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

My reading of the bash ref manual says this stuff is captured in BASH_ARGV, although it talks about "the stack" a lot.

#!/bin/bash

function argv {
    for a in ${BASH_ARGV[*]} ; do
      echo -n "$a "
    done
    echo
}

function f {
    echo f $1 $2 $3
    echo -n f ; argv
}

function g {
    echo g $1 $2 $3
    echo -n g; argv
    f
}

f boo bar baz
g goo gar gaz

Save in f.sh

$ ./f.sh arg0 arg1 arg2
f boo bar baz
farg2 arg1 arg0 
g goo gar gaz
garg2 arg1 arg0 
f
farg2 arg1 arg0