SICP 問題 3.24
assocをequal?以外を使ってテストできるようにする. make-table手続きはキーの等価性に使うsame-key?手続きを引数にとる.
(define (make-table same-key?) (define (assoc key value records) (cond ((null? records) #f) ((same-key? key (caar records)) (car records)) (else (assoc key (cdr records))))) (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 (cons key-2 value) (cdr subtable))))) (set-cdr! local-table (cons (list key-1 (cons key-2 value)) (cdr local-table))))) 'ok) (define (dispatch m) (cond ((eq? m 'lookup-proc) lookup) ((eq? m 'insert-proc!) insert!) (else (error "Unknown operation -- TABLE" m)))) dispatch))