haskell - Control.Arrow : Why "let (a,b) = (first,second)" fails?

Question

我想要的是写这样的东西:
let (a,b) = if *condition* then (first, second) else (second, first)
我发现我什至不能写这个:
let (a,b) = (first,second)
它失败并出现错误:

 <interactive>:7:5:                                                                                                                          
Could not deduce (Arrow a0)                                                                                                             
from the context (Arrow a)
  bound by the inferred type for `a':
             Arrow a => a b c -> a (b, d) (c, d)
  at <interactive>:7:5-26
The type variable `a0' is ambiguous
When checking that `a' has the inferred type
  a :: forall (a :: * -> * -> *) b c d.
       Arrow a =>
       a b c -> a (b, d) (c, d)
Probable cause: the inferred type is ambiguous

<interactive>:7:5:
Could not deduce (Arrow a0)
from the context (Arrow a)
  bound by the inferred type for `b':
             Arrow a => a b c -> a (d, b) (d, c)
  at <interactive>:7:5-26
The type variable `a0' is ambiguous
When checking that `b' has the inferred type
  b :: forall (a :: * -> * -> *) b c d.
       Arrow a =>
       a b c -> a (d, b) (d, c)
Probable cause: the inferred type is ambiguous

Answer 1

很快，您将尝试构建 GHC 无法推断的 Impredicative 类型。你可以做:

λ Control.Arrow > let (a,b) = (first, second) :: Arrow a => (a b b -> a (b, b) (b, b), a b b -> a (b, b) (b, b))
λ Control.Arrow > :t a
a :: Arrow a => a b b -> a (b, b) (b, b)
λ Control.Arrow > :t b
b :: Arrow a => a b b -> a (b, b) (b, b)

或者

:set -XImpredicativeTypes 
λ Control.Arrow > let (a,b) = (first, second) :: (Arrow a => a b b -> a (b, b) (b, b), Arrow a => a b b -> a (b, b) (b, b))
λ Control.Arrow > :t a
a :: Arrow a => a b b -> a (b, b) (b, b)
λ Control.Arrow > :t b
b :: Arrow a => a b b -> a (b, b) (b, b)

但你不能这样做:

λ Control.Arrow > let (a,b) = (first, second) :: (Arrow a, Arrow a') => (a b b -> a (b, b) (b, b), a' b b -> a' (b, b) (b, b))

为了隔离问题，这有效:

λ Control.Arrow > let p = (first, second) :: (Arrow a, Arrow a') => (a b b -> a (b, b) (b, b), a' b b -> a' (b, b) (b, b));
λ Control.Arrow > :t p
p :: (Arrow a', Arrow a) =>
     (a b b -> a (b, b) (b, b), a' b b -> a' (b, b) (b, b))

但是当您尝试将其绑定(bind)到模式时:

λ Control.Arrow > let (a, b) = p

它失败。约束在配对类型之外，并且对于配对的其他部分是多余的，因为

λ Control.Arrow > :set -XImpredicativeTypes 
λ Control.Arrow > let p = (first, second) :: (Arrow a => a b b -> a (b, b) (b, b), Arrow a => a b b -> a (b, b) (b, b))
λ Control.Arrow > let (a, b) = p

作品。

简单的例子:

λ Prelude Data.Monoid > :t (mappend, ())
(mappend, ()) :: Monoid a => (a -> a -> a, ())
λ Prelude Data.Monoid > let (a, b) = (mappend, ())

<interactive>:12:5:
    No instance for (Monoid a0)
      arising from the ambiguity check for ‘b’
    The type variable ‘a0’ is ambiguous
    When checking that ‘b’ has the inferred type ‘()’
    Probable cause: the inferred type is ambiguous

一个必须携带约束，但没有 a类型为 () ，即 Monoid a => ()是模棱两可的类型。

注: let (a,b) = ((+), (*))似乎工作。我不知道为什么以及如何 Num被特殊对待:

λ Prelude Data.Monoid > let x = () ::  Num a => ()
λ Prelude Data.Monoid > :t x
x :: ()
λ Prelude Data.Monoid > let x = () :: Monoid m => ()

<interactive>:12:9:
    No instance for (Monoid m0)
    ...