我有一个长期项目正在完成SICP的所有练习.我注意到最近的练习有些奇怪.我正在测试一个霍夫曼编码树.当我在DrScheme中执行以下代码时,我得到了预期的结果:
(a d a b b c a)
但是,如果我通过调用(加载"2.67.scm")或运行mzscheme -f 2.67.scm在mzscheme中执行相同的代码,它会报告:
symbols: expected symbols as arguments, given: (leaf D 1)
我的问题是:为什么?是因为mzscheme和drscheme使用不同的规则来加载程序定义吗?程序代码如下.
;; Define an encoding tree and a sample message ;; Use the decode procedure to decode the message, and give the result. (define (make-leaf symbol weight) (list 'leaf symbol weight)) (define (leaf? object) (eq? (car object) 'leaf)) (define (symbol-leaf x) (cadr x)) (define (weight-leaf x) (caddr x)) (define (make-code-tree left right) (list left right (append (symbols left) (symbols right)) (+ (weight left) (weight right)))) (define (left-branch tree) (car tree)) (define (right-branch tree) (cadr tree)) (define (symbols tree) (if (leaf? tree) (list (symbol-leaf tree)) (caddr tree))) (define (weight tree) (if (leaf? tree) (weight-leaf tree) (cadddr tree))) (define (decode bits tree) (define (decode-1 bits current-branch) (if (null? bits) '() (let ((next-branch (choose-branch (car bits) current-branch))) (if (leaf? next-branch) (cons (symbol-leaf next-branch) (decode-1 (cdr bits) tree)) (decode-1 (cdr bits) next-branch))))) (decode-1 bits tree)) (define (choose-branch bit branch) (cond ((= bit 0) (left-branch branch)) ((= bit 1) (right-branch branch)) (else (error "bad bit -- CHOOSE-BRANCH" bit)))) (define (test s-exp) (display s-exp) (newline)) (define sample-tree (make-code-tree (make-leaf 'A 4) (make-code-tree (make-leaf 'B 2) (make-code-tree (make-leaf 'D 1) (make-leaf 'C 1))))) (define sample-message '(0 1 1 0 0 1 0 1 0 1 1 1 0)) (test (decode sample-message sample-tree))
小智.. 6
默认情况下,MzScheme以一种现有定义的模式启动symbols
,并且它内联它所知道的函数 - 所以当它编译你的make-code-tree
定义时,它会使用它所知道的绑定.当它稍后编译时symbols
,它对先前的定义没有影响.
处理此问题的最简单方法是将代码添加到模块中,方法是将其添加到模块中#lang scheme
.
默认情况下,MzScheme以一种现有定义的模式启动symbols
,并且它内联它所知道的函数 - 所以当它编译你的make-code-tree
定义时,它会使用它所知道的绑定.当它稍后编译时symbols
,它对先前的定义没有影响.
处理此问题的最简单方法是将代码添加到模块中,方法是将其添加到模块中#lang scheme
.