Indirectly expand variables in shell

Question

I need to indirectly reference a variable in the bash shell.

I basically want to what you can do in make by writing $($(var)).

I have tried using ${$var} which would be the most straight forward solution in bash but then I get this error:

bash: ${$var}: bad substitution

Is there a way to do this?

What I am trying to do is to iterate over all the arguments ($1, $2, $3, ...) to a program using an iteration variable and I cannot do this without indirection.

Answer 1

If you have var1=foo and foo=bar, you can get bar by saying ${!var1}. However, if you want to iterate over the positional parameters, it's almost certainly better to do

for i in "$@"; do
    # something
done

Answer 2

Using /bin/bash:

foo=bar
test=foo
echo ${!test}

# prints -> bar

Answer 3

Using grep + awk

foo=bar
var_name=foo
set | grep -v var_name | grep ^"${var_name}=" | awk -F= '{ print $2 }'

the output will be

bar

To prove that this approach works, check the bash script:

#!/bin/bash

function show_variable_value() {
  name=$1
  value=$(set | grep -v name | grep ^"${name}="  | awk -F= '{ print $2 }' )
  echo "${value}"
}


foo='bar'

# getting the variable value indirectly from the variable name
value=$(show_variable_value foo)
if [ "${value}" == "${foo}" ]; then
  echo "The function returned indirectly the variable value from the variable name >>> '${value}' == '${foo}'"
else
  echo "This approach does not work"
fi