【SICP练习】111 练习3.24

2015/09/08 21:52
阅读数 7

练习3-24

原文

Exercise 3.24. In the table implementations above, the keys are tested for equality using equal? (called by assoc). This is not always the appropriate test. For instance, we might have a table with numeric keys in which we don’t need an exact match to the number we’re looking up, but only a number within some tolerance of it. Design a table constructor make-table that takes as an argument a same-key? procedure that will be used to test “equality” of keys. Make-table should return a dispatch procedure that can be used to access appropriate lookup and insert! procedures for a local table.

代码

(define (make-table same-key?) (let ((local-table (list '*table*))) (define (lookup key-1 key-2) (let ((subtable (assoc key-1 (cdr local-table)))) (if subtable (let ((record (assoc key-2 (cdr subtable))) (if record (cdr record) false)) false))) (define (insert! key-1 key-2 value) (let ((subtable (assoc key-1 (cdr local-table)))) (if subtable (let ((record (assoc key-2 (cdr subtable)))) (if record (set-cdr! record value) (set-cdr! subtable (cons (key-2 value) (cdr subtable))))) (set-cdr! local-table (cons (list key-1 (cons key-2 value)) (cdr local-table))))) 'ok) (define (assoc key records) (cond ((null? records) false) ((same-key? key (caar records)) (car records)) (else (assoc key (cdr records))))) (define (dispatch m) (cond ((eq? m 'lookup-proc) lookup) ((eq? m 'insert-proc!) insert!) (else (error "Unknown operation -- TABLE" m)))) dispatch))



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