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

In bash (but not in other sh implementations), indirection is done by: ${!arg}

Input

foo=bar
bar=baz

echo $foo
echo ${!foo}

Output

bar
baz