[英]Scheme: Avoid to invoke a procedure in order to grab a continuation?
所以我在下面的代碼:
(define escape
(lambda ()
(set! halt (call/cc (lambda (k) k)))
0))
(define multiply
(lambda (l)
(if (null? l)
1
(if (zero? (car l))
(halt halt)
(* (car l) (multiply (cdr l)))))))
並且我不得不重新設計“乘”以避免“逃脫”並獲取延續的值,以便“暫停”返回答案的值。我想出的解決方案看起來如何?
(define multiply
(lambda ()
(let ((result (call/cc (lambda (k) (set! halt k) '()))))
(if (procedure? halt)
(tester
((multiply (lambda (k)
(if (= k 0) (halt k)
(multiply (* k 1))))))
(result 1))
halt))))
您是否只是在嘗試使用call/cc
編寫返回早期的乘法函數?
(define multiply
(lambda (l)
(call/cc
(lambda (k)
(let loop ([l l])
(cond
[(null? l) 1]
[(= 0 (car l)) (k 0)]
[else (* (car l) (multiply (cdr l)))]))))))
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