无法理解 monad 的错误处理

Question

我正在尝试构建一个函数，该函数 return 将我作为列表中的单个元素。该列表是 Maybe (Int,[Int]) 元组的一部分。

如果列表不包含任何元素，我想return报错。如果列表恰好包含 1 个元素，我想 return 该元素作为 Monad。如果列表包含超过 1 个元素，我想 return 一个错误。

我有点迷茫，看不出如何让这个相当简单的东西发挥作用。这是我目前所拥有的：

import Control.Monad

test1 = Just (1,[2,3]) :: Maybe (Int,[Int])
test2 = Just (2,[1]) :: Maybe (Int,[Int])
test3 = Just (3,[]) :: Maybe (Int,[Int])

getValue :: Maybe Bool -> Bool
getValue (Just x) = x
getValue Nothing = False

singleElemOnly :: (MonadPlus m) => [a] -> m a
singleElemOnly x = let result = test2
                       value = fmap fst result
                       isEmpty = fmap null (fmap snd result)
                   in if (getValue isEmpty) then value else mzero

不幸的是，我在尝试编译时收到的错误消息对我这个初学者来说绝对没有用..

Playground.hs:15:50:
    Could not deduce (a ~ Int)
    from the context (MonadPlus m)
      bound by the type signature for
                 singleElemOnly :: MonadPlus m => [a] -> m a
      at Playground.hs:11:19-45
      `a' is a rigid type variable bound by
          the type signature for singleElemOnly :: MonadPlus m => [a] -> m a
          at Playground.hs:11:19
    Expected type: m a
      Actual type: Maybe Int
    Relevant bindings include
      x :: [a]
        (bound at Playground.hs:12:16)
      singleElemOnly :: [a] -> m a
        (bound at Playground.hs:12:1)
    In the expression: value
    In the expression: if (getValue isEmpty) then value else mzero

Playground.hs:15:50:
    Could not deduce (m ~ Maybe)
    from the context (MonadPlus m)
      bound by the type signature for
                 singleElemOnly :: MonadPlus m => [a] -> m a
      at Playground.hs:11:19-45
      `m' is a rigid type variable bound by
          the type signature for singleElemOnly :: MonadPlus m => [a] -> m a
          at Playground.hs:11:19
    Expected type: m a
      Actual type: Maybe Int
    Relevant bindings include
      singleElemOnly :: [a] -> m a
        (bound at Playground.hs:12:1)
    In the expression: value
    In the expression: if (getValue isEmpty) then value else mzero

非常感谢任何帮助！

Answer 1

首先，hoogle是你的朋友。您可以查找签名，发现 getFirst 只是 fst，getSecond 是 snd，而您实现的 getValue 可以写成 fromMaybe false.

就错误消息而言，value 的类型为 (Integral i) => Maybe i（或类似的东西，我现在没有编译器），这是 [=18] 的一个可能值=] return秒。但是签名 (MonadPlus m) => [a] -> m a 说它必须 return any MonadPlus 调用者想要的。

同理，如果你在GHCi中是运行这个，那么test2的类型默认为Maybe (Integer, [Integer])，但是singleElemOnly必须能够return 任何 Integral类型。

Answer 2

我会根据您的说明进行翻译：

If the list contains no elements, I want to return an error.

f [] = mzero

If the list contains exactly 1 element, I want to return that element as a Monad.

f [x] = return x

If the list contains more than 1 element, I want to return an error.

f (_:_:_) = mzero

因此，将所有内容放在一起：

singleElemOnly :: (MonadPlus m) => [a] -> m a
singleElemOnly []  = mzero
singleElemOnly [x] = return x
singleElemOnly _   = mzero
       -- a _ catches everything else, no need to write the exact pattern

或者更简单，因为第三个案例包括第一个：

singleElemOnly :: (MonadPlus m) => [a] -> m a
singleElemOnly [x] = return x
singleElemOnly _   = mzero

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