SICP:1.41 double过程
#lang racket (define (double f) (lambda (x) (f (f x)) );lambda );double (define ((double1 f) x) (f (f x)) ) (define (inc x) (+ x 1) );inc (((double (double double))inc )5) (((double1 (double1 double1)) inc) 0 )
最后的结果:21 16
posted on 2015-04-04 16:43 Zachary_wiz 阅读(205) 评论(0) 编辑 收藏 举报