Aagh - I did try it before I replied but like this:

print @name . "\n";
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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
[download]

(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.