in reply to regex substitution

But I can't for the life of me find an example that does this (substitutions within substitutions).

When doing substitutions like this, /e is your friend.   $text =~ s/\#\w+\#/$hash{$1}/eg; /e forces the right-hand side of the s/// to be evaluated as a Perl expression, and the result substituted for whatever the left-hand side matched. The boundary between interpolation and substitution can seem pretty hazy, but it makes a bit more sense if you imagine that the regexp is "compiled" immediately before it is executed. At that point, $1 either doesn't have a value, or holds a value from a prior regexp. Neither is what you want.

Consult perlre for full details.

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Re: Re: regex substitution
by coolmichael (Deacon) on Nov 05, 2001 at 10:40 UTC
    Perhaps it shouldn't work without /e, but it does, at least with activestate perl 5.6.1. (obviously, you should use strict and warnings...) 
    #!perl
$_="this is a #fish#\n";
$a{"fish"}="test";
print;
s/#(\w+)#/$a{$1}/;
print;

this is a #fish#
this is a test

C:\perl -v

This is perl, v5.6.1 built for MSWin32-x86-multi-thread
(with 1 registered patch, see perl -V for more detail)

Copyright 1987-2001, Larry Wall

Binary build 626 provided by ActiveState Tool Corp. http://www.ActiveS
+tate.com
Built 01:31:15 May  2 2001

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[d/l]
      Have you read what the /e switch does? I sugguest you do.

      Your argument basically boils down to s/#(\w+)#/$variable/; in which $variable gets expanded, as it will no matter what (if /e is there or not).

      Compare this to: 

      #!perl -w
use strict;
my %hashola = (fish => "test" );

sub f($){ return $hashola{shift @_};  }

my $rock = "this is a #fish#\n";
my $block = $rock;

print $rock;

$block =~ s/#(\w+)#/&f($1);/; # expands $1, doesn't call &f

print $block;

$block = $rock; # restoree

$block =~ s/#(\w+)#/$hashola{$1}/; # expands the variable

print $block;

$block = $rock; # restoree

$block =~ s/#(\w+)#/&f($1)/e; # calls the function

print $block;

$block = $rock; # restoree

$block =~ s/#(\w+)#/$hashola{$1}/e;

print $block;

$block = $rock; # restoree

__END__
F:\dev>perl f
this is a #fish#
this is a &f(fish);
this is a test
this is a test
this is a test

F:\dev>

      [download]
      Ok ok, from perlop (since s is an operator), I quote:
      e Evaluate the right side as an expression.
      Unless you escape $hashname{$1} like \$hashmane{$1}, the actual variable will be expanded independent of /e, as such is the nature of the s operator.

       
      ___crazyinsomniac_______________________________________
      Disclaimer: Don't blame. It came from inside the void
      perl -e "$q=$_;map({chr unpack qq;H*;,$_}split(q;;,q*H*));print;$q/$q;"

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