Question

我正在尝试构建一个函数，该函数从列表中返回单个元素。该列表是Maybe (Int,[Int]) tupel的一部分。

如果列表中没有元素，我想返回错误。如果列表中只包含1个元素，我想将该元素作为Monad返回。如果列表包含多个元素，我想返回错误。

我有点迷失，看不出如何让这个相当简单的事情发挥作用。以下是我到目前为止的情况：

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，我在尝试编译时遇到的错误信息对初学者来说绝对没用..

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

我将从您的规范中翻译出来：

如果列表中没有元素，我想返回错误。

f [] = mzero

如果 list只包含1个元素，我想将该元素作为一个返回单子。

f [x] = return x

如果列表包含多个元素，我想返回一个错误。

f (_:_:_) = mzero

所以，把所有东西放在一起：

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

或者甚至更简单，因为第三种情况包括第一种情况：

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

Answer 2

首先，hoogle是你的朋友。您可以查找签名，发现getFirst只是fst，getSecond是snd，您getValue的实现可以写为fromMaybe false }。

就错误消息而言，value的类型为(Integral i) => Maybe i（或接近，我现在没有编译器），这是{{1}的一个可能值返回。但签名singleElemOnly表示必须返回调用者想要的任何 (MonadPlus m) => [a] -> m a。

同样，如果您在GHCi中运行此功能，则MonadPlus的类型默认为test2，但Maybe (Integer, [Integer])必须能够返回任何 { {1}}输入。

无法理解monad的错误处理

2 个答案: