From: Lloyd Zusman <ljz@...>
Date: 2004-10-21T07:46:10+09:00
Subject: Re: Regexp equality

Jamis Buck <jgb3@email.byu.edu> writes:

> [ ... ]
>
> I realize that particular question is most likely impossible to answer 
> for the general case. What I was wanting, though, was to be able to know 
> whether /\/blah/ is equal to %r{/blah}. Surely that's not too hard?
>
> As I said, though, converting them both to strings gives the same 
> result, so the strings, at least, are comparable with expected results.

OK.  Now I better understand your question.  I agree with you that
something seems kind of counter-intuitive here:

  irb(main):001:0> x = /\/blah/
  => /\/blah/
  irb(main):002:0> y = %r{/blah}
  => /\/blah/
  irb(main):003:0> z = %r{\/blah}
  => /\/blah/
  irb(main):004:0> x == y
  => false
  irb(main):005:0> x == z
  => true
  irb(main):006:0> x.source
  => "\\/blah"
  irb(main):007:0> y.source
  => "/blah"
  irb(main):008:0> z.source
  => "\\/blah"

In line 006, why is the source of /\/blah/ returned as "\\/blah"?

It seems to me that it's incorrect to return the leading backslash
doubled.  Isn't the backslash in the initial definition of x just an
escape character for enabling the second forward slash to be treated as
part of the regexp?  Why is that initial escape character treated as a
bona fide part of the regexp source?

Also, consider this:

  irb(main):001:0> x = /\/blah/
  => /\/blah/
  irb(main):002:0> y = %r{/blah}
  => /\/blah/
  irb(main):003:0> z = %r{\/blah}
  => /\/blah/
  irb(main):004:0> x.to_s
  => "(?-mix:\\/blah)"
  irb(main):005:0> y.to_s
  => "(?-mix:\\/blah)"
  irb(main):006:0> z.to_s
  => "(?-mix:\\/blah)"
  irb(main):007:0> x.to_s == y.to_s
  => true
  irb(main):008:0> x.to_s == z.to_s
  => true

In other words, the string representation of the 3 regexp's (as opposed
to their _source_ representations) are identical.

So this brings up two questions:

1.  Why is the backslash in a statement like this /\/blah/ _not_ just
    treated as an escape character when determining the source?

2.  Why does the `==' operator compare the source and not the internal
    representations?  It seems to me that the internal representation is
    a much more meaningful piece of information for use in a comparison.


-- 
 Lloyd Zusman
 ljz@asfast.com
 God bless you.