模板Haskell编译错误
考虑以下代码:模板Haskell编译错误,haskell,template-haskell,hlist,Haskell,Template Haskell,Hlist,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.
{-# 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
这将很好地编译并按预期打印出“Hello1”
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
为什么添加f2
行会导致编译错误
注意:模板Haskell部分在这里可能看起来很傻,但它们是对更复杂的模板Haskell的简化,后者对元组有效。我已经发布了我可以构建的最简单的示例,它仍然显示了错误。我意识到删除Haskell模板可以解决这个问题,但在我的实际代码中这不是一个选项
编辑:
此外,以下代码无法编译。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)
我发现给顶级函数类型签名可以解决任何问题。请参阅下面的代码:
{-# LANGUAGE TemplateHaskell #-}
module X where
import Data.HList.GhcSyntax((.!.))
f = [| (\x r -> (r .!. x)) |]
添加了f2的定义后,您是否在任何地方使用它?如果没有,那么如果给
f2
一个显式的类型注释是否有效?@DanielWagner:使用f2
似乎有帮助。然而,我发现了其他问题。请看一个不涉及HList的问题的例子。@DanielWagner:再想想,也许这个问题是在转移注意力,尽管它可能是相关的。我已经找到了一个解决办法,并把它作为我自己问题的答案,它不是理想的,但对我的问题是可行的解决方案。
{-# 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)