There is a catch though. If 'B' contains 'a', (or the replacement of an 'a' to a 'B' creates a new 'a'), this doesn't quite do what you want. A simple case, assume 'a' is '\w' and 'B' is just 'B'. Then 
$_ = 'pqrstuvw';
my $i = 3;
while($i) {
  $i--;
  s/\w/B/;
}
print "$_\n";
__END__
Bqrstuvw

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In reply to Re^2: Does Perl support a given number of replacements ? by JavaFan
in thread Does Perl support a given number of replacements ? by Anonymous Monk

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