Re: regex doubt on excluding

[kcott]

If you look in "perlre: Metacharacters", you'll see that '$' matches "the end of the line (or before newline at the end)" (emphasis added). So, what you really want is to match any whitespace character except if it follows '$'. That makes your substitution "s/(?<!$)\s//gm", shown here:

#!/usr/bin/env perl -l

use strict;
use warnings;

my $string = "\t is a TAB\n is a SPACE\nTAB\t and SPACE";

print '*** BEFORE ***';
print $string;

$string =~ s/(?<!$)\s//gm;

print '*** AFTER ***';
print $string;

Output:

*** BEFORE ***
     is a TAB
 is a SPACE
TAB     and SPACE
*** AFTER ***
isaTAB
isaSPACE
TABandSPACE

However, if you know that you only want to match the whitespace characters space (" ") and tab ("\t"), then the transliteration "y/\t //d" will be faster. (See "Search and replace or tr" in "Perl Performance and Optimization Techniques" for a Benchmark example.) As you can see, the code is virtually identical (which makes replacing the s/// with y/// easy):

#!/usr/bin/env perl -l

use strict;
use warnings;

my $string = "\t is a TAB\n is a SPACE\nTAB\t and SPACE";

print '*** BEFORE ***';
print $string;

$string =~ y/\t //d;

print '*** AFTER ***';
print $string;

Output:

*** BEFORE ***
     is a TAB
 is a SPACE
TAB     and SPACE
*** AFTER ***
isaTAB
isaSPACE
TABandSPACE

[In case you didn't know, y/// and tr/// are synonymous. You'll find both forms used in different sections of the perlop documentation.]