From: Holden Glova <dsafari@...>
Date: 2002-11-16T07:46:48+09:00
Subject: Re: Regexp: What does //o do?

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On Sat, 16 Nov 2002 10:19, Michael Campbell wrote:
> > Yes. It's not completely clear to me when it will be recompiled
> > and when it won't be. Above it is not. Below it is.
> >
> > irb(main):001:0> s="a"
> > "a"
> > irb(main):002:0> /#{s}/o
> > /a/
> > irb(main):003:0> s="b"
> > "b"
> > irb(main):004:0> /#{s}/o
> > /b/
>
> This is 2 different regex's, even though they evaluate to the same,
> so each one is compiled once.  (2 different spots in the source code)
>
> > irb(main):016:0> def foo(s)
> > irb(main):017:1> re = /#{s}/o
> > irb(main):018:1> p re
> > irb(main):019:1> re = nil
> > irb(main):020:1> end
> > nil
> > irb(main):021:0> foo("a")
> > /a/
> > nil
> > irb(main):022:0> foo("b")
> > /a/
> > nil
> >
> > and I'm confused even more.
>
> That's the SAME regex (same spot in the source code) used twice.
> Your first example was 2 different regex's, used once each.

I'm not sure I can agree that this is the same regex, Jim is setting it to nil 
at the end of the function so that also is 2 different regex objects. What 
sort of voodoo is going on here? I too am confused. I didn't know about the 
/o modifier and I think it is pretty cool, now if it just made sense to 
me.... :)

- -- 
Signed,
Holden Glova
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