Question

我正在尝试在Scheme中实现反向跟踪搜索。到目前为止，我有以下内容：

(define (backtrack n graph assignment)  
    (cond (assignment-complete n assignment) (assignment) )

    (define u (select-u graph assignment))

    (define c 1)
    (define result 0)

    (let forLoop ()
        (when (valid-choice graph assignment c)
             (hash-set! assignment u c)

             (set! result (backtrack n graph assignment))

             (cond ((not (eq? result #f)) result))

             (hash-remove! assignment u)            
        )

        (set! c (+ c 1))
        (when (>= n c) (forLoop))
    )

   #f ; I believe this is where I'm having problems
)

我的函数assignment-complete和select-u通过单元测试。参数赋值是一个带有（make-hash）的哈希表make，所以应该没问题。

我认为我遇到的问题与在循环结束时返回false有关，如果没有递归返回非假值（应该是有效赋值）。是否有一个Scheme等效的显式返回语句？

Answer 1

您的问题的答案是是：

(define (foo ...)
  (call-with-current-continuation
    (lambda (return)
      ...... ; here anywhere inside any sub-expression 
      ...... ; you can call (return 42)
      ...... ; to return 42 from `foo` right away
    )))

这会设置退出 continuation，以便您可以从函数体内返回结果值。通常的Scheme方法是将返回表单作为最后一个表单，因此返回其值：

    (let forLoop ()
        (when (valid-choice graph assignment c)
             (hash-set! assignment u c)
             (set! result (backtrack n graph assignment))
             (cond
                 ((not (eq? result #f))
                   result))       ; the value of `cond` form is ignored
             (hash-remove! assignment u))
                                  ; the value of `when` form is ignored
        (set! c (+ c 1))
        (if (>= n c)     ; `if` must be the last form 
           (forLoop)     ; so that `forLoop` is tail-recursive
           ;; else:
           return-value) ; <<------ the last form's value 
    )                    ; is returned from `let` form

   ;; (let forLoop ...) must be the last form in your function
   ;;                   so its value is returned from the function
)

你这里也有问题：

(cond (assignment-complete n assignment) (assignment) )

此代码不拨打电话(assignment-complete n assignment)。相反，它检查变量assignment-complete是否具有非空值，如果不是，则检查assignment变量，但无论如何它的返回值都会被忽略。也许在那里缺少一些括号，和/或else条款。

如何在Scheme中返回循环结束时的默认值？

1 个答案: