这是一个返回指针对齐方式的简单函数:
{-# LANGUAGE ScopedTypeVariables #-}
import Foreign.Ptr (Ptr)
import Foreign.Storable (Storable, alignment)
main = return ()
ptrAlign1 :: (Storable a) => Ptr a -> Int
ptrAlign1 _ = alignment (undefined :: a)
但我收到以下错误:
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
如果我像这样以更困惑的方式重写 ptrAlign
:
ptrAlign2 :: (Storable a) => Ptr a -> Int
ptrAlign2 = ptrAlign3 undefined where
ptrAlign3 :: (Storable a) => a -> Ptr a -> Int
ptrAlign3 x _ = alignment x
它工作得很好(当然这个版本甚至不需要ScopedTypeVariables
)。
但我仍然好奇为什么第一个版本会抛出错误,以及可以采取什么措施来解决它?
最佳答案
即使打开了 ScopedTypeVariables
,类型变量也不会放入作用域中,除非您显式量化它们,即
ptrAlign1 :: forall a. (Storable a) => Ptr a -> Int
ptrAlign1 _ = alignment (undefined :: a)
关于haskell - ScopedTypeVariables 不会将类型变量带入作用域,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/34012115/