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Re: Counting the number of items returned by split without using a named array

by Tanalis (Curate)
on May 03, 2006 at 10:12 UTC ( #547106=note: print w/ replies, xml ) Need Help??


in reply to Counting the number of items returned by split without using a named array

Why not just count the instances of whitespace?

my $str = "this is a test string"; my $cnt = 1; # num tokens = whitespace + 1 ++$cnt while $str =~ /\s+/g; print $cnt, "\n"; # prints 5


Comment on Re: Counting the number of items returned by split without using a named array
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Re^2: Counting the number of items returned by split without using a named array
by blazar (Canon) on May 03, 2006 at 11:07 UTC

    The =()= does work with matches:

    $ perl -lpe '($_=()=/\s+/g)++' foo 1 bar baz 2 foo bar baz 3

    But I would use split, especially with the smart behaviour provided by the default ' ' argument.

      That fails when input has leading or trailing blanks:

      $ perl -lpe '($_=()=/\s+/g)++' ## leading foo bar 3 $ perl -lpe '($_=()=/\s+/g)++' ## trailing foo bar 3 $ perl -lpe '($_=()=/\s+/g)++' ## both foo bar 4 $ _

      It's better to count ocurrences of actual elements (\S+):

      $ perl -wle 'print scalar (()=/\S+/g) for "a b c", " a b c", "a b c ", + " a b c "' 3 3 3 3 $ _

      Anyway I prefer split too, like salva++'s 0e0 solution.

      --
      David Serrano

        I know. My entire point was not to maintain a counter manually a' la

        ++$cnt while $str =~ /\s+/g;

        Well done to point out about \S anyway, since one often forgets about \S, \D and \W.

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