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From 吴耀华 <wuyao...@38do.com.cn>
Subject one question
Date Mon, 20 Mar 2006 06:30:52 GMT
Advanced Issues: Escaping and !
When a reference is silenced with the ! character and the ! character preceded by an \ escape
character, the reference is handled in a special way. Note the differences between regular
escaping, and the special case where \ precedes ! follows it: 

#set( $foo = "bar" ) 
$\!foo 
$\!{foo} 
$\\!foo 
$\\\!foo 
This renders as: 

$!foo 
$!{foo} 
$\!foo 
$\\!foo 

So that I can assume if i get a clause like:
$\\\\\!foo
that will render as:
$\\\\!foo
is it right? just lose one backslash?




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