Indirect parameter substitution in shell script

Question

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

Answer 1

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