【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))))))) 感谢您的访问，希望对您有所帮助。 欢迎大家关注或收藏、评论或点赞。
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