Haskell Prelude.read：不解析字符串 [英] Haskell Prelude.read: no parse String

问题描述

来自haskell示例 http://learnyouahaskell.com/types-and-typeclasses

  ghci>阅读5:: Int 
 5 
 ghci>阅读5:: Float 
 5.0 
 ghci> （读取5:: Float）* 4 
 20.0 
 ghci>阅读[1,2,3,4]:: [Int] 
 [1,2,3,4] 
 ghci>阅读（3，'a'）::（Int，Char）
（3，'a'）
  

但是当我尝试时

 阅读asdf:: String 
  
 
 
 
 $ b $ p code读取asdf :: [Char] 
  

我得到异常

< blockquote>

Prelude.read No Parse

我在这里做错了什么？

解决方案

这是因为您拥有的字符串表示不是 String 的字符串表示形式，它需要引用嵌入字符串本身：

 >阅读\asdf \:: String 
asdf
  

是这样读取。显示=== id 用于字符串：

 >显示asdf
\asdf \
>阅读$ showasdf:: String 
asdf
  

使用 Text.Read 中的 readMaybe 函数总是一个好主意：

 > ：t readMaybe 
 readMaybe :: Read a =>字符串 - >也许一个
> readMaybeasdf:: Maybe String 
 Nothing 
> readMaybe\asdf \:: Maybe String 
只是asdf
  

这可以避免（在我看来）破坏读取函数，这会在解析失败时引发异常。

from haskell examples http://learnyouahaskell.com/types-and-typeclasses

ghci> read "5" :: Int  
5  
ghci> read "5" :: Float  
5.0  
ghci> (read "5" :: Float) * 4  
20.0  
ghci> read "[1,2,3,4]" :: [Int]  
[1,2,3,4]  
ghci> read "(3, 'a')" :: (Int, Char)  
(3, 'a')  

but when I try

read "asdf" :: String 

or

read "asdf" :: [Char]

I get exception

Prelude.read No Parse

What am I doing wrong here?

解决方案

This is because the string representation you have is not the string representation of a String, it needs quotes embedded in the string itself:

> read "\"asdf\"" :: String
"asdf"

This is so that read . show === id for String:

> show "asdf"
"\"asdf\""
> read $ show "asdf" :: String
"asdf"

As a side note, it's always a good idea to instead use the readMaybe function from Text.Read:

> :t readMaybe
readMaybe :: Read a => String -> Maybe a
> readMaybe "asdf" :: Maybe String
Nothing
> readMaybe "\"asdf\"" :: Maybe String
Just "asdf"

This avoids the (in my opinion) broken read function which raises an exception on parse failure.

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