Haskell：将辅助函数内联到do块中

lsmepo6l 于 2023-10-19 发布在其他

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通过阅读Graham赫顿的《Haskell编程》（第二版），我成功地解决了练习10.5（第138页）。任务是编写一个函数adder :: IO ()，它读取n数字（交互式定义），对它们求和，并打印结果，如下所示：

> adder
How many numbers? 3
5
4
6
The total is 15

函数readInt和readLine已经给出为：

readInt :: IO Int
readInt = do
  line <- readLine
  return (read line :: Int)
readLine :: IO String
readLine = do
  c <- getChar
  case c of
    '\n' -> return []
    _ -> do
      cs <- readLine
      return (c:cs)

所以我只需要写adder函数：

adder :: IO ()
adder = do
  putStr "How many numbers? "
  n <- readInt
  ns <- sequence [readInt | _ <- [1..n]]
  sum <- sumUp ns
  putStr $ "The total is " ++ (show sum) ++ "\n"
sumUp :: [Int] -> IO Int
sumUp xs = return $ foldl (+) 0 xs

我几乎对我的解决方案很满意，但我只想内联sumUp函数。但是，我不知道该怎么做。
如何将[a] -> IO a函数内联到do块中？

haskell

来源：https://stackoverflow.com/questions/77014124/haskell-inline-auxiliary-function-into-do-block

1条答案

按热度按时间

ugmeyewa1#

这里不需要用return，我们可以用**sum :: (Foldable f, Num a) => f a -> a**来求和：

adder :: IO ()
adder = do
  putStr "How many numbers? "
  n <- readInt
  ns <- sequence [readInt | _ <- [1 .. n]]
  putStr $ "The total is " ++ (show (sum ns)) ++ "\n"

我们也可以重复readInt到**replicateM :: Applicative m => Int -> m a -> m [a]**：

import Control.Monad(replicateM)
adder :: IO ()
adder = do
  putStr "How many numbers? "
  ns <- readInt >>= (`replicateM` readInt)
  putStrLn $ "The total is " ++ (show (sum ns))

对于readInt，这可以实现为：

readInt :: IO Int
readInt = readLn

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Haskell：将辅助函数内联到do块中

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