in reply to Re: new to perl, syntax question
in thread new to perl, syntax question

print evaluates its operands in list context, so print @name; does not print 3. (print scalar(@name); and print 0+@name; do, though)

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Re^3: new to perl, syntax question
by tangent (Parson) on Feb 20, 2012 at 13:50 UTC
    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.
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[d/l]
[select]
      Concatenation puts its arguments in scalar context, that is the array is in scalar context.
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