Dave Wilson wrote:
> this optimization is used when the regexp is a 
> a literal, which means you can get by this by not using a regexp 
> literal, for example:
> 
> pattern = /a/
> # or
> pattern = Regexp.new('a')
> 
> "abc" =~ pattern

I should clarify: you can use a literal regexp, but not "in-place".  the 
optimization only happens when you match against the literal regexp 
"in-place", but if you match against a variable that has been assigned a 
literal regexp, this optimization won't be used.

Dave