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

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.

#!/usr/bin/env bash

echo name of script is $0
echo first argument is $1
echo second argument is $2
echo seventeenth argument is $17
echo number of arguments is $#

Edit: please see my comment on question

Ravi's comment is essentially the answer. Functions take their own arguments. If you want them to be the same as the command-line arguments, you must pass them in. Otherwise, you're clearly calling a function without arguments.

That said, you could if you like store the command-line arguments in a global array to use within other functions:

my_function() {
    echo "stored arguments:"
    for arg in "${commandline_args[@]}"; do
        echo "    $arg"
    done
}

commandline_args=("$@")

my_function

You have to access the command-line arguments through the commandline_args variable, not $@, $1, $2, etc., but they're available. I'm unaware of any way to assign directly to the argument array, but if someone knows one, please enlighten me!

Also, note the way I've used and quoted $@ - this is how you ensure special characters (whitespace) don't get mucked up.

# Save the script arguments
SCRIPT_NAME=$0
ARG_1=$1
ARGS_ALL=$*

function stuff {
  # use script args via the variables you saved
  # or the function args via $
  echo $0 $*
} 


# Call the function with arguments
stuff 1 2 3 4

One can do it like this as well

#!/bin/bash
# script_name function_test.sh
function argument(){
for i in $@;do
    echo $i
done;
}
argument $@

Now call your script like

./function_test.sh argument1 argument2

This is @mcarifio response with several comments incorporated:

#!/bin/bash

shopt -s extdebug

function stuff() {
  local argIndex="${#BASH_ARGV[@]}"
  while [[ argIndex -gt 0 ]] ; do
    argIndex=$((argIndex - 1))
    echo -n "${BASH_ARGV[$argIndex]} "
  done
  echo
}

stuff

I want to highlight:

• The shopt -s extdebug is important. Without this the BASH_ARGV array will be empty unless you use it in top level part of the script (it means outside of the stuff function). Details here: Why does the variable BASH_ARGV have a different value in a function, depending on whether it is used before calling the function
• BASH_ARGV is a stack so arguments are stored there in backward order. That's the reason why I decrement the index inside loop so we get arguments in the right order.
• Double quotes around the ${BASH_ARGV[@]} and the @ as an index instead of * are needed so arguments with spaces are handled properly. Details here: bash arrays - what is difference between ${#array_name[*]} and ${#array_name[@]}

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

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

The simplest and likely the best way to get arguments passed from the command line to a particular function is to include the arguments directly in the function call.

# first you define your function 
function func_ImportantPrints() {
    printf '%s\n' "$1" 
    printf '%s\n' "$2" 
    printf '%s\n' "$3" 
} 
# then when you make your function call you do this:  
func_ImportantPrints "$@" 

This is useful no matter if you are sending the arguments to main or some function like func_parseArguments (a function containing a case statement as seen in previous examples) or any function in the script.

I do it like this:

#! /bin/bash

ORIGARGS="$@"

function init(){   
  ORIGOPT= "- $ORIGARGS -" # tacs are for sed -E 
  echo "$ORIGOPT"
}