> # 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>


Reply via email to