Watch out for that comparison you do! 
$a = $b;    # assigns the value of $b to $a
$a == $b;   # compares the two numerically

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Note that your code won't fail compiling (aside from the already answered sub issues ;-)) but it won't do what you expect it would.

janx

In reply to Re: Defining Subroutines by janx
in thread Defining Subroutines by @rocks

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