Haskell 如何创建 Word8？
我想编写一个简单的函数，使用 '\n' 作为分隔符将 ByteString 拆分为 [ByteString]。我的尝试：
import Data.ByteString
listize :: ByteString -> [ByteString]
listize xs = Data.ByteString.splitWith (=='\n') xs
这会引发错误，因为 '\n' 是 Char 而不是 Word8，这就是 Data.ByteString.splitWith< /a> 正在等待。
如何将这个简单的字符转换为 ByteString 可以使用的 Word8？
I want to write a simple function which splits a ByteString into [ByteString] using '\n' as the delimiter. My attempt:
import Data.ByteString
listize :: ByteString -> [ByteString]
listize xs = Data.ByteString.splitWith (=='\n') xs
This throws an error because '\n' is a Char rather than a Word8, which is what Data.ByteString.splitWith is expecting.
How do I turn this simple character into a Word8 that ByteString will play with?
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您可以只使用数字文字
10，但如果您想转换字符文字，可以使用fromIntegral (ord '\n')（fromIntegral<需要 /code> 将ord返回的Int转换为Word8）。您必须为ord导入Data.Char。您还可以导入
Data.ByteString.Char8，它提供了在相同ByteString数据类型上使用Char而不是Word8的函数。 （事实上，它有行函数完全符合您的要求。）但是，通常不推荐这样做，因为ByteString不 em> 存储 Unicode 代码点（这就是Char代表），而是原始八位字节（即Word8）。如果您正在处理文本数据，则应考虑使用
Text而不是ByteString。You could just use the numeric literal
10, but if you want to convert the character literal you can usefromIntegral (ord '\n')(thefromIntegralis required to convert theIntthatordreturns into aWord8). You'll have to importData.Charforord.You could also import
Data.ByteString.Char8, which offers functions for usingCharinstead ofWord8on the sameByteStringdata type. (Indeed, it has alinesfunction that does exactly what you want.) However, this is generally not recommended, asByteStrings don't store Unicode codepoints (which is whatCharrepresents) but instead raw octets (i.e.Word8s).If you're processing textual data, you should consider using
Textinstead ofByteString.