SICP 3.52 delayed cdr

Question

Exercise 3.52,

(define sum 0)

(define (accum x)
  (set! sum (+ x sum))
  sum)

;1: (define seq (stream-map accum (stream-enumerate-interval 1 20)))
;2: (define y (stream-filter even? seq))
;3: (define z (stream-filter (lambda (x) (= (remainder x 5) 0))
;                           seq))

;4: (stream-ref y 7)
;5: (display-stream z)

Step 1: ;1: ==> (cons-stream 1 (stream-map proc (stream-cdr s)) (Assume stream-cdr is evaluated only when we force the cdr of this stream)

sum is now 1

Step 2: 1 is not even, hence (also memoized so not added again), it calls (stream-filter pred (stream-cdr stream)). This leads to evaluation of cdr hence materializing 2 which is even, hence it should call: (cons-stream 2 (stream-cdr stream)).

According to this answer should be 1+2 = 3 , but it is 6

Can someone help with why the cdr's car is materialized before the current cdr is called?

Answer 1

Using Daniel P. Friedman's memoizing tail

#lang r5rs

(define-syntax cons-stream
  (syntax-rules () 
    ((_ h t) (cons h (lambda () t)))))

(define (stream-cdr s)
  (if (and (not (pair? (cdr s)))
           (not (null? (cdr s))))
      (set-cdr! s ((cdr s))))
  (cdr s))

we observe:

> sum
0
> (define seq (stream-map accum (stream-enumerate-interval 1 20)))
> sum
1
> seq
(mcons 1 #<procedure:friedmans-tail.rkt:21:26>)
> (define y (stream-filter even? seq))
> sum
6
> seq
(mcons
 1
 (mcons
  3
  (mcons 6 #<procedure:friedmans-tail.rkt:21:26>)))
> y
(mcons 6 #<procedure:friedmans-tail.rkt:21:26>)
> 

stream-filter? needs to get to the first element of the stream it is constructing in order to construct it. A stream has its head element already forced, calculated, so it must be already present.

In the list of accumulated sums of the enumerated interval from 1 to 20, the first even number is 6:

1      = 1
1+2    = 3
1+2+3  = 6
...

Run in Racket.

To see more, name the enumerating sequence as well: (define nums (stream-enumerate-interval 1 20)), (define seq (stream-map accum nums)), and inspect nums as well as seq etc., while trying the expressions one by one.