Hi,

 others have explained where the problem is ($foo is the scalar foo, not
 the function foo), but...

>> Carter and Cathleen <[EMAIL PROTECTED]> writes:

 > Now when I spit this through I get:
 > 
 > + = x

 Are you sure about this output?  In particular about the "x" there.
 Since you are doing

    $x = undef;
    $y = undef;
    $z = $x + $y;

 $z is 0.

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