$numb=@_;, since the left hand side is a scalar, automatically looks to the scalar invokation of the right-hand side, which when it is an array, will be the size of that array. Which is why you got '1' back.

$numb=$_[0] returns the scalar in the 1st position of the @_ array, which is what you wanted.

You could have also done ($numb)=@_;, which , since the LHS is now an array, does an array evaluation of the RHS, and thus would set $numb to the first scalar in @_.

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Dr. Michael K. Neylon - mneylon-pm@masemware.com || "You've left the lens cap of your mind on again, Pinky" - The Brain
It's not what you know, but knowing how to find it if you don't know that's important