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argument hysteresis?

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# 1  
Old 07-09-2008
argument hysteresis?
Hi there,

When I try to execute the following simple script "simple_cript.ksh":
PHP Code:
NB_ARG=$#
ARGS=$@ 
the number of arguments and the arguments themselves seem to be remembered. If I run it a first time with argument "arg1", and a second time without any arguments, it still displays "arg1" at the second run.

How can I reset the arguments? please help...
# 2  
Old 07-09-2008
I cannot reproduce this behavior:

Code:
zsh-4.3.4% cat simple_script.ksh
printf "%d\n" $#
printf "%s " "$@"
printf "\n"
zsh-4.3.4% ./simple_script.ksh arg1
1
arg1
zsh-4.3.4% ./simple_script.ksh
0

Could you post the entire test case?
# 3  
Old 07-09-2008
here it is, my file "arguments_try.ksh":

PHP Code:
#!/bin/ksh
# my shell arguments
NB_ARG=$#
ARGS=$@
echo nb arguments\: $NB_ARGS; echo
echo arguments\: $ARGS; echo 
commands in the terminal:
PHP Code:
st25219@mx73:ian$ arguments_try.ksh arg1
st25219@mx73:ian$ arguments_try.ksh 
and the screen output:

nb arguments: 1
arguments: arg1
**
nb arguments: 1
arguments: arg1


ps: the only way for me to modify the "arg1" in the screen output is to give another argument list ...
Last edited by MarkZWEERS; 07-09-2008 at 12:36 PM.. Reason: ps
# 4  
Old 07-09-2008
I don't get the same output, perhaps someone else can try to reproduce it:


Code:
$ arguments_try.ksh arg1
nb arguments: 1

arguments: arg1

$ arguments_try.ksh
nb arguments: 0

arguments:

$ cat arguments_try.ksh
#!/bin/ksh
# my shell arguments
NB_ARG=$#
ARGS=$@
echo nb arguments: $NB_ARG; echo
echo arguments: $ARGS; echo

The variable name $NB_ARGS is misspelled.
# 5  
Old 07-09-2008
that's a copy-paste error...
I'll try the script at home on my linux machine as well. Thanks anyways!
# 6  
Old 07-09-2008
Yes, it does do the same thing on my system (Debian Etch).

This is the entire script:
PHP Code:
#!/bin/ksh
#
# try-outs for script arguments from command line
#

echo You\'ve typed $# arguments, being\:
echo $*

echo
echo

echo You\'ve launched the script $0 with $# arguments, being\:
ARGS=$@
for arg_i in $ARGS
do
  set -- $arg_i
  echo $arg_i
done 
and this is the screen output:
PHP Code:
mark@zweers:~/progs/scripts$ . arguments.ksh 
You've typed 0 arguments, being:



You've launched the script bash with 0 arguments, being:
mark@zweers:~/progs/scripts$ . arguments.ksh tmp1
You've typed 1 arguments, being:
tmp1


You've launched the script bash with 1 arguments, being:
tmp1
mark@zweers:~/progs/scripts$ . arguments.ksh
You've typed 1 arguments, being:
tmp1

You've launched the script bash with 1 arguments, being:
tmp1 
Last edited by MarkZWEERS; 07-09-2008 at 03:46 PM.. Reason: correction 'php' tags
# 7  
Old 07-09-2008
This script is different ...
You're resetting the arguments in the for loop (why?):

Code:
$ . arguments.ksh
Youve typed 0 arguments, being:



Youve launched the script bash with 0 arguments, being:
$ . arguments.ksh 1
Youve typed 1 arguments, being:
1


Youve launched the script bash with 1 arguments, being:
1
$ . arguments.ksh 1 2
Youve typed 2 arguments, being:
1 2


Youve launched the script bash with 2 arguments, being:
1
2
$ . arguments.ksh
Youve typed 1 arguments, being:
2


Youve launched the script bash with 1 arguments, being:
2

In the end of the for loop your're setting $1 to the last argument.

Is this a joke?

Try commenting the set statement and rerun the script ...
Last edited by radoulov; 07-09-2008 at 04:56 PM.. Reason: corrected
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