Perl Guru Forums: Perl Programming Help: Intermediate: Re: creating directory list?: Edit Log

Hi,

I think you got me wrong. I wasn't saying you were looking only for the last character. What your regex does is look for the $path with the last character of $path being optional instead of the whole $path. Perhaps it's because I'm no native english speaker. Anyway, here's an example to emphasize my concern:

Code

#!/bin/perl -w 
use strict; 
  
my $path       = '/path/to/files'; 
my $other_path = '/another/path'; 
  
my $file       = 'myfile'; 
  
my $test       = "$path/$file"; 
my $other_test = "$other_path/$file"; 
  
  
# your original regex 
  
print $test       =~ /^$path?(.*)$/       # /myfile 
      ? "$1\n" : "no match\n"; 
print $other_test =~ /^$path?(.*)$/       # no match 
      ? "$1\n" : "no match\n"; 
  
  
# your regex without the ? 
  
print $test       =~ /^$path(.*)$/        # /myfile 
      ? "$1\n" : "no match\n"; 
print $other_test =~ /^$path(.*)$/        # no match 
      ? "$1\n" : "no match\n"; 
  
  
# my modified version of your regex 
  
print $test       =~ /^(?:$path)?(.*)$/   # /myfile 
      ? "$1\n" : "no match\n"; 
print $other_test =~ /^(?:$path)?(.*)$/   # /another/path/myfile 
      ? "$1\n" : "no match\n";

As far as I understand, you want to strip of the $path from the beginning of the string. But if the string doesn't begin with $path, you don't want to strip off anything.
If you look at the three regexes above, only my modified variant does this. For any other behaviour, I guess, the ? is completely useless, as you can see if you compare the first two regexes. It's useless because

In Reply To

Also, when a scalar is tossed into a regex, ? isn't looking for just the last character of the scalar's value -- it's looking for the entire value of the scalar.

is wrong. First, $path is interpolated, then the resulting string is passed to the regex engine. You can find this illustrated in Jeffrey Friedl's 'Mastering Regular Expression'. As a proof that your original regex has only the last character of $path optional, I've included the debug output of the regex engine for your original regex

Code

Compiling REx `^/path/to/files?(.*)$' 
size 18 first at 2 
   1: BOL(2) 
   2: EXACT </path/to/file>(7)    <- exact match 
   7: CURLY {0,1}(11)           <- {0,1} means optional 
   9:   EXACT <s>(0)              <- 's' is optional 
  11: OPEN1(13) 
  13:   STAR(15) 
  14:     REG_ANY(0) 
  15: CLOSE1(17) 
  17: EOL(18) 
  18: END(0)

and for my modified one

Code

Compiling REx `^(?:/path/to/files)?(.*)$' 
size 18 first at 2 
   1: BOL(2) 
   2: CURLYM[0] {0,1}(11) 
   4:   EXACT </path/to/files>(9)    <- this is optional 
   9:   SUCCEED(0) 
  10:   NOTHING(11) 
  11: OPEN1(13) 
  13:   STAR(15) 
  14:     REG_ANY(0) 
  15: CLOSE1(17) 
  17: EOL(18) 
  18: END(0)

Do you understand what I'm trying to express? Sorry if I cannot explain this understandable enough for you.

-- Marcus

(This post was edited by mhx on Jun 22, 2001, 6:22 PM)


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