bash process substitution while loop ♥ George Opritescu

I recently stumbled upon a problem in a bash script. I was doing something like this:

cat $1 | while read -r line; do
  line_with_new_value=$line
  echo $line_with_new_value | grep -vE '^#|^$' | grep -o -E '\${.*?}' | while read -r dollar_var; do
    # do something with line_with_new_value here
    #
  done
  echo $line_with_new_value
done

Basically, when I was echoing, after the while, I was expecting the modified value to be shown. However, piping into something will create a new subshell, and thus the modification is only visible inside that shell. This link gives some useful examples on process substitution. The fix was to rewrite that loop like this:

cat $1 | while read -r line; do
  line_with_new_value=$line
  while read -r interpolation; do
    # modify line_with_new_value here
  done < <(echo $interpolated_line | grep -vE '^#|^$' | grep -o -E '\${.*?}')
  echo $line_with_new_value
done