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Question

这是我的尝试：

getLineGraphic :: Point -> Point -> ColourName -> Graphic
getLineGraphic a b colourType = (Graphic (Line(a b) colourType (0,0)))

Shape由

定义

data Shape = Rectangle Side Side
       | Ellipse   Side Side
       | Polygon   [Point]
       | Line      Point Point
deriving (Show)

和ColourName是

data ColourName
= Magenta
| Black
| Green
| Yellow
| Orange
| Cyan
deriving (Show)

这是我得到的错误：

src\View.hs:54:34: error:
    * Couldn't match expected type `Graphic'
                  with actual type `ColourName -> Point -> Graphic'
    * Probable cause: `Graphic' is applied to too few arguments
      In the expression: (Graphic (Line (a b) colourType (0, 0)))
      In an equation for `getLineGraphic':
          getLineGraphic a b colourType
            = (Graphic (Line (a b) colourType (0, 0)))
   |
54 | getLineGraphic a b colourType = (Graphic (Line(a b) colourType (0,0)))

最重要的是我的尝试。不知道我哪里出错了？对于这部分，翻译必须是{​​{1}}，这就是为什么会出现这种情况。不知道为什么我会收到此错误？任何帮助，将不胜感激！感谢。

Answer 1

虽然您没有发布MCVE，但我只能猜出原因，

(Graphic (Line(a b) colourType (0,0)))

这似乎是错误的。你可能想要

(Graphic (Line a b) colourType (0,0))

说明：就像函数一样，Haskell中的数据构造函数是curry。所以

Line a b

与

相同

((Line a) b)

与

完全不同

Line (a b)