OpenEuphoria

RE: Uninitialized Variables

• Posted by Andy Serpa <renegade at earthling.net> Mar 23, 2002
• 443 views

No difference on my system.  It might also be you are being tripped up 
by an optimization:

atom x,y,nan,inf
inf = 1e300 * 1e300
nan = inf / inf

x = nan

Now equal(x,nan) will be TRUE, probably because Euphoria doesn't 
actually even compare them, since they both will share the same internal 
pointer at this point.

Now do:

y = x + 5   <= y is now also a nan (actually -nan)

? equal(y,nan)  

-- This is FALSE.  Once again, we can't determine a nan this way.
? x
? y

-nan
-nan


Chris Bensler wrote:
> You need to use equal(), instead of =
> 
> ? not equal(324876,NAN)
> 
> 
> Chris
> 
> Andy Serpa wrote:
> > 
> > 
> > > Apparently NAN is (silly me) Not A Number! :P
> > > using equal() compares NAN properly and consistently
> > > 
> > > Here is my revised uninitialized values for variables:
> > > 
> > >    integer  = -INF
> > >    atom     =  INF
> > >    sequence =  NAN
> > >    object   = -NAN
> > > 
> > 
> > So how do I test if something is a nan?  The "official" way is to use 
> > x!=x, but that is usually optimized away by most compilers (& Euphoria, 
> > apparently.)  Using something like if x=1 and x=2 will work in the 
> > interpreter, but not translated to C, even with Watcom.  (In fact, it is 
> > 
> > 
> > different depending on the compiler).
> > 
> > Am I stuck with "if x and compare(x/x,1)"?
> > 
> > For my genetic programming system this is a very real problem, as it 
> > comes up with random mathmatical expressions that sometimes are nan's.  
> > If you then take a predicted value (which is a nan) as output for a 
> > function that it has created and compare it with a target value, it will 
> > 
> > 
> > show as being equal (& therefore error = 0).  So functions with nan's as 
> > 
> > 
> > output get the highest fitness, which is a disaster...
> > 
> >