my $how_many = @array;
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my @a = (1, undef, 3);

# Now scalar @a = 3, but there are only two elements of interest.
# Brute force counting method:

my $c = 0;
$c++ for (grep { defined } @a);

# $c now holds the number of defined elements, in this case 2.
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Or simply

my $c = grep defined, @a;
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the lowliest monk

Ah, yes, or that :/

my $c1  = 0;
   $c1 += 1 for @array;

my $c2  = grep {$_ || !$_} @array;

my $c3 = 0;
L1: goto L2 unless @array % 2;
    $c3++;
    pop @array;
L2: goto L3 unless @array;
    $c3 *= 2; 
    splice @array, 0 @array / 2;
    goto L1;
L3:

my $c4 = do {local $" = "+"; eval "@{[map {1} @array]}"};

my $c5 = length join "", map " ", @array;

use POSIX 'ceil';
my $c6 = ceil((length join " ", map "", @array)/2);

my $c7 = 0;
$c7++, shift @array while @array;
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#!/usr/bin/perl -w
use strict;

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

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."

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