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тАО09-07-2004 02:02 AM
тАО09-07-2004 02:02 AM
i.e. :
"'onintr -;T0 path $planpath \!*;T1 start \!* $t2;onintr "
T0, T1, and t2 all are assigned value in intial declaration..
what I am interested to know is what does these commands do and especially
what does the "!*" serve as.. ?
Solved! Go to Solution.
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тАО09-07-2004 02:12 AM
тАО09-07-2004 02:12 AM
Re: what does "!*" do in shell script?
In bash it gives the arguments of the last executed command.
manish
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тАО09-07-2004 02:20 AM
тАО09-07-2004 02:20 AM
Re: what does "!*" do in shell script?
thank you all for your advise!
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тАО09-07-2004 02:22 AM
тАО09-07-2004 02:22 AM
Re: what does "!*" do in shell script?
If - is specified, all interrupts are ignored.
The end of the line reset the interrupts-command
Regards
Michael
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тАО09-07-2004 02:56 AM
тАО09-07-2004 02:56 AM
SolutionThe code you have turns off interupts, runs the program specified by T0 passing parameter "path $planpath" followed by the arguments from the command line that launched the script, then the program specified by T1 is run passing argument "start" followed by the command line arguments again, followed by "$t2".
The onintr at the end resets the "on interupt flag".
HTH
-- Rod Hills
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тАО09-07-2004 08:26 PM
тАО09-07-2004 08:26 PM