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

Question

With first 9 arguments being referred from $1-$9, $10 gets interpreted as $1 followed by a 0. How do I account for this and access arguments to functions greater than 10?

Thanks.

Answer 1

Use :

#!/bin/bash
echo ${10}

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

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

Then :

$ ./foo.sh first 2 3 4 5 6 7 8 9 10
first0
10

the same thing is true if you have :

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

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

my2c

Answer 2

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"
  shift
done

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

> cat script.sh
echo "$1"
shift
echo "$1"

> ./script.sh "first arg" "second arg"
first arg
second 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
   ;;
 esac
done
shift $( expr $OPTIND - 1 )
FILE="$1"

Answer 3

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}