1. brain gymnastics
Check out this perfectly legal (promise!) Euphoria
program, what value do you suppose will be printed?
integer i
i = 1
i = -i
i /= i+1
+i = +i
i *= i-1
-i = -i
? i
2. Re: brain gymnastics
This program sure behaves strangely on my system.
It should give divide by zero error at i /= i+1 but it doesn't - but it will
if you remove the line following.
Huh?
Bye
Martin
----- Original Message -----
From: Tor Gausen <tor.gausen at C2I.NET>
To: <EUPHORIA at LISTSERV.MUOHIO.EDU>
Sent: Thursday, June 01, 2000 1:38 AM
Subject: brain gymnastics
> Check out this perfectly legal (promise!) Euphoria
> program, what value do you suppose will be printed?
>
> integer i
>
> i = 1
>
> i = -i
>
> i /= i+1
>
> +i = +i
>
> i *= i-1
>
> -i = -i
>
> ? i
>
3. Re: brain gymnastics
And it will also give divide by zero error *at the final print i* if you
just insert a print i right after i /= i+1 !?!
Dan Moyer
-----Original Message-----
>This program sure behaves strangely on my system.
>It should give divide by zero error at i /= i+1 but it doesn't - but it
will
>if you remove the line following.
>Huh?
>
>Bye
>Martin
>
>----- Original Message -----
>From: Tor Gausen <tor.gausen at C2I.NET>
>To: <EUPHORIA at LISTSERV.MUOHIO.EDU>
>Sent: Thursday, June 01, 2000 1:38 AM
>Subject: brain gymnastics
>
>
>> Check out this perfectly legal (promise!) Euphoria
>> program, what value do you suppose will be printed?
>>
>> integer i
>>
>> i = 1
>>
>> i = -i
>>
>> i /= i+1
>>
>> +i = +i
>>
>> i *= i-1
>>
>> -i = -i
>>
>> ? i
>>
4. Re: brain gymnastics
This program is easier to understand if you format it like this:
integer i
i = 1
i = -i
i /= i+1+(i = +i) -- this statement originally on 2 lines
i *= i-1-(i = -i) -- this statement originally on 2 lines
? i
Colin Taylor
----- Original Message -----
From: Tor Gausen <tor.gausen at C2I.NET>
To: <EUPHORIA at LISTSERV.MUOHIO.EDU>
Sent: Thursday, June 01, 2000 3:38 AM
Subject: brain gymnastics
> Check out this perfectly legal (promise!) Euphoria
> program, what value do you suppose will be printed?
>
> integer i
>
> i = 1
>
> i = -i
>
> i /= i+1
>
> +i = +i
>
> i *= i-1
>
> -i = -i
>
> ? i
>
5. Re: brain gymnastics
On Fri, 2 Jun 2000 00:01:53 -0400, Tor Gausen
<tor.gausen at C2I.NET> wrote:
> Check out this perfectly legal (promise!) Euphoria
> program, what value do you suppose will be printed?
> integer i
> i = 1
> i = -i
> i /= i+1
You should get an error here, at run time - you're dividing by
zero.
> +i = +i
> i *= i-1
> -i = -i
> ? i
--
Jeff Zeitlin
jzeitlin at cyburban.com
6. Re: brain gymnastics
Whoops - Jeff is just a little red-faced, having forgotten that a
line break is _not_ a statement separator in Euphoria...
--
Jeff Zeitlin
jzeitlin at cyburban.com
7. Re: brain gymnastics
> Whoops - Jeff is just a little red-faced, having forgotten that a
> line break is _not_ a statement separator in Euphoria...
> i /= i+1
You should get an error here, at run time - you're dividing by
zero.
> +i = +i
Now I understand!
This shouldnt be like this!?
8. Re: brain gymnastics
> > integer i
> > i = 1
> > i = -i
> > i /= i+1 [note a]
> > +i = +i [note b]
> > i *= i-1 [note c]
> > -i = -i
> > ? i
integer i
i = 1
i = -1
i = i / ( i + 1 + ( i = +i ) )
i = i * ( ( i -1 ) - (i = -i) )
? i
Does precize the same.
notice: a sign is first considered to be part of a calculating (i.e.
statement contiuator: 3 + 4; 5 - 7) secondly part of the number (i.e.
sign: -3; + 4) ..
notice: self reflecting calculations (such as *=, /=, +=, -=) convert to
this form:
[left] [*/+-]= [right]
[left] = [left] [*/+-] ( [right] )
Everything on the right is used on the calculation.
So:
i /= i -1
+i = +i
Converts to:
i = i / ( i -1 + (i = +i) )
And will therefor not crash.
More cynically explained: euphoria can be just as messy as C.
Ralf N.
9. Re: brain gymnastics
On Sun, 4 Jun 2000 15:10:33 +0200, Ralf Nieuwenhuijsen <nieuwen at XS4ALL.NL>
wrote:
>> > integer i
>> > i =3D 1
>> > i =3D -i
>> > i /=3D i+1 [note a]
>> > +i =3D +i [note b]
>> > i *=3D i-1 [note c]
>> > -i =3D -i
>> > ? i
>
>integer i
>i =3D 1
>i =3D -1
>i =3D i / ( i + 1 + ( i =3D +i ) )
>i =3D i * ( ( i -1 ) - (i =3D -i) )
>? i
>
>Does precize the same.
Run the two programs as they stand, Ralf. Yours returns '2'. The first
returns '0'... 
I made the same mistake right after I'd figured out the linebreak trick...
I've noticed those who have tried to solve Tor's problem have all been
doing the same thing:
There's a tendency to see (things like):
a *=3D a + b + c =3D d
-- Uh-oh.
...as:
a =3D a * (a + b + (c =3D d))
-- Multiply 'a' by (a + b) if c !=3D d, and (a + b + 1) if c =3D d.
...when we should be reading it as:
a =3D a * ((a + b + c) =3D d)
-- Make 'a' zero if (a + b + c) =3D d, leave it alone otherwise.
Yet in situations like this:
if a + b + c =3D d then...
...we know exactly what the expression means!
I reckon that this is some kind of "codical" illusion, and also a darn good
argument for using parentheses to make expressions more readable.
Carl
--
Insert clich=E9 about inserting witty comment here, here.
