From: Alan Moore <alan_moore@...>
Date: 2001-10-07T03:39:58+09:00
Subject: [ruby-talk:22163] Re: Match expressions

matz wrote:

>I often confuse left and right even after I've grown up.

That _was_ one of my theories... ;-)


>This should be:
>
>      case NODE_MATCH3:
>        {
>            VALUE r = rb_eval(self,node->nd_recv);
>            VALUE l = rb_eval(self,node->nd_value);
>            if (TYPE(l) == T_STRING) {
>                result = rb_reg_match(r, l);
>            }
>            else {
>                result = rb_funcall(l, match, 1, r);
>            }
>        }
>        break;


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?

--Alan