sicp每日一题[2.33]

合集 - sicp(77)
1.sicp每日一题[1.42]2.sicp每日一题[1.43]3.sicp每日一题[1.44]4.sicp每日一题[1.45]5.sicp每日一题[1.46]6.sicp每日一题[2.1]7.sicp每日一题[2.2]8.sicp每日一题[2.3]9.sicp每日一题[2.4]10.sicp每日一题[2.5]11.sicp每日一题[2.6]12.sicp每日一题[2.7]13.sicp每日一题[2.8]14.sicp每日一题[2.9]15.sicp每日一题[2.11]16.sicp每日一题[2.10]17.sicp每日一题[2.12]18.sicp每日一题[2.13-2.16]19.sicp每日一题[2.17]20.sicp每日一题[2.18]21.sicp每日一题[2.19]22.sicp每日一题[2.20]23.sicp每日一题[2.21]24.sicp每日一题[2.22-2.23]25.sicp每日一题[2.24-2.27]26.sicp每日一题[2.28]27.sicp每日一题[2.29]28.sicp每日一题[2.30]29.sicp每日一题[2.31]30.sicp每日一题[2.32]
31.sicp每日一题[2.33]
32.sicp每日一题[2.34]33.sicp每日一题[2.35]34.sicp每日一题[2.36-2.37]35.sicp每日一题[2.38-2.39]36.sicp每日一题[2.40]37.sicp每日一题[2.41]38.sicp每日一题[2.42]39.sicp每日一题[2.43]40.sicp每日一题[2.44]41.sicp每日一题[2.45]42.sicp每日一题[2.46]43.sicp每日一题[2.47]44.sicp每日一题[2.48]45.sicp每日一题[2.49]46.sicp每日一题[2.50]47.sicp每日一题[2.51]48.sicp每日一题[2.52]49.sicp每日一题[2.53-2.55]50.sicp每日一题[2.56]51.sicp每日一题[2.57]52.sicp每日一题[2.58]53.sicp每日一题[2.59]54.sicp每日一题[2.60]55.sicp每日一题[2.61]56.sicp每日一题[2.62]57.sicp每日一题[2.63-2.64]58.sicp每日一题[2.65]59.sicp每日一题[2.66]60.sicp每日一题[2.67-2.68]61.sicp每日一题[2.69]62.sicp每日一题[2.70]63.sicp每日一题[2.71]64.sicp每日一题[2.72]65.sicp每日一题[2.73]66.sicp每日一题[2.74]67.sicp每日一题[2.75]68.sicp每日一题[2.76]69.sicp每日一题[2.77]70.sicp每日一题[2.78]71.sicp每日一题[2.79]72.sicp每日一题[2.80]73.sicp每日一题[2.81]74.sicp每日一题[2.82]75.sicp每日一题[2.83]76.sicp每日一题[2.84]77.sicp每日一题[2.85-2.86]
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Exercise 2.33

Fill in the missing expressions to complete the following definitions of some basic list-manipulation operations as accumulations:

; p 表示一个函数，sequence 表示一个列表
; 这个函数将对列表中每一个元素进行 p 操作
(define (map p sequence)
  (accumulate (lambda (x y) <??>) nil sequence))

(define (append seq1 seq2)
  (accumulate cons <??> <??>))

(define (length sequence)
  (accumulate <??> 0 sequence))

这道题还是有一定难度的，尤其是第一个 (map p sequence)，我开始没明白它的意思，不知道 lambda 两个参数要干嘛，所以先做的后面2个，做到第三个的时候才发现是 accumulate 函数中 op 函数需要2个参数，明白了这一点，剩下的就比较好处理了，x 就表示当前要处理的项，y 就是原函数中的递归项。

(define (map p sequence)
  (accumulate
   (lambda (x y) (if (null? x) y (cons (p x) y)))
   nil
   sequence))

(define (append seq1 seq2)
  (accumulate cons seq2 seq1))

(define (length sequence)
  (accumulate
   (lambda (x y) (if (null? x) y (+ y 1)))
   0
   sequence))


(define odds (list 1 3 5 7 9))
(define evens (list 2 4 6 8 10))

(map square odds)
(map sqrt evens)
(append odds evens)
(length odds)

; 执行结果
'(1 9 25 49 81)
'(1.4142156862745097 2.0000000929222947 2.4494897427875517 2.8284271250498643 3.162277665175675)
'(1 3 5 7 9 2 4 6 8 10)
5
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