将函数应用程序的结果存储在DO块内的元组中

Question

虽然我可以应用一个函数两次并将结果绑定到元组:

  let foo :: Num a => a -> a
      foo x = x + 1

  let (x,y) = (foo 10, foo 20)

这不能在一个do块内完成(至少我不知道如何正确地做到这一点):

  let bar :: Num a => a -> IO a
      bar x = do
                  let y = x + 1
                  return y

  let test :: Num a => IO a
      test = do
                 (x,y) <- (bar 10, bar 20)
                 return y

键入GHCI REPL时出现以下错误:

:29:15:
    Couldn't match expected type ‘IO a1’ with actual type ‘(t0, a)’
    Relevant bindings include
      test :: IO a (bound at :28:5)
    In the pattern: (x, y)
    In a stmt of a 'do' block: (x, y) <- (bar 10, bar 20)
    In the expression:
      do { (x, y) <- (bar 10, bar 20);
           return y }

:29:24:
    Couldn't match type ‘(,) (IO a0)’ with ‘IO’
    Expected type: IO (IO a1)
      Actual type: (IO a0, IO a1)
    In a stmt of a 'do' block: (x, y) <- (bar 10, bar 20)
    In the expression:
      do { (x, y) <- (bar 10, bar 20);
           return y }
    In an equation for ‘test’:
        test
          = do { (x, y) <- (bar 10, bar 20);
                 return y }

我显然可以用更详细的方法来解决它:

  let test' :: Num a => IO a
      test' = do
                 x <- bar 10
                 y <- bar 20
                 return y

是否有正确的表达方式test而不是像它一样test'？

Answer 1

import Control.Applicative

test = do (x,y) <- (,) <$> bar 10 <*> bar 20
          return y

阿卡(x,y) <- liftA2(,) (bar 10) (bar 20).

当然,对于这个特定的例子(在哪里x被抛弃),它将是相同的,并且更好地写

test = bar 20

Answer 2

我将自由地建议您对代码进行一些更改.这是我的版本:

import Control.Monad

-- no need for the do and let
bar :: Num a => a -> IO a
bar x = return $ x + 1 -- or: bar = return . (1+)

-- liftM2 to make (,) work on IO values
test :: Num a => IO a
test = do (x,y) <- liftM2 (,) (bar 10) (bar 20) -- or: (,) <$> bar 10 <*> bar 20
          return y

-- show that this actually works
main :: IO ()
main = test >>= print

您的类型不匹配:您的(bar 10, bar 20)评估类型,Num a => (IO a, IO a)但您将其视为Num a => IO (a, a).通过解除(,)我们使它在IO值上工作并返回一个IO值.

看看这个(GHCi,import Control.Monad得到liftM2):

:t (,)
-- type is :: a -> b -> (a, b)

:t liftM2 (,)
-- type is :: Monad m => m a -> m b -> m (a, b)

在我们的例子中,Monad是IOmonad.因此,最终输出liftM2 (,)将在IOdo-block 内很好地工作,因为它返回一个正确的IO值.

而且,当然你可以用较少的详细解决这个特殊的问题:

test'' = bar 20

PS:请不要IO没有任何理由将东西归还给monad.你完全纯粹的操作看起来不合适,没有合理的方法.