How to access command line arguments of the caller inside a function?

Question

I\'m attempting to write a function in bash that will access the scripts command line arguments, but they are replaced with the positional arguments to the function. Is ther

Answer 1

You can use the shift keyword (operator?) to iterate through them. Example:

#!/bin/bash
function print()
{
    while [ $# -gt 0 ]
    do
        echo $1;
        shift 1;
    done
}
print $*;

Answer 2

If you want to have your arguments C style (array of arguments + number of arguments) you can use $@ and $#.

$# gives you the number of arguments.
$@ gives you all arguments. You can turn this into an array by args=("$@").

So for example:

args=("$@")
echo $# arguments passed
echo ${args[0]} ${args[1]} ${args[2]}

Note that here ${args[0]} actually is the 1st argument and not the name of your script.