Control.Arrow：为什么 "let (a,b) = (first,second)" 失败了？

Question

我想要的是这样写：

let (a,b) = if *condition* then (first, second) else (second, first)

我发现我连这个都不会写:

let (a,b) = (first,second)

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

看起来你是运行 monomorphism restriction。这只是 Haskell 类型推断的一个限制，您可以通过添加显式类型签名来绕过它。

import Control.Arrow

foo :: (Arrow a, Arrow a1) => (a b c -> a (b, d) (c, d), a1 b1 c1 -> a1 (d1, b1) (d1, c1))
foo = (first, second)

此代码可以通过 foo 的类型签名进行类型检查，但如果删除它，则会出现 "ambiguous variable" 编译错误。

顺便说一句，我使用的类型签名是 :t (first, second) 在 GHCI 中推断出的类型签名。由于您希望 (first, second) 和 (second, first) 具有相同的类型，因此您可能希望在注释中使用更具体的类型，例如以下类型：

foo :: (Arrow a) => (a b b -> a (b, b) (b, b), a b b -> a (b, b) (b, b))

Answer 2

很快，您尝试构造 GHC 无法推断的命令类型。你可以这样做：

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

: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，这个有效：

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到模式时：

λ Control.Arrow > let (a, b) = p

它失败了。约束在对类型之外，并且对于对的另一半是多余的，如

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

有效。

简单示例：

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

要带constraints过来，但是()的类型里没有a，即Monoid a => ()是有歧义的类型

注意：let (a,b) = ((+), (*)) 似乎有效。我不知道 Num 为什么以及如何被特殊对待：

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)
    ...

Control.Arrow：为什么 "let (a,b) = (first,second)" 失败了？

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

haskell

functional-programming

arrows