Haskell：由于Regex库的普通用法而导致'没有实例'

问题描述

遵循的（接受）答案，我期待以下工作：

  Prelude Text.Regex.Posix Text.Regex.Base.RegexLike Text.Regex.Posix.String> makeRegex。*
  

（ makeRegex 是 makeregexOpts 带有预定义选项的快捷方式）

然而，它并没有：

 < interactive>：1：0：
没有用于（RegexMaker正则表达式compOpt execOpt [Char]）
的实例< interactive> 1：0-13 
可能的修正：
为
添加实例声明（RegexMaker正则表达式compOpt execOpt [Char]）
在&表达式：makeRegex。*
在`it'的定义中：it = makeRegex。*
 Prelude Text.Regex.Posix Text.Regex.Base.RegexLike Text.Regex.Posix.String> ; make 
 Regex。*:: Regex 
 
< interactive>：1：0：
（RegexMaker正则表达式compOpt execOpt [Char]）
在< interactive> 1：0-13 
处使用`makeRegex'引起的可能的修正：
为
添加实例声明（RegexMaker Regex compOpt execOpt [Char]）
在表达式中：makeRegex。*:: Regex 
在`it'的定义中：it = makeRegex。*:: Regex 
  

我真的不明白为什么。

编辑

Haskell Platform 2009.02.02（GHC 6.10.4）on Windows

EDIT2

  Prelude Text.Regex.Base.RegexLike Text.Regex.Posix.String> ：i RegexMaker 
 class（RegexOptions regex compOpt execOpt）=> RegexMaker正则表达式compOpt execOpt源|正则表达式 - > compOpt execOpt，compOpt  - >正则表达式execOpt，execOpt  - >正则表达式compOpt其中
 makeRegex :: source  - >正则表达式
 makeRegexOpts :: compOpt  - > execOpt  - >来源 - >正则表达式
 makeRegexM ::（Monad m）=>来源 - > m正则表达式
 makeRegexOptsM :: 
（Monad m）=> compOpt  - > execOpt  - >来源 - > m regex 
  - 在Text.Regex.Base.RegexLike中定义
  

解决方案

您的第一次尝试不起作用，因为makeRegex具有多态返回类型（称为 regex ）。由于RegexMaker没有用于任何类型的实例，因此您会收到消息。

为了使它工作，您需要指定返回类型。看起来你已经想出了自己的想法，因为这就是你在第二次尝试时所做的事情，当我在我的ghci中尝试时偶然发生了。

编辑：我应该添加使用regexen的最直接的方法是只使用 =〜〜，而不用为makeRegex而烦恼。例如：

 > lale=〜。*:: Bool 
 True 
 
> lale=〜lo:: Bool 
 False 
 
> lale=〜l。 :: String 
la
 
> lale=〜l。 :: [String] 
 [la，le] 
  

Following the (accepted) answer from this question, I am expecting the following to work:

Prelude Text.Regex.Posix Text.Regex.Base.RegexLike Text.Regex.Posix.String> makeRegex ".*"

(makeRegex is a shortcut for makeRegexOpts with predefined options)

However, it doesn't:

<interactive>:1:0:
    No instance for (RegexMaker regex compOpt execOpt [Char])
      arising from a use of `makeRegex' at <interactive>:1:0-13
    Possible fix:
      add an instance declaration for
      (RegexMaker regex compOpt execOpt [Char])
    In the expression: makeRegex ".*"
    In the definition of `it': it = makeRegex ".*"
Prelude Text.Regex.Posix Text.Regex.Base.RegexLike Text.Regex.Posix.String> make
Regex ".*"::Regex

<interactive>:1:0:
    No instance for (RegexMaker Regex compOpt execOpt [Char])
      arising from a use of `makeRegex' at <interactive>:1:0-13
    Possible fix:
      add an instance declaration for
      (RegexMaker Regex compOpt execOpt [Char])
    In the expression: makeRegex ".*" :: Regex
    In the definition of `it': it = makeRegex ".*" :: Regex

And I really don't understand why.

EDIT

Haskell Platform 2009.02.02 (GHC 6.10.4) on Windows

EDIT2

Prelude Text.Regex.Base.RegexLike Text.Regex.Posix.String> :i RegexMaker
class (RegexOptions regex compOpt execOpt) => RegexMaker regex compOpt execOpt source | regex -> compOpt execOpt, compOpt -> regex execOpt, execOpt -> regex compOpt where
  makeRegex :: source -> regex
  makeRegexOpts :: compOpt -> execOpt -> source -> regex
  makeRegexM :: (Monad m) => source -> m regex
  makeRegexOptsM ::
    (Monad m) => compOpt -> execOpt -> source -> m regex
        -- Defined in Text.Regex.Base.RegexLike

解决方案

Your first try does not work because makeRegex has a polymorphic return type (called regex). Since there is no instance of RegexMaker for abitrary types, you get the message you do.

To make it work you need to specify a return type. It seems like you figured that out yourself because that's what you did in your second try, which incidentally works when I try it out in my ghci.

Edit: I should add that the most straight-forward way to use regexen is to just use =~ and not bother with makeRegex at all. For example:

> "lale" =~ ".*" :: Bool
True

> "lale" =~ "lo" :: Bool
False

> "lale" =~ "l." :: String
"la"

> "lale" =~ "l." :: [String]
["la","le"]

这篇关于Haskell：由于Regex库的普通用法而导致'没有实例'的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！