2
考虑下面的代码:模板哈斯克尔编译错误
{-# 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
编译没有问题,并打印出“Hello1”预期。
然而,添加以下行(GHC 7.4.1)给出一个编译错误:
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
线导致编译错误?
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,它可以在元组上工作。我发布了我可以构建的最简单的示例,但仍然显示错误。我意识到在这种情况下删除模板Haskell解决了这个问题,但这不是我真正的代码中的一个选项。
编辑:
此外,以下编译失败。为什么是这种情况:
{-# 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)
添加定义'f2'后,你使用它的任何地方?如果不是,如果给'f2'一个明确的类型注解,它是否工作? –
@DanielWagner:使用'f2'似乎有所帮助。但是,我发现了其他问题。请参阅http://stackoverflow.com/questions/12144250/template-haskell-compile-error-when-calling-with-different-parameters关于不涉及HList的问题的示例。 – Clinton
@DanielWagner:再想一想,或许这个问题是一个红鲱鱼,尽管它可能是相关的。我找到了一个解决方法,并将其作为我自己问题的答案,这不是理想的,但对我的问题可行的解决方案。 – Clinton