【SICP练习】149 练习4.5

2015/09/08 21:49
阅读数 5

练习4-5

原文

Exercise 4.5. Scheme allows an additional syntax for cond clauses, ( => ). If evaluates to a true value, then is evaluated. Its value must be a procedure of one argument; this procedure is then invoked on the value of the , and the result is returned as the value of the cond expression. For example

(cond ((assoc 'b '((a 1) (b 2))) => cadr) (else false))

returns 2. Modify the handling of cond so that it supports this extended syntax.

分析

代码

 (define (extended-cond-syntax? clause) (eq? (cadr clause) '=>)) 
 (define (extended-cond-test clause) (car clause)) 
 (define (extended-cond-recipient clause) (caddr clause)) 
 (define (cond->if expr) (expand-clauses (cond-clauses expr))) 

 (define (expand-clauses clauses) (if (null? clauses) 'false (let ((first (car clauses)) (rest (cdr clauses))) (cond ((cond-else-clause? first) (if (null? rest) (sequence->exp (cond-actions first)) (error "ELSE clause isn't last -- COND->IF" clauses))) ((extended-cond-syntax? first) (make-if (extended-cond-test first) (list (extended-cond-recipient first) (extended-cond-test first)) (expand-clauses rest))) (else (make-if (cond-predicate first) (sequence->exp (cond-actions first)) (expand-clauses rest))))))) 



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