如何在 zip 生成的列表上折叠 io 操作以打印 2 列？

Question

我看到了几个关于使用 foldl 和 IO 的问题，但其中 none 似乎为我的案例提供了解决方案。

我只是想输出两列数字。

我尝试了以下方法，但我只能打印一行而不是整列：

zipMerge :: [a] -> [b] -> [(a,b)]
zipMerge [] _      = []
zipMerge _  []     = error "second list cannot be empty"
zipMerge (a:as) bs = (zip (replicate (length bs) a) bs) ++ (zipMerge as bs)

-- foldl :: (a -> b -> a) -> a -> [b] -> a
printPixel :: IO() -> (Int, Int) -> IO()
printPixel _ (i, j) = putStrLn (show i) ++ " " ++ (show j)

printPixels :: [(Int, Int)] -> IO()
printPixels colors = foldl printPixel (return ()) colors

printGradient :: IO()
printGradient = do
    putStrLn "255"
    let is = [0..256]
        js = [256,255..0]
    printPixels (zipMerge js is)

我做错了什么？

Answer 1

弃牌在这里绝对是大材小用。 fold 的想法是在你遍历列表时保持某种状态，但在你的情况下没有状态要保持：你只需要按顺序对每个元素执行一个效果，并且独立于其他元素。

为此，use mapM_:

printPixel :: (Int, Int) -> IO()
printPixel (i, j) = putStrLn $ (show i) ++ " " ++ (show j)

printPixels :: [(Int, Int)] -> IO()
printPixels colors = mapM_ printPixel colors

或 its twin for_，这是同一回事，但参数的顺序相反（和较弱的约束），这使您可以将效果作为 lambda 表达式提供，但没有括号，使其看起来差不多像 C 类语言中的 for 结构：

printPixels :: [(Int, Int)] -> IO()
printPixels colors = for_ colors $ \(i, j) -> putStrLn $ (show i) ++ " " ++ (show j)

如何在 zip 生成的列表上折叠 io 操作以打印 2 列？

How to foldl an io operation on a list generated by a zip in order to print 2 columns?

io

haskell

functional-programming

fold