Re: Re: regex substitution

[coolmichael]

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

[crazyinsomniac]

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>

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;"