Nope, the regex is fine as it isn't re-entrant. If there was regex *within* the regex then that would be re-entrant, but as far as perl is concerned you can do as you please in the replace part of a s///.

So this code will break^[1]

my $str = "foo bar baz quux";
   $str =~ s<([a-z]+)
             (?{ s| \1|\L$&|; })><\U$1>xg;
[download]

But this is fine

$str =~ s<([a-z]+ ?)><local $_ = $1; s/$1/\U$&/g; $_>eg;
[download]

HTH

_________
broquaint

^[1] in theory, I'll be happy to change it to a proper re-entrant regex if someone will provide me with one