2.24-2.26没什么代码量，所以跟 2.27 一起发吧。
Exercise 2.24
Suppose we evaluate the expression
(list 1 (list 2 (list 3 4))). Give the result printed by the interpreter, the corresponding box-and-pointer structure, and the interpretation of this as a tree (as in Figure 2.6).
解释器执行结果为：'(1 (2 (3 4)))
另外两种表示方法见下图：
Exercise 2.25
Give combinations of cars and cdrs that will pick 7 from each of the following lists:
(1 3 (5 7) 9)
((7))
(1 (2 (3 (4 (5 (6 7))))))
纯看耐心。。
; 先创建这些要求的列表
(list 1 3 (list 5 7) 9)
(list (list 7))
(list 1 (list 2 (list 3 (list 4 (list 5 (list 6 7)))))); 根据它们的结构依次定位到 7
(car (cdr (car (cdr (cdr (list 1 3 (list 5 7) 9))))))
(car (car (list (list 7))))
(car (cdr (car (cdr (car (cdr (car (cdr (car (cdr (car (cdr (list 1 (list 2 (list 3 (list 4 (list 5 (list 6 7)))))))))))))))))); 执行结果
'(1 3 (5 7) 9)
'((7))
'(1 (2 (3 (4 (5 (6 7))))))
7
7
7
Exercise 2.26
Suppose we define x and y to be two lists:
(define x (list 1 2 3))
(define y (list 4 5 6))
What result is printed by the interpreter in response to evaluating each of the following expressions:
(append x y)
(cons x y)
(list x y)
结果如下所示：
'(1 2 3 4 5 6)
'((1 2 3) 4 5 6)
'((1 2 3) (4 5 6))
Exercise 2.27
Modify your reverse procedure of Exercise 2.18 to produce a deep-reverse procedure that takes a list as argument and returns as its value the list with its elements reversed and with all sublists deep-reversed as well. For example,
(define x (list (list 1 2) (list 3 4)))
x
((1 2) (3 4))
(reverse x)
((3 4) (1 2))
(deep-reverse x)
((4 3) (2 1))
这道题难度还是挺大的，我开始想用 reverse 实现，比起原来的 reverse 函数，deep-reverse 需要在 items 不为空的情况下，额外判断
(car items)是否为 pair，虽然也能做，但是步骤有点麻烦，最后在网上发现一个使用 append 的版本，程序非常简单，而且逻辑也很清楚。
(define (deep-reverse-by-reverse items) (define (iter items result) (if (null? items) result (if (pair? (car items)) (let ((x (iter (car items) nil))) (iter (cdr items) (cons x result))) (iter (cdr items) (cons (car items) result))))) (iter items nil))(define (deep-reverse items)(if (pair? items)(append (deep-reverse (cdr items))(list (deep-reverse (car items))))items))(define x (list (list 1 2) (list 3 4)))x
(reverse x)
(deep-reverse x)
