Perl regex to find keywords and not variables - regex

Question

I'm trying to create a regex as following :
print $time . "\n"; --> match only print because time is a variable ($ before)
$epoc = time(); --> match only time
My regex for the moment is /(?-xism:\b(print|time)\b)/g but it match time in $time in the first example.
Check here.
I tried things like [^\$] but then it doesn't match print anymore.
(I will have more keyword like print|time|...|...)
Thanks

Answer 1

Parsing perl code is a common and useful teaching tool since the student must understand both the parsing techniques and the code that they're trying to parse.
However, to do this properly, the best advice is to use PPI
The following script parses itself and outputs all of the barewords. If you wanted to, you could compare the list of barewords to the ones that you're trying to match. Note, this will avoid things within strings, comments, etc.
use strict;
use warnings;
use PPI;
#my $src = do {local $/; <DATA>}; # Could analyze the smaller code in __DATA__ instead
my $src = do {
local #ARGV = $0;
local $/;
<>;
};
# Load a document
my $doc = PPI::Document->new( \$src );
# Find all the barewords within the doc
my $barewords = $doc->find( 'PPI::Token::Word' );
for (#$barewords) {
print $_->content, "\n";
}
__DATA__
use strict;
use warnings;
my $time = time;
print $time . "\n";
Outputs:
use
strict
use
warnings
use
PPI
my
do
local
local
my
PPI::Document
new
my
find
for
print
content
__DATA__

Answer 2

What you need is a negative lookbehind (?<!\$), it's zero-width so it doesn't "consume" characters.
(?<!\$)a means match a if not preceded with a literal $. Note that we escaped $ since it means end of string (or line depending on the m modifier).
Your regex will look like (?-xism:\b(?<!\$)(print|time)\b).
I'm wondering why you are turning off the xism modifiers. They are off by default.So just use /\b(?<!\$)(?:print|time)\b/g as pattern.
Online demo
SO regex reference