Indirect parameter substitution in shell script

I'm having a problem with a shell script (POSIX shell under HP-UX, FWIW). I have a function called print_arg into which I'm passing the name of a parameter as $1. Given the name of the parameter, I then want to print the name and the value of that parameter. However, I keep getting an error. Here's an example of what I'm trying to do:

#!/usr/bin/sh

function print_arg
  {
  # $1 holds the name of the argument to be shown

  arg=$1

  # The following line errors off with
  #   ./test_print.sh[9]: argval=${"$arg"}: The specified substitution is not valid for this command.

  argval=${"$arg"}

  if [[ $argval != '' ]] ; then
    printf "ftp_func: $arg='$argval'\n"
  fi
  }

COMMAND="XYZ"

print_arg "COMMAND"

I've tried re-writing the offending line every way I can think of. I've consulted the local oracles. I've checked the online "BASH Scripting Guide". And I sharpened up the ol' wavy-bladed knife and scrubbed the altar until it gleamed, but then I discovered that our local supply of virgins has been cut down to, like, nothin'. Drat!

Any advice regarding how to get the value of a parameter whose name is passed into a function as a parameter will be received appreciatively.

Solution

You could use eval, though using direct indirection as suggested by SiegeX is probably nicer if you can use bash.

#!/bin/sh

foo=bar
print_arg () {
    arg=$1
    eval argval=\"\$$arg\"
    echo "$argval"
}
print_arg foo