From: seijin@...
Date: 2007-07-25T04:05:00+09:00
Subject: Re: Regex Question

Oh, I see.  So using the "*" with just one ... errr... thing will
never match it?  "banana" =~ /a*/ will always be 0 but "banana" =~ /
a*n/ will be 1 (one) ?  It's just confusing because I think of
"greedy" as it trying to match as many characters as possible.  So
that although "*" tell it to find 0 or more that it would work hard to
return the maximum number of matches.  Perhaps the difference between
"Find as many a's as possible" and "Find as many a's or nothing as
possible" ?  If you see what I'm trying to say.  No, I think I just
figured out what my problem is.  I'm taking it out of context.  The
"*" was created for use other arguments.  I keep getting stuck on
using it with just one argument which is not what it was intended for.

And for Mississippi - is that method of matching only used when using
back references via the \1-9 ?  If so, I can set my mind at ease as
that makes sense.  The thing that was bothering me was that I didn't
see why it would work so hard to find the match.  I thought it was a
simple... (\w+) matches "Mississippi" but there is no
"MississippiMississippi" so I'll return nil... type of thing.  I
didn't realize it'd start scanning the string aggresively.

Thanks for the reply!


On Jul 24, 10:27 am, Rob Biedenharn <R...@AgileConsultingLLC.com>
wrote:
> On Jul 24, 2007, at 12:50 PM, sei...@gmail.com wrote:
>
>
>
>
>
> > I recently started studying Ruby (1.8.6) and things are going well.
>
> > However, something bothered me when I got to the regular expression
> > section.  This is my first time deal with regular expressions directly
> > so I'm completely new to things.  I understand the idea behind regular
> > expressions and most of the usage but two of them stuck out for me.
>
> > I'm studying the 2nd edition Pragmatic book and using their
> > show_regexp method...
>
> > def show_regexp(a, re)
> >    if a =~ re
> >       "#{$`}<<#{$&}>>#{$'}"
> >    else
> >       "no match"
> >    end
> > end
>
> > ...hopefully I don't get in trouble posting that method here.  So here
> > are my two problems...
>
> > First, why does ... show_regexp("banana", /(an)*/) ... not match
> > "anan" ?  I thought it was a greedy algorithm that tried to match as
> > much as possible?
>
> Well, it would but the * is 'zero or more times' and it can match  
> zero "an"s before it ever needs to get past the 'b'.  If you change  
> the '*' to a '+' and force 'one or more times', then it has to find  
> the first "an" and will then greedily consume the next "an" also.
>
> > Second, how does ... show_regexp("Mississippi", /(\w+)\1/) ... work?
> > Why in the world does it match "ississ" rather than returning no
> > match?  I think most of my problem with this one is not understanding
> > the underlying logic used when doing pattern matching with back
> > references.  Does it go through and check "Mississippi", "Mississipp"
> > down to "M" and then "i", "is", "iss", etc?  Like trying all possible
> > combinations in a lock?  I would be extremely grateful if someone
> > would do a short step-by-step of how it matches "ississ".  I
> > understand what the "\1" does, I just don't understand how the first
> > part even gets to the first "iss".
>
> > Thanks a lot!
>
> The '+' is greedy so I think it tries \w+ as Mississippi and finds  
> that there's no following copy, backs up to try Mississipp and finds  
> that 'i' isn't a copy, and so on back to 'M' before moving past 'M'  
> to 'ississippi', 'ississipp', 'ississip', etc. until 'iss' has a  
> following copy and a match is declared.  Think of it this way:  Until  
> the regexp gets to the end, it doesn't know that the string isn't  
> "MississippiMississipp".
>
> -Rob
>
> Rob Biedenharn          http://agileconsultingllc.com
> R...@AgileConsultingLLC.com- Hide quoted text -
>
> - Show quoted text -