <address@hidden> writes:
> Bernard Urban <address@hidden> writes:
>
> > Debian woody on i386.
> >
> > $ guile
> > guile> (version)
> > "1.6.4"
> > guile> (/ 0)
> > +#.#
> > guile> (/ 1.0 0)
> > +#.#
> > guile> (/ 1 0.0)
> > +#.#
> > guile>(/ 1 0)
> > standard input:3:1: In procedure / in expression (/ 1 0):
> > standard input:3:1: Numerical overflow
> > ABORT: (numerical-overflow)
> >
> > Type "(backtrace)" to get more information or "(debug)" to enter the
> > debugger.
> > guile>
> >
> > Problem happens in numbers.c, function scm_divide(), where the test
> > #line 3274 should not be made.
>
> The 1.7 series should be handling this more correctly. From NEWS:
>
> ** There is support for Infinity and NaNs.
>
> Following PLT Scheme, Guile can now work with infinite numbers, and
> 'not-a-numbers'.
>
> There is new syntax for numbers: "+inf.0" (infinity), "-inf.0"
> (negative infinity), "+nan.0" (not-a-number), and "-nan.0" (same as
> "+nan.0"). These numbers are inexact and have no exact counterpart.
>
> Dividing by an inexact zero returns +inf.0 or -inf.0, depending on the
> sign of the dividend. The infinities are integers, and they answer #t
> for both 'even?' and 'odd?'. The +nan.0 value is not an integer and is
> not '=' to itself, but '+nan.0' is 'eqv?' to itself.
>
> For example
>
> (/ 1 0.0)
> => +inf.0
>
> (/ 0 0.0)
> => +nan.0
>
> (/ 0)
> ERROR: Numerical overflow
Is (/ 1 x) always equal to (/ x) in 1.7 ?
This is actually my problem. It originates in the fact that hobbit
converts (/ x) to (/ 1 x), and for x = 0, it fails for 1.6.
Why would I want to divide by 0 ? To obtain... nan !
In the interpreter, you can have:
(define nan (- (/ 0) (/ 0)))
For hobbit, you must do:
(define nan (eval '(- (/ 0) (/ 0)) (interaction-environment)))
>
> Two new predicates 'inf?' and 'nan?' can be used to test for the
> special values.
--
Bernard Urban