如何处理“推断类型的多态性低于预期”?

时间:2009-06-11 20:58:12

标签: haskell types numerical fad

我需要Numeric.FAD库,尽管仍然被存在类型完全困惑。

这是代码:

error_diffs :: [Double] -> NetworkState [(Int, Int, Double)]
error_diffs desired_outputs = do diff_error <- (diff_op $ error' $ map FAD.lift desired_outputs)::(NetworkState ([FAD.Dual tag Double] -> FAD.Dual tag Double))
                                 weights <- link_weights
                                 let diffs = FAD.grad (diff_error::([FAD.Dual tag a] -> FAD.Dual tag b)) weights

                                 links <- link_list
                                 return $ zipWith (\link diff ->
                                                       (linkFrom link, linkTo link, diff)
                                                  ) links diffs

error'在一个Reader monad中运行,由diff_op运行,后者又生成一个匿名函数来获取当前NetworkState和来自FAD.grad的差分输入,并将它们填充到Reader中。

Haskell对我抱有以下困惑:

Inferred type is less polymorphic than expected
  Quantified type variable `tag' is mentioned in the environment:
    diff_error :: [FAD.Dual tag Double] -> FAD.Dual tag Double
      (bound at Operations.hs:100:33)
In the first argument of `FAD.grad', namely
    `(diff_error :: [FAD.Dual tag a] -> FAD.Dual tag b)'
In the expression:
    FAD.grad (diff_error :: [FAD.Dual tag a] -> FAD.Dual tag b) weights
In the definition of `diffs':
    diffs = FAD.grad
              (diff_error :: [FAD.Dual tag a] -> FAD.Dual tag b) weights

2 个答案:

答案 0 :(得分:4)

此代码会出现与您相同的错误:

test :: Int
test =
  (res :: Num a => a)
  where
    res = 5

编译器认为res始终属于Int类型,并且由于某些原因您认为res是多态的而感到困扰。

但是,这段代码运行良好:

test :: Int
test =
  res
  where
    res :: Num a => a
    res = 5

此处,res被定义为多态,但仅用作Int。当您以这种方式键入嵌套表达式时,编译器只会受到打扰。在这种情况下,res可以重复使用,并且可能其中一个用途不会将其用作Int,与您键入嵌套表达式相反,后者无法自行重复使用。

答案 1 :(得分:3)

如果我写,

bigNumber :: (Num a) => a
bigNumber = product [1..100]

然后评估bigNumber :: Int时,
它正在评估(product :: [Int] -> Int) [(1 :: Int) .. (100 :: Int)]

当评估bigNumber :: Integer时,
它正在评估(product :: [Integer] -> Integer) [(1 :: Integer) .. (100 :: Integer)]

两者之间没有任何共享。

error_diffs只有一种类型,即:[Double] -> NetworkState [(Int, Int, Double)]。它必须以一种方式进行评估。

然而,你有什么内容:

... :: NetworkState ([FAD.Dual tag Double] -> FAD.Dual tag Double)

可以用不同的方式进行评估,具体取决于tag是什么。

看到问题?