spe*_*ool 3 scheme sicp racket
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)
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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)
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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))
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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.