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我是 Racket 的新手，正在尝试为我的编程语言课程编写 Mastermind 游戏。我已经完成了代码，它也运行了。在下面的代码中，“解决方案”是 4 种随机生成的颜色的列表。我想问如何修改我的“constructSolution”函数，以便我可以运行循环/递归而不是编写集合！4个不同的时间。我尝试像这样修改constructSolution函数：

    (define i 0)
(define (constructSolution)
  (cond
    ((if (< i 4)
         (set! solution (append solution (list (random-color))))
         (set! i (+ i 1))))
    (else #f))
)

但它没有用。理想情况下，用户将输入列表的长度，该函数将检查“（如果（< i 长度））”而不是“（如果（< i 4））”。任何帮助，将不胜感激。这是我的代码：

    #lang racket
(require racket/gui/base)

(define solution1 '(yellow red black blue))
(define solution '())
(define blackPegs 0)
(define totalPegs 0)
(define whitePegs 0)
(define countPegs 0)
(define numBlackPegs 0)
(define guessText "")
(define displayGuess "Guess ")
(define displayScore "Black Pegs = ")
(define displaySolution "Solution: ")
(define numGuess 1)


(define f (new frame% [label "Mastermind"]
                      [width 300]
                      [height 500]
                      [alignment '(center center)]))
(send f show #t)
(define c (new editor-canvas% [parent f]))
(define t (new text%))
(send c set-editor t)
(send t insert "Welcome to Mastermind!\n\n")

(define (constructSolution)
  (set! solution (append solution (list (random-color))))
  (set! solution (append solution (list (random-color))))
  (set! solution (append solution (list (random-color))))
  (set! solution (append solution (list (random-color))))
)

(define (random-color)
  (choose '(red blue green yellow orange purple black)))

(define (choose xs)
  (list-ref xs (random (length xs))))

(define (alist->string alst)
  (string-join (map symbol->string alst) " "))

(constructSolution)
(set! displaySolution (string-append displaySolution (alist->string solution)))
(send t insert displaySolution)
(send t insert "\n\n")

(define (try guess)
  (compareBlackPegs guess solution blackPegs)
  (printf "(~a, " numBlackPegs)
  (compareWhitePegs guess solution)
  (printf "~a) ~n" whitePegs)
  (set! guessText (alist->string guess))
  (set! totalPegs 0)
  (set! displayGuess (string-append displayGuess (number->string numGuess)))
  (set! displayGuess (string-append displayGuess ": "))
  (set! displayGuess (string-append displayGuess guessText))
  (set! displayGuess (string-append displayGuess "\n"))
  (send t insert displayGuess)
  (set! displayGuess "Guess ")
  (set! displayScore (string-append displayScore (number->string numBlackPegs)))
  (set! displayScore (string-append displayScore ", White Pegs = "))
  (set! displayScore (string-append displayScore (number->string whitePegs)))
  (set! displayScore (string-append displayScore "\n\n"))
  (send t insert displayScore)
  (set! displayScore "Black Pegs = ")
  (set! numGuess (+ numGuess 1))
)

(define (compareBlackPegs guess solution blackPegs)
  (if (null? guess) (set! numBlackPegs blackPegs)
      (if (equal? (car guess) (car solution)) (compareBlackPegs (cdr guess) (cdr solution) (+ blackPegs 1))
          (compareBlackPegs (cdr guess) (cdr solution) blackPegs)))
)

(define (compareWhitePegs guess solution)
  (cond ((null? solution) 
         (set! whitePegs (- totalPegs numBlackPegs)))
         ;(printf "~a) ~n" totalPegs))
        (else
         (let ((a (count (car solution) guess countPegs))
               (b (count (car solution) solution countPegs)))
         (if (equal? (< a b) #t) (set! totalPegs (+ totalPegs a))
             (set! totalPegs (+ totalPegs b))))
         (compareWhitePegs guess (cdr solution))))
)

(define (count x alist countPegs)
  (if (null? alist) countPegs
      (if (equal? x (car alist)) (count x (cdr alist) (+ countPegs 1))
          (count x (cdr alist) countPegs)))
)


(try '(red red red red))
(try '(red yellow yellow yellow))
(try '(yellow red green green))
(try '(yellow red blue blue))
(try '(yellow red blue white))
(try '(yellow red black blue))
(send t insert "\nGoodbye!\n")
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1 回答 1

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Please avoid using set! for implementing a loop in Scheme, that's not the way to go - use recursion or the built-in looping constructs. If you want to repeat a function call a fixed number of times (say, 4) here's an example of how to do it using explicit recursion:

(define (repeat f n)
  (cond ((zero? n) 'done)
        (else (f)
              (repeat f (sub1 n)))))

The above will call f a given number of times. If necessary modify it to pass additional parameters to f. For example:

(repeat (lambda () (display 'xo)) 4)
=> xoxoxoxo
   'done
于 2013-03-22T04:41:38.633 回答