Aagh - I did try it before I replied but like this:
print @name . "\n";
then I took the "\n" out to make it clearer (but didn't run it!). I take it the concatenation makes print evaluate in scalar context. So should be:
print "@name"; # prints x y z print @name; # prints xyz print @name . "\n"; # prints 3
(lidden: I was still fixing this when you replied)

Update - then I found this:
If you believe in Lists in Scalar Context, Clap your Hands
I'll be back in a few weeks.

In reply to Re^3: new to perl, syntax question by tangent
in thread new to perl, syntax question by tics

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