诚然，我是 Haskell 新手。为了探索惰性，我在 ghci 中创建了一个函数，它返回它的第二个参数:
Prelude> let latter x y = y
latter :: t -> t1 -> t1
我可以用
Char 类型的参数调用它,
[Char] ,
Num ,
Floating , 和
Fractional (以小数表示):
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
为什么我在尝试应用
latter 时会收到一个可怕的错误(这是什么意思)到
Fractional以比率表示:
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
请您参考如下方法:
Haskell 中的函数应用程序绑定(bind)比其他任何东西都更紧密。所以
latter 1/2 1/2
读作
((latter 1) / (2 1)) / 2
申请
2至
1不是一个热门的想法，因为
latter接受两个参数，
latter 1实际上是一个函数。将功能除以某物也不是一个好主意。您可以使用一些括号来解决所有这些问题:
latter (1/2) (1/2)
