Re: division by 0

[Bernard Urban]

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