Question

我无法使用derivingUnbox将类型转换为未装箱类型。我尝试了下面的代码，但是在[t | Color -> Word32 |]行

上给出错误“在输入上解析错误' - ＆gt;'”

    type Color = (Word8,Word8,Word8)


    colorToWord32 :: Color -> Word32
    colorToWord32 (r,g,b) = 0 .|. 
                           (shift (fromIntegral r) 24) .|. 
                           (shift (fromIntegral g) 16) .|. 
                           (shift (fromIntegral b) 8)

    word32ToColor :: Word32 -> Color
    word32ToColor color = (r,g,b)
        where
            r = fromIntegral (shift (color .&. 0xFF000000) (-24))
            g = fromIntegral (shift (color .&. 0x00FF0000) (-16))
            b = fromIntegral (shift (color .&. 0x0000FF00) (-8))


    derivingUnbox "Color"
        [t | Color -> Word32 |]
        colorToWord32
        word32ToColor

Answer 1

首先，让我们修复引用语法：

{-# LANGUAGE
  MultiParamTypeClasses, TemplateHaskell, TypeFamilies, FlexibleInstances #-}

derivingUnbox "Color"
    [t| Color -> Word32 |]
    [| colorToWord32 |]
    [| word32ToColor |]

其次，请注意，这仍然不对，因为unbox已经存在(Word8, Word8, Word8)个实例。如果我们要定义新的data实例，我们必须创建新的newtype或unbox。

import Data.Vector.Unboxed.Deriving
import Data.Word
import Data.Bits

data Color = Color !Word8 !Word8 !Word8

colorToWord32 :: Color -> Word32
colorToWord32 (Color r g b) = 0 .|. 
                       (shift (fromIntegral r) 24) .|. 
                       (shift (fromIntegral g) 16) .|. 
                       (shift (fromIntegral b) 8)

word32ToColor :: Word32 -> Color
word32ToColor color = Color r g b where
  r = fromIntegral (shift (color .&. 0xFF000000) (-24))
  g = fromIntegral (shift (color .&. 0x00FF0000) (-16))
  b = fromIntegral (shift (color .&. 0x0000FF00) (-8))

我们依赖unbox的预先存在的Word32实例。或者，我们可以使用unbox的{{1}}实例：

(Word8, Word8, Word8)

现在我们可以简单地使用import qualified Data.Vector.Unboxed as UV type Color = (Word8, Word8, Word8)。

请注意，元组的默认UV.Vector Color实例对字段使用多个未装箱的向量，而不是使用一个向量并将所有字段打包在一起（正如我们之前所做的那样）。这取决于查找模式，在实践中配置更快。

将类型转换为未装箱类型

1 个答案: