[英]How to foldl an io operation on a list generated by a zip in order to print 2 columns?
我已经看到了几个关于使用 foldl 和 IO 的问题,但它们似乎都没有为我的案例提供解决方案。
我只是想 output 两列数字。
0 256
1 256
2 256
...
256 256
0 255
1 255
2 255
3 255
4 255
我尝试了以下方法,但我只能打印一行而不是一整列:
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)
我究竟做错了什么?
弃牌在这里绝对是矫枉过正。 fold 的想法是在遍历列表时保留某种 state ,但在您的情况下,没有要保留的 state :您只需要按顺序为每个元素执行效果,并且独立于其他元素。
为此,请使用mapM_
:
printPixel :: (Int, Int) -> IO()
printPixel (i, j) = putStrLn $ (show i) ++ " " ++ (show j)
printPixels :: [(Int, Int)] -> IO()
printPixels colors = mapM_ printPixel colors
或者它的双胞胎for_
,这是同一件事,但 arguments 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,使其看起来几乎像 C 中的for
构造-喜欢语言:
printPixels :: [(Int, Int)] -> IO()
printPixels colors = for_ colors $ \(i, j) -> putStrLn $ (show i) ++ " " ++ (show j)
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