编程范式23 笔记

memory model

> '(1 2 3)

有一个1 2 3的链表 返回首地址

>(define seq '(1 2 3))
>(car seq)
1
>(cdr seq)//跳到指针域
2 3
>(cons '(1 2 3)'(4 5 6))

cons的memory model

指向数据域两次的list

((lambda(x)(cons x x))'(1 2))

泛型map

>(map car '((1 2)(3 4)(5 6 7)))
(1 3 5)
>(map + '(1 2)'(10 20)'(100 400))
(111 422)

(define (unary-map fn seq)
    (if(null?seq)()
        (cons(fn(car seq))
            (unary-map fn
                (cdr sq)))))

(define (mymap fn first-list . other-lists)
    （if(null?first-list)'()
        (cons (apply fn
                (cons (car first-list)
                        (unary-map car
                            other-lists)))