title: How to pass arguments to a script invoked by source command?
body: <p>I am invoking a script through <code>source</code> command and want to pass arguments to the script.</p>
  
  <p>I have checked <code>man source</code>, the bash returns:</p>
  
  <blockquote>
    <p>: [arguments]<br>
                  No effect; the command does nothing beyond expanding arguments and performing any  specified redirections.  A zero exit code is returned.  </p>
    
    <p>source filename [arguments]<br>
                  Read  and  execute commands from filename in the current shell   environment and return the exit status of the last command executed from     filename.  If filename does not contain a slash, file names  in  PATH  are used   to find the directory containing filename.  The file searched for in PATH need   not be executable.  When bash is  not  in  posix  mode,  the  current  directory    is searched  if  no file is found in PATH.  If the sourcepath option to the shopt   builtin command is turned off, the PATH is not searched.  If any arguments are   supplied, they become the positional  parameters  when  filename  is  executed.     Otherwise  the  positional  parameters are unchanged.  The return status is the   status of the last command exited within the script (0 if no commands are executed), and false if filename is not found or cannot be read.</p>
  </blockquote>
  
  <p>It has no examples, so I don't understand it .</p>