Question

请考虑以下代码：

{-# LANGUAGE TemplateHaskell #-}
{-# LANGUAGE NoMonomorphismRestriction #-}

import Data.HList.GhcSyntax((.!.),(.=.),(.*.))
import Data.HList.Record(emptyRecord)
import Data.HList.TypeCastGeneric1
import Data.HList.TypeEqGeneric1
import Data.HList.Label5

data Hello1 = Hello1
data Hello2 = Hello2

record = (Hello1 .=. "Hello1") .*. (Hello2 .=. "Hello2") .*. emptyRecord

f1 = $([| (\r1 -> (r1 .!. Hello1)) |]) 

main = print $ f1 record

编译好并打印出来＆＃34; Hello1＆＃34;如预期的那样。

但是，添加以下行（GHC 7.4.1）会产生编译错误：

f2 = $([| (\r2 -> (r2 .!. Hello2)) |])

给出的错误是：

error.hs:16:1:
    Could not deduce (Data.HList.Record.HasField Hello2 r0 v0)
      arising from the ambiguity check for `main'
    from the context (Data.HList.Record.HasField Hello2 r v)
      bound by the inferred type for `main':
                 Data.HList.Record.HasField Hello2 r v => IO ()
      at error.hs:(16,1)-(20,38)
    Possible fix:
      add an instance declaration for
      (Data.HList.Record.HasField Hello2 r0 v0)
    When checking that `main'
      has the inferred type `forall r v.
                             Data.HList.Record.HasField Hello2 r v =>
                             IO ()'
    Probable cause: the inferred type is ambiguous

error.hs:16:1:
    Could not deduce (Data.HList.Record.HasField Hello2 r0 v0)
      arising from the ambiguity check for `f1'
    from the context (Data.HList.Record.HasField Hello2 r v)
      bound by the inferred type for `f1':
                 Data.HList.Record.HasField Hello2 r v =>
                 Data.HList.Record.Record
                   (Data.HList.HListPrelude.HCons
                      (Data.HList.Record.LVPair Hello1 [Char])
                      (Data.HList.HListPrelude.HCons
                         (Data.HList.Record.LVPair Hello2 [Char])
                         Data.HList.HListPrelude.HNil))
                 -> [Char]
      at error.hs:(16,1)-(20,38)
    Possible fix:
      add an instance declaration for
      (Data.HList.Record.HasField Hello2 r0 v0)
    When checking that `f1'
      has the inferred type `forall r v.
                             Data.HList.Record.HasField Hello2 r v =>
                             Data.HList.Record.Record
                               (Data.HList.HListPrelude.HCons
                                  (Data.HList.Record.LVPair Hello1 [Char])
                                  (Data.HList.HListPrelude.HCons
                                     (Data.HList.Record.LVPair Hello2 [Char])
                                     Data.HList.HListPrelude.HNil))
                             -> [Char]'
    Probable cause: the inferred type is ambiguous

为什么在编译错误中添加f2行结果？

注意：模板Haskell部分在这里可能看起来很愚蠢，但它们是更复杂的模板Haskell的简化，它对元组起作用。我发布了我能构建的最简单的例子，但仍然表现出错误。我意识到删除模板Haskell解决了这种情况下的问题，但这不是我真实代码中的一个选项。

修改

另外，以下编译失败。为什么会这样：

{-# LANGUAGE TemplateHaskell #-}
{-# LANGUAGE NoMonomorphismRestriction #-}

import Data.HList.GhcSyntax((.!.),(.=.),(.*.))
import Data.HList.Record(emptyRecord)
import Data.HList.TypeCastGeneric1
import Data.HList.TypeEqGeneric1
import Data.HList.Label5

data Hello1 = Hello1
data Hello2 = Hello2
data Hello3 = Hello3

record1 = (Hello1 .=. "Hello1") .*. (Hello2 .=. "Hello2") .*. emptyRecord
record2 = (Hello1 .=. "Hello1") .*. (Hello2 .=. "Hello2") .*. (Hello3 .=. "Hello3") .*. emptyRecord

f1 = $([| (\r1 -> (r1 .!. Hello1)) |]) 

main = print $ (f1 record1, f1 record2)

Answer 1

我发现给你的顶级功能类型签名可以修复任何问题。请参阅以下代码：

{-# LANGUAGE TemplateHaskell #-}

module X where
  import Data.HList.GhcSyntax((.!.))

  f = [| (\x r -> (r .!. x)) |]

{-# LANGUAGE TemplateHaskell #-}
{-# LANGUAGE NoMonomorphismRestriction #-}
{-# LANGUAGE FlexibleContexts #-}

import Data.HList.GhcSyntax((.!.),(.=.),(.*.))
import Data.HList.Record(emptyRecord)
import Data.HList.TypeCastGeneric1
import Data.HList.TypeEqGeneric1
import Data.HList.Label5
import X
import Data.HList.Record (HasField)

data Hello1 = Hello1
data Hello2 = Hello2
data Hello3 = Hello3

record1 = (Hello1 .=. "Hello1") .*. (Hello2 .=. "Hello2") .*. emptyRecord
record2 = (Hello1 .=. "Hello1") .*. (Hello2 .=. "Hello2") .*. (Hello3 .=. "Hello3") .*. emptyRecord

g1 :: (HasField Hello1 a b) => a -> b -- Type signature here
g1 = $(f) Hello1

g2 :: (HasField Hello2 a b) => a -> b -- Type signature here
g2 = $(f) Hello2

main = print $ (g1 record1, g2 record1, g1 record2, g2 record2)

模板Haskell编译错误

1 个答案: