Haskell Boomerang编译错误

所以我觉得我会用Boomerang解析一些AIS数据有一些乐趣，而且我在第一个障碍上磕磕绊绊。编译错误令人困惑。看到我在Boomerang解析类似的东西之前，我试图解决这个问题。Haskell Boomerang编译错误

该库很简单。我定义一些基本类型和它们的解析器/语法：

import   Control.Category  (id, (.)) 
import   Control.Monad   (forever) 
import   Prelude    hiding (id, (.)) 
import   System.IO    (hFlush, stdout) 
import   Text.Boomerang 
import   Text.Boomerang.String 
import   Text.Boomerang.TH 

data MessageType = AIVDM | AIVDO deriving (Enum, Eq, Show) 

data AIS = AIS { 
       msgType :: MessageType 
      } deriving (Eq, Show) 

$(makeBoomerangs ''MessageType) 
$(makeBoomerangs ''AIS) 

messageTypeP :: StringBoomerang() (MessageType :-()) 
messageTypeP = rAIVDM . "!AIVDM" <> rAIVDO . "!AIVDO" 

aisP :: StringBoomerang() (AIS :-()) 
aisP = rAIS . messageTypeP . lit ","

我现在希望支持的句子计数值，其自带的消息类型后;我添加Int到AIS：

data AIS = AIS { 
       msgType :: MessageType, sCount :: Int 
      } deriving (Eq, Show)

和更改解析器/打印机：

aisP :: StringBoomerang() (AIS :-()) 
aisP = rAIS . messageTypeP . lit "," . int

，但它无法编译：

• Couldn't match type ‘()’ with ‘Int :-()’ 
    Expected type: Boomerang 
        StringError String() (MessageType :- (Int :-())) 
    Actual type: Boomerang StringError String() (MessageType :-()) 
• In the second argument of ‘(.)’, namely 
    ‘messageTypeP . lit "," . int’ 
    In the expression: rAIS . messageTypeP . lit "," . int 
    In an equation for ‘aisP’: 
     aisP = rAIS . messageTypeP . lit "," . int

哎哟。请帮助？

来源

2017-05-05 λlεx

回飞镖应该是多态的。

messageTypeP :: StringBoomerang r (MessageType :- r) 

aisP :: StringBoomerang r (AIS :- r)

解释是，r是类型的叠层，和飞镖弹出/推从类型/进去。将r设置为()将强制输入堆栈为空，这会伤害这些回飞镖的重用性。

来源

2017-05-05 13:59:05

完美。谢谢：） –

Haskell Boomerang编译错误

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