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10-19-2003
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pass by reference in shell
Hi,
I would like to write a function which takes one integer argument and adds 10 to it. But, I'd like to return the result in the data that was passed in. Is this possible? function1() { 1=$(($1+10)) } my_number=1 function "$my_number" echo $my_number I'd like the echo statement to print 11 in this case. Is that the right syntax? I don't have a shell to test with right now... I appreciate your help. Thanks. |
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