catfish1116:   Further to toolic's reply:   Another way to evaluate an array in list context when printing is to interpolate the array into a string: 

c:\@Work\Perl\monks>perl -wMstrict -le
"my @lines = qw(one two three);
 ;;
 print qq{\@lines array (@lines): number of elements: }, scalar @lines
+;
 print qq{\@lines array (@lines): number of elements: } . @lines;
"
@lines array (one two three): number of elements: 3
@lines array (one two three): number of elements: 3

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Note that in the first print statement, evaluating the array in scalar context to get the number of elements in the array had to be forced with the scalar built-in due to the  , (comma) operator (see perlop). In the second print statement, the  . (string concatenation) operator (perlop again) supplies scalar context. (See also Context tutorial.)

And yes, please edit your OP so the janitors don't have to clean up the mess. Please see How do I change/delete my post?

Give a man a fish:  <%-{-{-{-<

In reply to Re: 3 member list ouput by AnomalousMonk
in thread 3 member list ouput by catfish1116

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