Question

Functor, Applicative, and Monad 中的一个示例稍有变化：

{-# LANGUAGE ApplicativeDo #-}

import Safe (readMay)
-- import Control.Applicative ((<$>), (<*>))

displayAge maybeAge =
    case maybeAge of
        Nothing -> putStrLn "You provided invalid input"
        Just age -> putStrLn $ "In that year, you will be: " ++ show age

yearDiff futureYear birthYear = futureYear - birthYear

maybeAge fS bS = do 
   fI <- readMay fS
   bI <- readMay bS
   pure $ yearDiff fI bI       

main = do
    putStrLn "Please enter your birth year"
    birthYearString <- getLine
    putStrLn "Please enter some year in the future"
    futureYearString <- getLine
    displayAge $ maybeAge birthYearString futureYearString

我使用maybeAge和do代替

maybeAge fS bS = yearDiff <$> readMay fS <*> readMay bS

我有两个问题：

在这种情况下，如何检查maybeAge是否使用Applicative Functor语义或Monad？
如果使用Applicative Functor，在这种情况下有什么优势？

关于：ApplicativeDo。

Answer 1

我从您身上做了一个完整的例子：

{-# LANGUAGE ApplicativeDo #-}

import Text.Read (readMaybe)

displayAge :: Maybe Int -> IO ()
displayAge maybeAge =
    case maybeAge of
        Nothing -> putStrLn "You provided invalid input"
        Just age -> putStrLn $ "In that year, you will be: " ++ show age

yearDiff :: Int -> Int -> Int
yearDiff  = (-)

maybeAge :: String -> String -> Maybe Int
maybeAge fS bS = do 
   fI <- readMaybe fS
   bI <- readMaybe bS
   pure $ yearDiff fI bI

main :: IO ()
main = do
    putStrLn "Please enter your birth year"
    birthYearString <- getLine
    putStrLn "Please enter some year in the future"
    futureYearString <- getLine
    displayAge $ maybeAge futureYearString birthYearString

此外，在最后一行，我交换了参数，因为它们在示例中的顺序似乎不正确。另外，我根据@Redu的评论改进了yearDif的定义。

这是您问题的答案。

您可以按照GHC User Guide中的建议，即使用-ddump-ds编译器开关，检查是否确实应用了应用（和函子）操作。我在下面添加了几个开关，以使输出更简洁。我还仅显示有关maybeAge函数的摘录。

$ ghc appdo.hs -ddump-ds -dsuppress-type-applications -dsuppress-module-prefixes 
[1 of 1] Compiling Main             ( appdo.hs, appdo.o )

==================== Desugar (after optimization) ====================
Result size of Desugar (after optimization)
  = {terms: 75, types: 75, coercions: 0, joins: 0/0}
...
-- RHS size: {terms: 17, types: 13, coercions: 0, joins: 0/0}
maybeAge :: String -> String -> Maybe Int
[LclId]
maybeAge
  = \ (fS_a1h3 :: String) (bS_a1h4 :: String) ->
      <*>
        $fApplicativeMaybe
        (fmap
           $fFunctorMaybe
           (\ (fI_a1h5 :: Int) (bI_a1h6 :: Int) -> yearDiff fI_a1h5 bI_a1h6)
           (readMaybe $fReadInt fS_a1h3))
        (readMaybe $fReadInt bS_a1h4)
...

最肯定的是，这里没有加速。 Maybe的适用操作符具有恒定的复杂度（O(1)），就像单子步操作一样。

在original paper中，ApplicativeDo的作者给出了一些更复杂的单子类型（Haxl，Data.Seq，解析等）的示例，从而使渐近效率更高应用操作。请参阅本文的第6节。

Haskell中的ApplicativeDo编译指示和Applicative函子

1 个答案: