$input =~ s{\Q.\E}{\Q\.\E}g;
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Kev\.has\.a\.stupid\.perl\.question
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perl -le '$_=q[Kev.has.a.stupid.perl.question];s/\./\\./g;print'
Kev\.has\.a\.stupid\.perl\.question
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when I do print, yes it does give me that,
looking through perl debug it gives me
Kev\\.has\\.a\\.stupid\\.perl\\.question'

maybe it's just how it shows up in debug. Cheers

viffer:

How are you displaying it under the debugger? I'm seeing the expected value (using perl v5.10.1 under cygwin):

roboticus@Boink:~$ cat 854238.pl 
my $text = "Kev.has.a.stupid.perl.question";
$text =~ s/\./\\./g;
$text .= ".";
print $text;

roboticus@Boink:~$ perl 854238.pl
Kev\.has\.a\.stupid\.perl\.question.

roboticus@Boink:~$ perl -d 854238.pl
  ...snip...
main::(854238.pl:1):    my $text = "Kev.has.a.stupid.perl.question";
  DB<1> b 3
  DB<2> r
main::(854238.pl:3):    $text .= ".";
  DB<2> p $text
Kev\.has\.a\.stupid\.perl\.question
  DB<3>
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...roboticus

use strict;
use warnings;
my $string = "Kev.has.a.stupid.perl.question";
my @array = split /\./, $string;
my $fixed = join('\.',@array);
print $fixed;
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update: added links and a bit of details

One regex-less way to do this is to split

Uhm, you do know split, don't you? And you do know the type of its first argument, don't you?

Indeed, you do. Your "regexp-less" way uses the same regexp as you're trying to avoid. All you did was to move the regexp from s/// to split. You "regexp-less" way does not use less regexpes, or even a simpler one.

the original regex didn't worked on my system too. I'm using Strawberry Perl 5.10 on WinXP.

I'm going to assume escaping the dots is so that the string can be used inside a regex, in which case, quotemeta will be perfect.

print quotemeta('Kev.has.a.stupid.perl.question')."\n";
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Produces:

Kev\.has\.a\.stupid\.perl\.question
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Well, if this is driving you crazy, you're not alone. I tried your code on my system (ActiveState Perl 5.12) and used Data::Dumper. When I print the variable (print "$text\n";), it appears just as you want it. When I do print Dumper($text);, it has an extra \ in it just like you're seeing.

As for why Data::Dumper is not matching the print statement, I'm stumped.

As for why Data::Dumper is not matching the print statement, I'm stumped.

Because the output of Data::Dumper can be evaled into the original string, hence the "escaping" of the backslashes.

This means that the original string only had single backslashes, otherwise Data::Dumper would have shown 4 backslashes in a row!

CountZero

A program should be light and agile, its subroutines connected like a string of pearls. The spirit and intent of the program should be retained throughout. There should be neither too little or too much, neither needless loops nor useless variables, neither lack of structure nor overwhelming rigidity." - The Tao of Programming, 4.1 - Geoffrey James

Cool! Thanks for the explanation. Learned something new.