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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