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From: sherman@unx.sas.com (Chris Sherman)
Newsgroups: comp.lang.perl
Subject: Re: pattern matching question
Message-ID: <sherman.675057031@foster>
Date: 24 May 91 03:50:31 GMT
References: <1991May22.193037.12166@cherokee.uswest.com> <1991May22.221507.12660@cherokee.uswest.com> <49420@ut-emx.uucp>
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In <49420@ut-emx.uucp> dboles@ccwf.cc.utexas.edu (David Boles) writes:

>Warning: PERL NOVICE approaching !!!

Warning:  PERL NOVICE answering!!!

>I am trying to do the following file manipulation:
>p220 p220   20    5
>p220 p235   20    7
>etc.
> ==>
>p220 -20 -5
>p220 20 5
>p220 -20 -7
>p235 20 7
>etc.
>I am using:
>while (<>) {
>    s/(p\d*) (p\d*) (\d*) (\d*)/$1 -$3 -$4\n$2 $3 $4\n/;
>    print;
>}
>and I get:
>p220 - -
>p220 20 5
>Why aren't $3 and $4 "alive" in the first half of the replacement
>string?  What am I missing?

I think I got it.  Perl is taking your test string literally, space for space.
The synchronization is lost.

I used the following input file with your code:

p220 p220 20 5
p220 p235 20 7

and got:

p220 -20 -5
p220 20 5

p220 -20 -7
p235 20 7


So then I tried the following code:

#!/usr/local/bin/perl
while (<>) {
  s/(p\d*) *(p\d*) *(\d*) *(\d*)/$1 -$3 -$4\n$2 $3 $4\n/;
  print;
}

With the following input:

p220 p220   20   5
p220 p235   20   7

and got 

p220 -20 -5
p220 20 5

p220 -20 -7
p235 20 7

Maybe perl pro's can tell me what the '*'s meant exactly, why they are working,
and if they would work in every case.  (I have my ideas, but they are probably
wrong, and I just got lucky.  I was hoping to set up a one-or-more-number-of
spaces type thing, but I don't think I did that right).
--
Chris Sherman .................... sherman@unx.sas.com   |
              ,-----------------------------------------'
             /  Q:  How many IBM CPU's does it take to execute a job?
            |   A:  Four; three to hold it down, and one to rip its head off.