Haskell Boomerang編譯錯誤

所以我覺得我會用Boomerang解析一些AIS數據有一些樂趣，而且我在第一個障礙上磕磕絆絆。編譯錯誤令人困惑。看到我在Boomerang解析類似的東西之前，我試圖解決這個問題。Haskell Boomerang編譯錯誤

該庫很簡單。我定義一些基本類型和它們的解析器/語法：

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 ","

我現在希望支持的句子計數值，其自帶的消息類型後;我添加Int到AIS：

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

和更改解析器/打印機：

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

，但它無法編譯：

• 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

哎喲。請幫助？

來源

2017-05-05 λlεx

回飛鏢應該是多態的。

messageTypeP :: StringBoomerang r (MessageType :- r) 

aisP :: StringBoomerang r (AIS :- r)

解釋是，r是類型的疊層，和飛鏢彈出/推從類型/進去。將r設置爲()將強制輸入堆棧爲空，這會傷害這些回飛鏢的重用性。

來源

2017-05-05 13:59:05

完美。謝謝：） –

Haskell Boomerang編譯錯誤

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