Question

你能看出这有什么问题：

(define (box d x1 y1 x2 y2)  (
                              (graphics-draw-line d x1 y1 x1 y2)
                              (graphics-draw-line d x1 y2 x2 y2)
                              (graphics-draw-line d x2 y2 x2 y1)
                              (graphics-draw-line d x2 y1 x1 y1) ))

当我这样称呼时：

( begin 
     (define w (make-graphics-device 'x))
     (box  w .10 .10 .20 .20) )

我得到一个错误：

;The object #!unspecific is not applicable.
;To continue, call RESTART with an option number:
; (RESTART 2) => Specify a procedure to use in its place.
; (RESTART 1) => Return to read-eval-print level 1.

2 error>

这有效：

(begin
    (define w (make-graphics-device 'x))
    (graphics-draw-line w .1 .1 .1 .2)
    (graphics-draw-line w .1 .2 .2 .2)
    (graphics-draw-line w .2 .2 .2 .1)
    (graphics-draw-line w .2 .1 .1 .1) )

我看不出差异！

Answer 1

不要只用() s对表达式进行分组 - 它会尝试将第一个结果用作函数，但值为#!unspecific - 绝对不是函数。

使用此：

(define (box d x1 y1 x2 y2)
  (graphics-draw-line d x1 y1 x1 y2)
  (graphics-draw-line d x1 y2 x2 y2)
  (graphics-draw-line d x2 y2 x2 y1)
  (graphics-draw-line d x2 y1 x1 y1))

使用图形绘制线定义过程

1 个答案: