这里有一个返回指针对齐的简单函数:
{-# LANGUAGE ScopedTypeVariables #-}
import Foreign.Ptr (Ptr)
import Foreign.Storable (Storable, alignment)
main = return ()
ptrAlign1 :: (Storable a) => Ptr a -> Int
ptrAlign1 _ = alignment (undefined :: a)
但我得到以下错误:
Could not deduce (Storable a0) arising from a use of `alignment'
from the context (Storable a)
bound by the type signature for
ptrAlign1 :: Storable a => Ptr a -> Int
at prog.hs:8:14-41
The type variable `a0' is ambiguous
如果我在一个更混乱的派别中重写ptrAlign
,比如
ptrAlign2 :: (Storable a) => Ptr a -> Int
ptrAlign2 = ptrAlign3 undefined where
ptrAlign3 :: (Storable a) => a -> Ptr a -> Int
ptrAlign3 x _ = alignment x
它运行良好(当然,这个版本甚至不需要ScopedTypeVariables
)。
但我仍然很好奇为什么第一个版本会出现错误,以及可以做些什么来解决它?
即使打开了ScopedTypeVariables
,类型变量也不会放在作用域中,除非您明确量化它们,即
ptrAlign1 :: forall a. (Storable a) => Ptr a -> Int
ptrAlign1 _ = alignment (undefined :: a)