memoized parameters: call by need

puzzle

• how will this procedure behave
  (define (doubler x)(if (zero? x) (display ‘done)  (begin (display (* 2 x)) (doubler x))))
  
• when applied like this?
  (doubler (read))
  

not always what we want…

• consider absolute value procedure
  (define (abs i) (if (< i 0) (- i) i))
  
• does (abs …) always evaluate to a non-negative number? no!
  (define (down) (begin (set! x (- x 2)) x))
  
  (define x 3)
  (abs (down))
• so want to memoize the value of the argument

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