How do I access arguments to functions if there are more than 9 arguments?

Use :

#!/bin/bashecho ${10}

To test the difference with $10, code in foo.sh :

#!/bin/bashecho $10echo ${10}

Then :

$ ./foo.sh first 2 3 4 5 6 7 8 9 10first010

the same thing is true if you have :

foobar=42foo=FOOecho $foobar # echoes 42echo ${foo}bar # echoes FOObar

Use {} when you want to remove ambiguities ...

my2c

If you are using bash, then you can use ${10}.

${...} syntax seems to be POSIX-compliant in this particular case, but it might be preferable to use the command shift like that :

while [ "$*" != "" ]; do  echo "Arg: $1"  shiftdone

EDIT: I noticed I didn't explain what shift does. It just shift the arguments of the script (or function). Example:

> cat script.shecho "$1"shiftecho "$1"> ./script.sh "first arg" "second arg"first argsecond arg

In case it can help, here is an example with getopt/shift :

while getopts a:bc OPT; do case "$OPT" in  'a')   ADD=1   ADD_OPT="$OPTARG"   ;;  'b')   BULK=1   ;;  'c')   CHECK=1   ;; esacdoneshift $( expr $OPTIND - 1 )FILE="$1"

In general, to be safe that the whole of a given string is used for the variable name when Bash is interpreting the code, you need to enclose it in braces: ${10}