sicp每日一题[2.84]
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]Exercise 2.84
Using the raise operation of Exercise 2.83, modify the apply-generic procedure so that it coerces its arguments to have the same type by the method of successive raising, as discussed in this section.
You will need to devise a way to test which of two types is higher in the tower. Do this in a manner that is “compatible” with the rest of the system and will not lead to problems in adding new levels to the tower.
这道题难度也不大，而且 tower 结构更简单，参考 2.81 和 2.83 再针对参数类型分别进行处理就行了
(define (apply-generic op . args)
(define (no-method type-tags)
(error "No method for these types"
(list op type-tags)))
(define (type-tags args)
(map type-tag args))
; 将当前的类型升级为上一级，如果不存在对应的转换函数，返回 false
(define (tower-raise origin target)
(let ((o-type (type-tag origin))
(t-type (type-tag target)))
(cond ((eq? o-type t-type)
origin)
((get 'raise (list o-type))
(tower-raise ((get 'raise (list o-type)) (contents origin) target)))
(else false))))
(let ((proc (get op (type-tags args))))
(if proc
(apply proc (map contents args))
(if (= (length args) 2)
;; 假设类型以简单的 tower 形式排序，从低到高
(let ((curr (car args))
(next (cadr args)))
(cond ((eq? (type-tag curr) (type-tag next))
(no-method (type-tags args)))
((tower-raise curr next)
(apply-generic op (tower-raise curr next) next))
((tower-raise curr next)
(apply-generic op (tower-raise next curr) curr))
(else (no-method (type-tags args)))))
(no-method (type-tags args))))))
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