【SICP练习】149 练习4.5

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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