编译qtHaskell时发生错误

Question

在qmake和mingw32-make从这些指令成功后,我执行runhaskell Setup.hs build,我收到以下错误:

[651 of 662] Compiling Qtc.Core.Attributes ( Qtc\Core\Attributes.hs, dist\build\Qtc\Core\Attributes.o )
Qtc\Core\Attributes.hs:584:13:
Could not deduce (Qstt a (QDialogSc b))
  arising from a use of `slotReject''
from the context (Qstt a (QDialogSc b1))
  bound by the instance declaration
  at Qtc\Core\Attributes.hs:582:10-52
Possible fix:
  add (Qstt a (QDialogSc b)) to the context of
    the instance declaration
  or add an instance declaration for (Qstt a (QDialogSc b))
In the expression: slotReject'
In an equation for `reject'': reject' = slotReject'
In the instance declaration for `QsaSlotReject a'

Attributes.hs文件(第578 - 583行):

class QsaSlotReject w where
  slotReject', reject' ::  (Qslot w (w -> ()), (w -> ()))

instance (Qstt a (QDialogSc b)) => QsaSlotReject (a) where
  slotReject' = (Qslot "reject()", \_ -> ())
  reject' = slotReject'

环境 :

• Windows 7的
• Haskell平台2011.2.0
• Qt sdk 4.7

顺便说一句,我在这个过程中遇到了两次内存不足,但我猜这没关系.

Answer 1

问题来自于这个事实

data Qslot x f = Qslot String

所以从Qslot"Blah blah"形式的给定项目推断x和f可能有点困难.自从去年秋天qthaskell的最后一个版本上升以来,GHC使用的推理机制也许发生了微妙的变化.

在任何情况下,它似乎编译,带有一些好奇的警告,并且示例工作,如果你替换

 instance (Qstt a (QDialogSc b)) => QsaSlotReject (a) where
    slotReject' = (Qslot "reject()", \_ -> ())
    reject' = slotReject'

同

 instance (Qstt a (QDialogSc b)) => QsaSlotReject (a) where
   slotReject' = (Qslot "reject()", \_ -> ())
   reject' = (Qslot "reject()", \_ -> ())

这样GHC简化版,不得不惊叹相当之多......

必须有一些东西可以使事情更精确.我不知道eta减少警告是否会在以后系统地开始出现,这与此线路有关.