为什么简单的Haskell函数拒绝表示为比率的分数参数？

Question

我当然是Haskell的新手.为了探索懒惰,我在ghci中创建了一个返回其第二个参数的函数:

Prelude> let latter x y = y
latter :: t -> t1 -> t1

我能够用的类型参数来调用它Char,[Char],Num,Floating,和Fractional(表示为小数):

Prelude> latter 'x' 'y'
'y'
it :: Char

Prelude> latter "foo" "bar"
"bar"
it :: [Char]

Prelude> latter 1 2
2
it :: Num t1 => t1

Prelude> latter pi pi
3.141592653589793
it :: Floating t1 => t1

Prelude> latter 0.5 0.7
0.7
it :: Fractional t1 => t1

为什么我得到了一个可怕的错误(什么意思)当我尝试应用latter到一个Fractional表示为一个比值:

Prelude> 1/2
0.5
it :: Fractional a => a

Prelude> latter 1/2 1/2

<interactive>:62:1:
    Could not deduce (Num (a0 -> t1 -> t1))
      arising from the ambiguity check for ‘it’
    from the context (Num (a -> t1 -> t1),
                      Num a,
                      Fractional (t1 -> t1))
      bound by the inferred type for ‘it’:
                 (Num (a -> t1 -> t1), Num a, Fractional (t1 -> t1)) => t1 -> t1
      at <interactive>:62:1-14
    The type variable ‘a0’ is ambiguous
    When checking that ‘it’
      has the inferred type ‘forall t1 a.
                             (Num (a -> t1 -> t1), Num a, Fractional (t1 -> t1)) =>
                             t1 -> t1’
    Probable cause: the inferred type is ambiguous

Answer 1

Haskell中的函数应用程序比其他任何东西都紧密绑定.所以

latter 1/2 1/2

被读为

((latter 1) / (2 1)) / 2

应用2到1不是这么热的想法,因为latter需要两个参数,latter 1实际上是一个功能.用某种东西划分功能也不是一个好主意.您可以使用一些括号来解决所有这些问题:

latter (1/2) (1/2)