GHCI可以用来解释强迫行为吗？

Question

如何要求GHCI解释以下原因：

*Lib> sum Nothing
0

甚至编译？有没有和Monoid有牵连？这不在签名里！

*Lib> :i Foldable
class Foldable (t :: * -> *) where
  ...
  maximum :: Ord a => t a -> a
  minimum :: Ord a => t a -> a
  sum :: Num a => t a -> a
  product :: Num a => t a -> a
        -- Defined in ‘Data.Foldable’
instance Foldable [] -- Defined in ‘Data.Foldable’
instance Foldable Maybe -- Defined in ‘Data.Foldable’
instance Foldable (Either a) -- Defined in ‘Data.Foldable’
instance Foldable ((,) a) -- Defined in ‘Data.Foldable’
*Lib> :i Num
class Num a where
  (+) :: a -> a -> a
  (-) :: a -> a -> a
  (*) :: a -> a -> a
  negate :: a -> a
  abs :: a -> a
  signum :: a -> a
  fromInteger :: Integer -> a
        -- Defined in ‘GHC.Num’
instance Num Word -- Defined in ‘GHC.Num’
instance Num Integer -- Defined in ‘GHC.Num’
instance Num Int -- Defined in ‘GHC.Num’
instance Num Float -- Defined in ‘GHC.Float’
instance Num Double -- Defined in ‘GHC.Float’
*Lib> sum Nothing
0

Answer 1

您可以使用键入的孔：

> sum (Nothing :: _)
<interactive>:4:17:
    Found hole `_' with type: Maybe a
    Where: `a' is a rigid type variable bound by
               the inferred type of it :: Num a => a at <interactive>:4:1
    To use the inferred type, enable PartialTypeSignatures
    Relevant bindings include it :: a (bound at <interactive>:4:1)
    In an expression type signature: _
    In the first argument of `sum', namely `(Nothing :: _)'
    In the expression: sum (Nothing :: _)

这说明a是一个严格类型的变量，由it :: Num a => a的推断类型绑定，并且由于Maybe是Foldable的一个实例(正如您已经在输出到:i Foldable中看到的那样，尽管您也可以在到:i Maybe的输出中看到)，它将假定Nothing :: Num a => Maybe a进行编译，因为sum对它施加了Num约束。

所以它编译的原因是，sum接受一个包含代码值的代码，而Maybe是一个代码，Nothing本身具有类型Maybe a，并且sum对a必须实现< Num a => a >d19Foldable >进行了约束。在GHCi中时，默认设置为Integer，因此您可以看到0的输出。