这是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)

但是我收到以下错误:
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

如果我在像这样的更混乱的派系中重写ptrAlign:
ptrAlign2 :: (Storable a) => Ptr a -> Int
ptrAlign2 = ptrAlign3 undefined where
  ptrAlign3 :: (Storable a) => a -> Ptr a -> Int
  ptrAlign3 x _ = alignment x

它工作正常(当然,此版本甚至不需要ScopedTypeVariables)。

但是我仍然很好奇为什么第一个版本会引发错误,以及如何解决该错误?

最佳答案

即使打开ScopedTypeVariables,类型变量也不会放在范围内,除非您明确量化它们,即

ptrAlign1 :: forall a. (Storable a) => Ptr a -> Int
ptrAlign1 _ = alignment (undefined :: a)

07-24 09:46