Perhaps using a negative lookahead?

$foo =~ /$users_regex(?!,v$)/
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From perldoc perlre:

"(?!pattern)"

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 can- not be "foo"--and it's not, it's a "bar", so "foobar" will match. You would have to do something like "/(?!foo)...bar/" for that. We say "like" because there's the case of your "bar" not having three characters before it. You could cover that this way: "/(?:(?!foo)...|^.{0,2})bar/". Sometimes it's still easier just to say:

if (/bar/ && $` !~ /foo$/)

Update: Fixed formatting, and adding a comment that the documentation does suggest, as Fletch suggests, that it's sometimes still easier to do what Melly is originally doing.

s**lil*;  $*=join'',sort split q**;  s;.*;grr; &&s+(.(.)).+$2$1+; $; =
qq-$_-;s,.*,ahc,;$,.=chop for split q,,,reverse;print for($,,$;,$*,$/)
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