> No, a seg fault in a shell is always a bug. The above bug is fixed in
> the latest ksh version.

My ksh is Version M 1993-12-28 r. 
Has it been fixed in ksh93 s+?

Regards
Chuli

> -----Original Message-----
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Bernd Eggink
> Sent: Tuesday, December 25, 2007 7:49 PM
> To: [email protected]
> Subject: Re: [ast-users]Segment fault in for command
> 
> Jerry Rocteur schrieb:
> > * chuli <[EMAIL PROTECTED]> [2007-12-25 09:42]:
> >> Hello, everyone
> >>
> >>    When I use such script with ksh, it will produced "Segment fault!"
> >> But when I use it with bash, it will output normally. Why does ksh print
> "Segment fault"?
> >> for((;i<10;i++))
> >> do
> >>   echo $i -----> or just use "$i" here
> >> done
> >> "
> >
> > The man page says:
> >
> >   for (( [expr1] ; [expr2] ; [expr3] )) ;do list ;done
> >               The  arithmetic  expression expr1 is evaluated first
> >
> > So you need to initialize your i
> 
> No, you don't. An unitialized variable in an arithmetic expression is
> treated as zero.
> 
> >     (( i=0; i < 10, i++ ))
> >
> > If you do:
> >
> > while (( z < 10 ))
> > do
> >     ...
> >
> > And you don't initialize your z you'll get the same thing!
> >
> > ksh prints "Segment fault" or "Bus error" on Version M 1993-12-28 p on Mac
> OS X because i is not initialized!
> 
> No, a seg fault in a shell is always a bug. The above bug is fixed in
> the latest ksh version.
> 
> Greetings,
> Bernd
> 
> --
> Bernd Eggink
> [EMAIL PROTECTED]
> http://sudrala.de
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