Question

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

如果我在一个更混乱的派系中重写ptrAlign，就像这样：

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抛出错误，以及如何解决它？

Answer 1

即使启用了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)

ScopedTypeVariables不会将类型变量带入范围

1 个答案: