SICP Exercise 3.35

sicp

Question

Ben Bitdiddle tells Louis that one way to avoid the trouble in Exercise 3.34 is to define a squarer as a new primitive constraint. Fill in the missing portions in Ben’s outline for a procedure to implement such a constraint:

(define (squarer a b)
  (define (process-new-value)
    (if (has-value? b)
        (if (< (get-value b) 0)
            (error "square less than 0:
                    SQUARER"
                   (get-value b))
            ⟨alternative1⟩)
        ⟨alternative2⟩))
  (define (process-forget-value) ⟨body1⟩)
  (define (me request) ⟨body2⟩)
  ⟨rest of definition⟩
  me)

Answer

(define (squarer a b)
  (define (process-new-value)
    (if (has-value? b)
        (if (< (get-value b) 0)
            (error "square less than 0:
                    SQUARER"
                   (get-value b))
            (set-value! a (sqrt (get-value b)) me))
        (set-value! b (* a a) me)))
  (define (process-forget-value)
    (forget-value! A 'user)
    (forget-value! B 'user))
  (define (me request)
    (cond ((eq? request 'I-have-a-value)
           (process-new-value))
          ((eq? request 'I-lost-my-value)
           (process-forget-value))
          (else (error "Unknown request:
                        SQUARER" request))))
  (connect a me)
  (connect b me)
  me)