Re: Regex replace in context

FWIW, it also means that $& does not include a.

I find the negative zero-width-lookbehind assertion rather handy (e.g. s/(?<!z)b/d/), which means, roughly, "substitute b for d whenever it's not preceded by z".

This trick reminds me of the reason we have \b -- sometimes a boundary condition can be met by no character at all, e.g., we'd like to match on both "b" and "ab".

In both cases, I've taken advantage of the not-modifying-$& effects in my code. I wish you could add the quantifiers, though, even though I can imagine just how complicated look-behinds with quantifiers might become...

jyust my $0.02,

--jeremy

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Re: Re: Regex replace in context by John M. Dlugosz (Monsignor) on Dec 11, 2002 at 21:32 UTC
I suppose it would be clearer than `(?<!(?i-x:pattern))` Idle thought: `sub import { shift; return unless @_; die "unknown import: @_" unless @_ == 1 and $_[0] eq ':constant'; overload::constant qr => sub { my $s= shift; $s =~ s/(<!\\)\x{ab}/(?:/g && $s =~ s/(<!\\)\x{bb}/)/g; return eval qr/$s/; }; }` [download] —John	[reply] [d/l] [select]

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Re: Re: Regex replace in context
by John M. Dlugosz (Monsignor) on Dec 11, 2002 at 21:32 UTC

(?<!(?i-x:pattern))

Idle thought:

sub import {
   shift;
   return unless @_;
   die "unknown import: @_" unless @_ == 1 and $_[0] eq ':constant';
   overload::constant qr => sub {
      my $s= shift;
      $s =~ s/(<!\\)\x{ab}/(?:/g && $s =~ s/(<!\\)\x{bb}/)/g;
      return eval qr/$s/;
      };
   }
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