Returning to this example below…

…assume “look” and “ahead” are on separate lines one below the other. Then the line becomes this…

Is there some restriction, as the two words do not get joined with a dash? (At least for me)

Plus, there is no error message; the command just doesn’t work. My one-liner is below. I’ve ensured my apostrophe’s are correct after I copy/paste; plus, I’ve key-stroked every character, and tried numerous variations, etc., etc. Using Mac OS Sequoia 15.0 and perl 5.34.

You're reading your input line-by-line but you're trying to match across lines. That won't work.

The following should work, reading the input all at once (see perlrun):

perl -g -i -pe ’s/(?<=look)\n(?=ahead)/-/g;’ myfile.txt
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Thank you very, very much for setting me on the right path! While (in my environment) the -g switch isn't allowed for (what I learned to be was) slurping, instead using the -0777 switch did work nicely. I had seen this switch before, but until now wasn't aware of its usage until I dug deeper, prompted by your input.

I was stuck on this for a long time - thanks again for the quick response!

P.S. I just found where version 5.36 would allow the use of the -g switch, which is apparently a shortcut for using -0777.

Just one little comment: The code in the last paragraph did not work because by default regular expressions are greedy. (Did this change with the Perl versions in between?) The only right-substring that comes out is the full string:

$_ = "Hello";
print "$1\n" while /(?=(.*))/g;
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Hello
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$_ = "Hello";
print "$1\n" while /(?=(.*)?)/g;
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Hello
ello
llo
lo
o
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Note that this bug has been fixed in 5.22.

$ 5.20t/bin/perl -e'$_ = "Hello"; print "<$1>\n" while /(?=(.*))/g;'
<Hello>

$ 5.22t/bin/perl -e'$_ = "Hello"; print "<$1>\n" while /(?=(.*))/g;'
<Hello>
<ello>
<llo>
<lo>
<o>
<>
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I think you have found a bug in 5.10's regex handling. The lookahead's greediness or non-greediness should not matter, because it does not consume any characters. When used in a global match, patterns that do not consume characters should advance one character on each match. At least that's how I read the documentation.

The really interesting thing about your version is that you didn't make the capture non-greedy, you made it optional. You probably meant (.*?), which (in pre-5.10) will output empty strings every time. I haven't installed 5.10 myself, so I can't play with it right now.

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...which (in pre-5.10) will output empty strings every time.

With 5.10.0, /(?=(.*?))/g; outputs one empty string.  And I can confirm the behavior reported by jds17 with /(?=(.*))/g.

You are right, my change did not affect greediness. The bad thing is: now I don't understand why my proposed solution worked at all. Maybe someone can explain? I don't think the question is too important, but I like to use regular expressions and it bugs me a little if I cannot understand one (especially such a tiny one).

I have read the documentation you have cited and it helped, so I played around some more and tried out the following, which only exchanges "+" for "*" in your original expression, really works as one would think and therefore would be my preferred solution, at least for Perl 5.10:

$_ = "Hello";
print "$1\n" while /(?=(.+))/g;
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Output:

Hello
ello
llo
lo
o
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my %hash = (
    1        => 'equity, private equity',
    2        => 'equity',
    3        => 'private equity',
    4        => 'private equity,equity',
    5        => 'private equity, equity',
    6        => 'equity,private equity',
    7        => 'private equity',
    8        => 'mutual funds',
    9        => 'cds'
);

while (my ($k, $v) = each %hash) {
    next unless $v =~ m/(?!private\s+)equity/;
    printf("%d -> %s\n", $k, $v);
}
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Hi, new questions go in Seekers Of Perl Wisdom because

Roy Johnson, whom you asked a question, hasn't been here in 6 weeks.

You used code tags and put your code in between, that is awesome :)

Welcome, see How do I post a question effectively?, Where should I post X?

The regex which is not working for you, contains A zero-width negative look-ahead assertion, and like perlre#(?!pattern) says

A zero-width negative look-ahead assertion. For example /foo(?!bar)/ matches any occurrence of "foo" that isn't followed by "bar". Note however that look-ahead and look-behind are NOT the same thing. You cannot use this for look-behind.

If you are looking for a "bar" that isn't preceded by a "foo", /(?!foo)bar/ will not do what you want. That's because the (?!foo) is just saying that the next thing cannot be "foo"--and it's not, it's a "bar", so "foobar" will match. Use look-behind instead (see below).

