Re: Re: regex substitution

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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(crazyinsomniac) Re^3: regex substitution by crazyinsomniac (Prior) on Nov 05, 2001 at 13:03 UTC
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;;,qH*));print;$q/$q;"`	[reply] [d/l] [select]

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(crazyinsomniac) Re^3: regex substitution
by crazyinsomniac (Prior) on Nov 05, 2001 at 13:03 UTC

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>
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perlop

e Evaluate the right side as an expression.

___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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