Exercise 4.51 · 习题

Exercise 4.51: Implement a new kind of
assignment called permanent-set! that is not undone upon failure. For
example, we can choose two distinct elements from a list and count the number
of trials required to make a successful choice as follows:

(define count 0)
(let ((x (an-element-of '(a b c)))
(y (an-element-of '(a b c))))
(permanent-set! count (+ count 1))
(require (not (eq? x y)))
(list x y count))

;;; Starting a new problem
;;; Amb-Eval value:
(a b 2)

;;; Amb-Eval input:
try-again

;;; Amb-Eval value:
(a c 3)

What values would have been displayed if we had used set! here

rather than permanent-set!?

练习 4.51：实现一种新的赋值形式，称为 permanent-set!，它在失败时不会被撤销。例如，我们可以从一个表中选取两个不同的元素，并统计成功完成选取所需的尝试次数，如下所示：

(define count 0)
(let ((x (an-element-of '(a b c)))
(y (an-element-of '(a b c))))
(permanent-set! count (+ count 1))
(require (not (eq? x y)))
(list x y count))

;;; Starting a new problem
;;; Amb-Eval value:
(a b 2)

;;; Amb-Eval input:
try-again

;;; Amb-Eval value:
(a c 3)

如果在这里使用 set! 而不是 permanent-set!，会显示什么值？