我认为Lazy Racket对于处理无限列表应该很有用。根据维基百科Lazy Racket的文章，fibs（斐波那契数的无限列表）可以定义为：
;; An infinite list:
(define fibs (list* 1 1 (map + fibs (cdr fibs))))
我们如何定义一个无限的自然数列表？
#lang lazy
(define Nat (cons 1 (map (lambda (x) (+ x 1)) Nat)))
感谢Will Ness。
我还发现了
#lang lazy
;;; infinite sequences represented by a_(n+1) = f(a_n)
(define inf-seq (lambda (a0 f) (cons a0 (inf-seq (f a0) f))))
(define Nat (inf-seq 1 (lambda (x) (+ x 1))))
输出
(define (outputListData list)
(cond
[(null? list) #f] ; actually doesn't really matter what we return
[else
(printf "~sn" (first list)) ; display the first item ...
(outputListData (rest list))] ; and start over with the rest
)
)
(outputListData Nat)
#lang lazy
(define nats (cons 1 (map 1+ nats)))
显然这不起作用，应该用(lambda (x) (+ x 1))替换1+。感谢@kenokabe的测试。（add1是正确的名称。）