Haskell 类型错误 : Inferred type is less polymorphic than expected

Question

我现在正在与 Haskell 混在一起，对于我的一生，我无法弄清楚为什么以下内容有效......

square :: (Num a) => a -> a 
square x = x * x 
dx = 0.0000001
deriv1 :: (Fractional a) => (a -> a) -> (a -> a)
deriv1 g = (\x -> ((g (x + 2) - (g x)) / 0.0000001 ))   
main = printf "res==%g %g\n" (square 5.12::Double) ((deriv1 square) 2::Float)

但这并没有......

square :: (Num a) => a -> a 
square x = x * x 
dx = 0.0000001
deriv1 :: (Fractional a) => (a -> a) -> (a -> a)
deriv1 g = (\x -> ((g (x + 2) - (g x)) / dx ))          
main = printf "res==%g %g\n" (square 5.12::Double) ((deriv1 square) 2::Float)

注意我用过 dx这次在derv1函数中。我是 Haskell 的新手，所以任何关于类型的深入讨论都可能很快从我身边溜走，我会死的旋转。我必须有一些类似于命令式答案的东西，否则在我的 Haskell 职业生涯早期，它几乎肯定会丢失。

我收到的错误消息是:

Inferred type is less polymorphic than expected
  Quantified type variable `a' is mentioned in the environment:
    dx :: a (bound at sicp-1.40.hs:12:0)
When trying to generalise the type inferred for `deriv1'
  Signature type:     forall a. (Fractional a) => (a -> a) -> a -> a
  Type to generalise: (a -> a) -> a -> a
In the type signature for `deriv1'
When generalising the type(s) for `deriv1'

Answer 1

您收到错误是因为 monomorphism restriction .由于您没有为 dx 提供类型签名它最终推断为 Double在这种情况下。你可以给出一个明确的、多态的签名，比如

dx :: Fractional a => a
dx = 0.0000001

或者您可以通过在源文件顶部包含此行来禁用单态限制

{-# LANGUAGE NoMonomorphismRestriction #-}