> # trim right space away. I'm shure there is a shorter > # and more elegant solution > # > @parts= map { do {$_=~s/\s*$//; $_ } } @parts;
Regular expressions have a tendency to be slower, than functions that achieve aquivalent results; but, chop() cannot be considered being used in above statement, since chops returns the trimmed character. s/// operates by default on $_, as many other built-in functions do, thus: @parts = map { do { s/\s*$//; $_ } @parts; Also, consider using the tr operator instead of s/// for speed-up purposes. -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] <http://learn.perl.org/> <http://learn.perl.org/first-response>