Question

我正在尝试制作一个以各种步骤接受用户输入的程序。在程序开始时，玩家通过外部函数命名自己，该函数应该返回一个玩家对象，稍后会多次引用它。

#lang racket


(define player
 (class object%
  (init-field
   name
   money
   lives
   health)
   (super-new)

(define/public (get-money)
  money)
)
)
;None of this function runs/the user is never prompted...
(define (start-game)
  (print "What do you want to name yourself?")
  (define name (read-line))
  (define start-mon (random 1000 5000))
  (print "You're starting with ")
  (print start-mon)
  (print " dollars.")
  (new player [name name]
             [money 0]
             [lives 2]
             [health 100])
)

(define (main)
  (define player start-game)
  (let/ec break
    (let loop ()
      (print ":")
      (define input (read-line))
  (cond [(string=? input "health") (send player get-health)]
        [(string=? input "exit")  (break)]
        [else  (print "sorry, command not recognized. Type help for commands.")])
  (loop)))
  (print "exiting..."))

(main)

我的问题是，如何正确地将玩家变量分配给函数start-game的返回对象，同时让整个函数执行并提示用户？

谢谢！

如何将变量设置为从Racket中的用户输入的函数？

0 个答案: