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Old 01-03-2006, 05:05 AM   #1
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insmod command variables in a script

What does the "$*" and "|| exit 1" in the command:

insmod ./$module.o $* || exit 1

mean? Google-ing for "linux $*" or "insmod $*" doesn't seem to uncover any answers. This is part of a script to load a module in 2.4.


Last edited by fpb; 01-03-2006 at 05:15 AM.
Old 01-03-2006, 05:33 AM   #2
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$* are the positional parameters (or command line arguments) to the script.
example: script arg1 arg2
arg1 and arg2 get put in $*

|| is the 'or' operator. If the insmod fails, the script exits with a code of 1, which is the standard exit code for failure. If the insmod succeeds the exit 1 is skipped.
Old 01-03-2006, 07:29 PM   #3
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Cool. That was easy to understand & very helpful.

Originally I'd thought it was the Makefile special macro $* that is the prefix shared by target & dependent files, but that doesn't make sense in the script of course.

Thanks a lot vls! Happy New Year!


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