in reply to how many elements in array?

or 
#!/usr/bin/perl -w
use strict;

my @one=(1,2,3,4,5,6);
print scalar(@one);

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or 
print $#one;

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depends on what you want.
UPDATE
you should also check out the the basic datatypes, three node

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Re^2: how many elements in array?
by Limbic~Region (Chancellor) on Apr 18, 2005 at 12:43 UTC
    tcf03,
    $#one and scalar @one do not produce the same thing. Using scalar @one gives the number of elements, which is what was asked for, while $#one gives is the index of the last element.

    Cheers - L~R

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      Although, technically, they actually could produce the same thing, if one were to modify $[ to be 1

      Update: Just a caveat that this is not recommended in the slightest :) as the docs for index say:
      The return value is based at 0 (or whatever you've set the $[ variable to--but don't do that).
      --------------
      "But what of all those sweet words you spoke in private?"
      "Oh that's just what we call pillow talk, baby, that's all."
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        RazorbladeBidet,
        While perldoc perlvar highly discourages the modification of $[, I recognize that in some very odd situation it may be the most natural way to solve a problem. In this case, it would be doing something very un-natural for the sake of being correct.

        A little bit of knowledge is a dangerous thing. If you are going to give someone a loaded gun, please give them the instruction manual too.

        Cheers - L~R

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[d/l]

      Helpful in a loop as in:

      for ( 0 .. $#one ) {
   ...do stuff to each element...
}

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      —Brad
      "The important work of moving the world forward does not wait to be done by perfect men." George Eliot
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        bradcathey,
        There are lots of ways in which $#array can be useful otherwise it wouldn't exist in the language. My point was more about correctness. For instance, if I am doing the following: 
        my @ucase = grep { $_ eq uc() } @words;
my $cnt = @ucase;

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        I have the count of uppercase words, making all appropriate assumptions about @words. I would need to use $#ucase + 1* since it doesn't produce a count of elements otherwise. This was not clear from the post that I was replying to.

        Cheers - L~R

        * As RazorbladeBidet points out, modifying $[, is one highly discouraged way of accomplishing the same thing.
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