in reply to Perl Idioms Explained - my $count = () = /.../g

Just to have it mentioned: another way to do it is given in perldoc -f scalar:
#!perl -l my $str = "here are some words"; my $count = @{[ $str =~ /\S+/g ]}; print $count;
Personally, I think this makes the meaning clearer, but I'm sure others may disagree.

Replies are listed 'Best First'.
Re^2: Perl Idioms Explained - my $count = () = /.../g
by japhy (Canon) on Feb 07, 2006 at 15:40 UTC
    I'd say "but that builds up and then destroys an array referencce", but I don't know which takes more time, and then I'd say "benchmark it", but this would be a case of micro-optimizing. So I will agree with your intent: use whichever construct is clearer to you.

    Jeff japhy Pinyan, P.L., P.M., P.O.D, X.S.: Perl, regex, and perl hacker
    How can we ever be the sold short or the cheated, we who for every service have long ago been overpaid? ~~ Meister Eckhart