Haskell如何将Char转换成Word8 [英] Haskell How to convert Char to Word8

问题描述

我想将 ByteString 分割为如下所示的单词：

 将合格的Data.ByteString导入为BS 
 
 main = do 
 input<  -  BS.getLine 
 let xs = BS.split''input 
  

但GHC似乎无法将字符文字本身转换为 Word8 ，所以我得到了：

 无法匹配预期类型`GHC.Word.Word8'
，实际类型为` char'
在'BS.split'的第一个参数中，即''
在表达式：BS.split''中输入
  

Hoogle没有发现任何类型签名为 Char - > Word8 和 Word.Word8''是无效的类型构造函数。如何解决它的任何想法？ Data.ByteString.Char8 模块允许您将 Word8 <字节串中的code>值作为 Char 。只需



 将限定的Data.ByteString.Char8导入为C 
  

然后参考例如 C.split 。它是相同的字节串，但为方便的字节/ ASCII解析提供了 Char 函数。

I want to split ByteString to words like so:

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:

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. Any ideas on how to fix it?

解决方案

The Data.ByteString.Char8 module allows you to treat Word8 values in the bytestrings as Char. Just

import qualified Data.ByteString.Char8 as C

then refer to e.g. C.split. It's the same bytestring under the hood, but the Char-oriented functions are provided for convenient byte/ascii parsing.

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