So, use a look-behind

But, that probably won't work either, because you can't have variable length length lookbehind , so you need to use a fixed width lookbehind.

#!/usr/bin/perl --
use strict; use warnings; 
use Test::More qw' no_plan ';
Main(@ARGV);
exit(0);
sub Main {
    my @yesWant = (
        'equity, private equity',
        'equity',
        'private equity,equity',
        'private equity, equity',
        'equity,private equity',
    );
    my @notWant = (
        'private equity',
        'private equity',
        'mutual funds',
        'cds',
    );
    for my $not ( @notWant ){
        ok(  (not TestEquity($not)), "not '$not'" );
    }
    for my $yes ( @yesWant ){
        ok(  TestEquity($yes), "yes '$yes'" );
    }
}
sub TestEquity {
    return 1 if $_[0] =~ m/(?<!private\s)equity/;
    return 0;
}

__END__
$ prove -v pm911357.lookbehind.pl
pm911357.lookbehind.pl ..
ok 1 - not 'private equity'
ok 2 - not 'private equity'
ok 3 - not 'mutual funds'
ok 4 - not 'cds'
ok 5 - yes 'equity, private equity'
ok 6 - yes 'equity'
ok 7 - yes 'private equity,equity'
ok 8 - yes 'private equity, equity'
ok 9 - yes 'equity,private equity'
1..9
ok
All tests successful.
Files=1, Tests=9,  0 wallclock secs ( 0.06 usr +  0.01 sys =  0.08 CPU
+)
Result: PASS
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If fixed width lookbehind doesn't work for you, simply do TWO tests

Here's a solution that exactly matches the phrases specified in AnonyMonk's Re: Using Look-ahead and Look-behind post (which the code of Re^2: Using Look-ahead and Look-behind does not quite do), and also shows how to use the newfangled backtracking control verbs of 5.10 to emulate variable-width negative look-behind. Variable-width positive look-behind is emulated by 5.10's  \K assertion.

Explanation:

• Any 'equity' that is preceded by
  • either a character that is not a comma or whitespace, or
  • by the 'private' phrase
  FAILS and is skipped over (this test has first precedence);
• Otherwise, any 'equity' that is not followed by a comma that is then followed by any non-whitespace SUCCEEDS.

>perl -wMstrict -le
"use Test::More 'no_plan';
 ;;
 for my $ar_vector (
   [ YES => 'equity, private equity', ],
   [ YES => 'equity',                 ],
   [ no  => 'private equity',         ],
   [ YES => 'private equity,equity',  ],
   [ YES => 'private equity, equity', ],
   [ no  => 'equity,private equity',  ],
   [ no  => 'private equity',         ],
   [ no  => 'mutual funds',           ],
   [ no  => 'cds'                     ],
   ) {
   my ($expected, $string) = @$ar_vector;
   is match($string), $expected, qq{'$string'};
   }
 ;;
 sub match {
   my ($string) = @_;
   ;;
   my $char_not_comma_or_space = qr{ [^,\s]      }xms;
   my $private                 = qr{ private \s+ }xms;
   return 'YES' if $string =~
     m{ (?: $char_not_comma_or_space | $private) equity (*SKIP)(*FAIL)
        |
        equity (?! , \S)
      }xms;
   return 'no',
   }
"
ok 1 - 'equity, private equity'
ok 2 - 'equity'
ok 3 - 'private equity'
ok 4 - 'private equity,equity'
ok 5 - 'private equity, equity'
ok 6 - 'equity,private equity'
ok 7 - 'private equity'
ok 8 - 'mutual funds'
ok 9 - 'cds'
1..9
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Nice. Very nice! You nailed. It's working. Thanks a bunch!

I changed the sub TestEquity to allow for any text between Private and Equity, but I can't get it to work. What have I done wrong?

sub TestEquity {
    return 1 if $_[0] =~ m/(?<!private).*equity/;
    return 0;
}
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Generally, the decision about using ?: is not about whether it's necessary, but that capturing whatever you're grouping isn't necessary. The ?: modifier makes parentheses not capture, which is somewhat more efficient and might make the task of counting left parentheses less onerous.

By the way, it's not a lookaround feature.

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Thank you!