title: How to wrap another shell still passing $OPTIND as-is?
body: <p>I'm trying to wrap a bash script b with a script a. 
  However I want to pass the options passed to a also to b as they are. </p>
  
  <pre><code>#!/bin/bash
  
  # script a
  ./b ${@:$OPTIND}
  </code></pre>
  
  <p>This will also print $1 (if any). What's the simplest way not to?</p>
  
  <p>So calling:</p>
  
  <pre><code>./a -c -d 5 first-arg
  </code></pre>
  
  <p>I want b to execute:
  ./b -c -d 5 # WITHOUT first-arg</p>