bash passing positional parameters in function

Question

I have a small problem, please look next code and output, and could you tell me what I miss. #!/bin/bash

foo () {

echo "Parameter #1 is $1" 

}

foo

my output: [user@host]$ ./test.scr test

Parameter #1 is

Etan Reisner · Accepted Answer · 2015-04-14 17:41:08Z

1

functions get their own positional parameters from when they are called.

If you need to pass the scripts arguments to the function use foo "$@" (the quotes are important do not forget them).

answered Apr 14, 2015 at 17:41

Etan Reisner

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