Though I would not recomend using the second on big datastructures as it builds a list of all the elements in the arrays in the hash. Use either the first solution or 

use List::Util qw(sum);

my $count = sum map scalar(@$_), values %h;

#or
# my $count = sum( map scalar(@$_), values(%h));
# or
# my $count = sum( map {scalar(@$_)} values(%h));
# if that syntax looks more readable to you.

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Regarding the update: the first solution uses O(N) of memory where N is the number of elements of the hash, the second O(N*M) where M is the average length of the array in the hash. The List::Util solution is also O(N) memorywise.

Jenda
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In reply to Re^2: Count of HoA elements by Jenda
in thread Count of HoA elements by sweetblood

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