From: Alan Moore <alan_moore@...>
Date: 2001-10-08T00:58:02+09:00
Subject: [ruby-talk:22198] Re: Match expressions

Okay, thanks for clearing that up.

--Alan

matz wrote:
>Hi,
>
>In message "[ruby-talk:22163] Re: Match expressions"
>    on 01/10/07, Alan Moore <alan_moore@gmx.net> writes:
>
>|So, given the expression "node1 =~ node2", if node1 is a literal 
>|regex (i.e., /xyz/ or %{xyz}), Ruby translates the expression to  
>|"node1.=~(node2)".  If node2 is a literal regex, and node1 evaluates 
>|to a string, Ruby does a "node2.=~(node1)".  In both cases, it calls 
>|the rb_reg_match function directly, as a speed optimization.  But if 
>|node1 doesn't evaluate to a string, Ruby turns it around again and 
>|does a "node1.=~(node2)", under the assumption that node1 is a user-
>|defined object that defines its own "=~" method.
>|
>|Is that right?  Is that what you meant, Guy?
>
>Yes.  And it's just an ugly hack.  I don't think you have to follow.
>
>                                                        matz.