Haskell：将辅助函数内联到 do 块中

Question

通过 Graham Hutton 的Haskell 编程（第二版），我刚刚成功解决了练习 10.5（第 138 页）。任务是编写一个函数

adder :: IO ()

读取

n

数字（以交互方式定义），对它们进行求和，然后打印结果，如下所示：

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

函数

readInt

和

readLine

已给出：

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)

所以我只需要编写

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

我对我的解决方案几乎满意，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？

Answer 1

这里不需要使用

return

，我们可以使用

sum :: (Foldable f, Num a) => f a -> a

来对数字进行求和：

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))

至于

readInt

，可以这样实现：

readInt :: IO Int
readInt = readLn