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-ne 0 doesn't work -le does

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# 1  
-ne 0 doesn't work -le does


I am using korn shell.

until [[ $# -ne 0 ]]
echo "\$# = " $#
echo "$1"

To the above script, I passed 2 parameters and the program control doesn't enter inside "until" loop. If I change it to until [[ $# -le 0 ]] then it does work.

Why numeric comparison is not working with -ne and works with -le?

thank you

Last edited by Scrutinizer; 06-04-2015 at 01:42 PM.. Reason: code tags
# 2  
-ne means not equal to
-le means less than or equal to

If you run the script and do not pass any parameters, it will enter the loop as it currently it is waiting 'until' $# is not equal to 0, which it will be set to if you do not pass any parameters (it will then fail on the shift command).

If you run the script and do pass some parameters, it will not enter the loop because $# is already not equal to 0, so it has nothing to wait 'until' for.

I hope that has made some sense.
# 3  
It is a matter of logic. Until the number of parameters is not equal to 0. If the number of parameters is 2 then it will not enter the loop, since the condition is satisfied ( 2 is unequal to 0 ).
# 4  
You probably want a while loop rather than an until loop:

while [[ $# -ne 0 ]]
  echo "\$# = " $#
  echo "$1"

It can also be written: while (($#))

until (( $# == 0 ))

Last edited by cfajohnson; 06-04-2015 at 06:11 PM..

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