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 $*;