[英]Haskell How to convert Char to Word8
I want to split ByteString
to words like so:我想将
ByteString
拆分为这样的单词:
import qualified Data.ByteString as BS
main = do
input <- BS.getLine
let xs = BS.split ' ' input
But it appears that GHC can't convert a character literal to Word8
by itself, so I got:但似乎 GHC 无法自行将字符文字转换为
Word8
,所以我得到了:
Couldn't match expected type `GHC.Word.Word8'
with actual type `Char'
In the first argument of `BS.split', namely ' '
In the expression: BS.split ' ' input
Hoogle doesn't find anything with type signature of Char -> Word8
and Word.Word8 ' '
is invalid type constructor. Hoogle 没有找到任何具有
Char -> Word8
类型签名的Char -> Word8
并且Word.Word8 ' '
是无效的类型构造函数。 Any ideas on how to fix it?关于如何解决它的任何想法?
The Data.ByteString.Char8 module allows you to treat Word8
values in the bytestrings as Char
. Data.ByteString.Char8模块允许您将字节字符串中的
Word8
值视为Char
。 Just 只是
import qualified Data.ByteString.Char8 as C
then refer to eg C.split . 然后参考例如C.split 。 It's the same bytestring under the hood, but the
Char
-oriented functions are provided for convenient byte/ascii parsing. 它是相同的字节串,但提供了面向
Char
的函数,可方便地进行字节/ ASCII解析。
In case you really need Data.ByteString (not Data.ByteString.Char8), you could do what Data.ByteString itself does to convert between Word8 to Char: 如果您确实需要Data.ByteString(而不是Data.ByteString.Char8),则可以执行Data.ByteString本身所做的操作以在Word8和Char之间进行转换:
import qualified Data.ByteString as BS
import qualified Data.ByteString.Internal as BS (c2w, w2c)
main = do
input <- BS.getLine
let xs = BS.split (BS.c2w ' ') input
return ()
People looking for a simple Char -> Word8
with base library: 人们在寻找具有基本库的简单
Char -> Word8
:
import Data.Word
charToWord8 :: Char -> Word8
charToWord8 = toEnum . fromEnum
I want to directly address the question in the subject line, which led me here in the first place. 我想直接在主题栏中解决这个问题,这首先导致了我的到来。
You can convert a single Char
to a single Word8
with fromIntegral.ord
: 您可以使用
fromIntegral.ord
将单个Char
转换为单个Word8
:
λ> import qualified Data.ByteString as BS
λ> import Data.Char(ord)
λ> BS.split (fromIntegral.ord $ 'd') $ BS.pack . map (fromIntegral.ord) $ "abcdef"
["abc","ef"]
Keep in mind that this conversion will be prone to overflows as demonstrated below. 请记住,这种转换很容易发生溢出 ,如下所示。 You have to assure that your
Char
fits in 8 bits , if you do not want this to occur. 你必须确保你的
Char
符合8位 ,如果你不希望出现这种情况。
λ> 260 :: Word8
4
Of course, for your particular problem, it is preferable to use the Data.ByteString.Char8 module as already pointed out in the accepted answer. 当然,对于您的特定问题,最好使用Data.ByteString.Char8模块,该模块已在接受的答案中指出。
Another possible solution is the following:另一种可能的解决方案如下:
charToWord8 :: Char -> Word8
charToWord8 = fromIntegral . ord
{-# INLINE charToWord8 #-}
where ord :: Chat → Int
and the rest one can infer.其中
ord :: Chat → Int
和其他人可以推断。
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