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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;"


In reply to (crazyinsomniac) Re^3: regex substitution by crazyinsomniac
in thread regex substitution by michellem

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