title: How does IFS work in Bash?
body: <pre><code>#!/bin/bash
  # This question is from advanced bash scripting guide section 5.1
  
  echo
  var="'(]\\{}\$\""
  IFS='\'
  echo $var
  
      # output is '(] {}$"     
      # \ converted to space. Why?
  
  echo "$var"      
  
      # output is '(]\{}$"      
      # special meaning of \ used, \ escapes \ $ and " RIGHT?
  
  echo
  var2="\\\\\""
  echo $var2       
  
      # output is   "         
      # \ converted to space. Why?
  
  echo
  
      # But ... var2="\\\\"" is illegal. Why?
  
  var3='\\\\'
  echo "$var3"     # \\\\
  
      # Strong quoting works, though. Why?
  </code></pre>