# [2021-Fall] Lab10 of CS61A of UCB

## Q2: Over or Under

Define a procedure over-or-under which takes in a number num1 and a number num2 and returns the following:

• -1 if num1 is less than num2
• 0 if num1 is equal to num2
• 1 if num1 is greater than num2

Challenge: Implement this in 2 different ways using if and cond!

(define (over-or-under num1 num2)
'YOUR-CODE-HERE
)


1. (if <predicate> <consequent> <alternative>)
2. (cond (<condition> <consequent>) ...)
(define (over-or-under num1 num2)
(if (< num1 num2)
(print -1))
(if (= num1 num2)
(print 0))
(if (> num1 num2)
(print 1))
)

(define (over-or-under num1 num2)
(cond ( (< num1 num2) (print -1) )
( (= num1 num2) (print 0)  )
( (> num1 num2) (print 1)  ))
)


Write the procedure make-adder which takes in an initial number, num, and then returns a procedure. This returned procedure takes in a number inc and returns the result of num + inc.

Hint: To return a procedure, you can either return a lambda expression or define another nested procedure. Remember that Scheme will automatically return the last clause in your procedure.

You can find documentation on the syntax of lambda expressions in the 61A scheme specification!

(define (make-adder num)
(lambda (inc) (+ num inc))
)


### Q4: Compose

Write the procedure composed, which takes in procedures f and g and outputs a new procedure. This new procedure takes in a number x and outputs the result of calling f on g of x.

(define (composed f g)
(lambda (x) (f (g x) ) )
)


### Q5: Make a List

In this problem you will create the list with the following box-and-pointer diagram:

Challenge: try to create this list in multiple ways, and using multiple list constructors!要求

1. cons 的方式, 这个方式很容易眼花, 最好是写完之后在这里 验证一下. 这里我真的写得头有点晕 😢
2. list 的方式, 这个代码会比较短, 注意我们每次在调用 (list ...) 相当于在链表中多创建了一个方向
(define lst
(cons (cons 1 nil)
(cons 2 (cons (cons 3 (cons 4 nil))
(cons 5 nil))))
)

(define lst
(list (list 1)
2 (list 3 4)
5)
)


### Q6: Remove

Implement a procedure remove that takes in a list and returns a new list with all instances of item removed from lst. You may assume the list will only consist of numbers and will not have nested lists.

Hint: You might find the built-in filter procedure useful (though it is definitely possible to complete this question without it).

You can find information about how to use filter in the 61A Scheme builtin specification!

1. 当前节点的值 = item, 我们删除它, 递归处理子链表
2. 当前节点的值 != item, 我们保留它, 递归处理子链表
(define (remove item lst)
(cond ( (null? lst) '() )             ; base case
( (= item (car lst)) (remove item (cdr lst)))           ; exclude item
( else (cons (car lst) (remove item (cdr lst)))))       ; include item
)

posted @ 2022-02-27 12:52  MartinLwx  阅读(125)  评论(3编辑  收藏  举报