From: ts <decoux@...>
Date: 2001-10-07T19:52:14+09:00
Subject: [ruby-talk:22188] Re: Match expressions

>>>>> "A" == Alan Moore <alan_moore@gmx.net> writes:

A> So, given the expression "node1 =~ node2", if node1 is a literal 
A> regex (i.e., /xyz/ or %{xyz}), Ruby translates the expression to  
A> "node1.=~(node2)".  If node2 is a literal regex, and node1 evaluates 
A> to a string, Ruby does a "node2.=~(node1)".  In both cases, it calls 
A> the rb_reg_match function directly, as a speed optimization.  But if 
A> node1 doesn't evaluate to a string, Ruby turns it around again and 
A> does a "node1.=~(node2)", under the assumption that node1 is a user-
A> defined object that defines its own "=~" method.

 Well, both String and Regexp has the method =~. The preferred way in ruby
 is to use Regexp#=~, i.e to write Regexp =~ Object

 The case Object =~ Regexp is rewritten in Regexp.=~(Object). If Object is
 evaluated to a String this is optimized to rb_reg_match() otherwise this
 is Object.=~(Regexp) to follow the normal call

 In the case String =~ String, the second string is the pattern to
 follow the tradition.

 NODE_MATCH2 and NODE_MATCH3 are just optimizations, when ruby know that it
 has a Regexp object.


Guy Decoux