Haskell回跳编译错误

Question

所以我想我会有一些有趣的分析一些AIS数据与回旋镖，我是绊脚石在第一个障碍。编译错误令人困惑。在我试图解决这个问题之前，我已经解析了类似的东西。

图书馆很简单。我定义了一些基本类型及其解析器/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 ","

现在，我希望支持在消息类型之后出现的句子计数值；我将一个Int添加到AIS

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

并更改解析器/打印机：

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

但它未能编译：

• 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

唉哟。救命求你了？

Answer 1

回旋镖应该是多态的。

messageTypeP :: StringBoomerang r (MessageType :- r)

aisP :: StringBoomerang r (AIS :- r)

其解释是，r是一堆类型，而回旋镖则是从/推到它的pop/push类型。将r设置为()强制输入堆栈为空，这会损害这些回旋镖的可重用性。