提问人：Patrick Bucher 提问时间：8/31/2023 更新时间：9/1/2023 访问量：52
Haskell：将内联辅助功能导入 do 块
Haskell: inline auxiliary function into do block
问：
通过Graham Hutton的Programming in Haskell（第二版），我刚刚设法解决了练习10.5（第138页）。任务是编写一个函数来读取数字（以交互方式定义），将它们相加，然后打印结果，如下所示：adder :: IO ()n
> adder
How many numbers? 3
5
4
6
The total is 15
函数 和 已经给出为：readIntreadLine
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)
所以我只需要编写函数：adder
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很满意，但我只想内联该函数。但是，我不知道该怎么做。sumUp
如何将函数内联到块中？[a] -> IO ado
答：
2赞
willeM_ Van Onsem
8/31/2023
#1
这里没有必要用，我们可以用 sum ：： （Foldable f， Num a） => f a -> a 来总结数字：return
adder :: IO ()
adder = do
putStr "How many numbers? "
n <- readInt
ns <- sequence [readInt | _ <- [1 .. n]]
putStr $ "The total is " ++ (show (sum ns)) ++ "\n"我们也可以通过 replicateM ：： Applicative m => Int -> m a -> m [a]：readInt
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 ，这可以实现为：readInt
readInt :: IO Int
readInt = readLn
评论
sumUp ns = return $ foldl (+) 0 nssumUp nssum <- sumUp nssum <- return $ foldl (+) 0 ns