Question

我知道它很冗长，但这就是我学习语法的方法。这条线在

（abCombo'a 2 lst）...我想返回'list'和/或打印列表但我无法使用此返回类型'Writer [String] [Int]'提取列表。

-- Loop through several integer values
-- and calculate the power of a^b, append to list
abCombo' :: Int -> Int -> [Int] -> Writer [String] [Int]
abCombo' a b lst
    | b == maxB = do
        tell [ "  ... x-Done(1): a^b = " ++ show (a^b) ++ " // " ++ show lst  ]
        return ((a^b):lst)
    | otherwise = do
        tell [ "  ... x-Processing: a^b = " ++ show (a^b) ++ " // " ++ show lst ]
        abCombo' a (b+1) ((a^b):lst)

-- Loop through several integer values
-- and calculate the power of a^b, append to list
abCombo :: Int -> [Int] -> Writer [String] [Int]
abCombo a lst
    | a == maxA = do              
        tell [ "- Done(2): a=" ++ show a ]
        abCombo' a 2 lst
    | otherwise = do
        (abCombo' a 2 lst) <<<<<<<<<<<<<<<<<<<<<< line of interest, here
        tell ["- Processing: a=" ++ show a]
        abCombo (a + 1) lst

...

这是上面的当前代码，我想将其更改为：

abCombo :: Int -> [Int] -> Writer [String] [Int]
abCombo a lst
    | a == maxA = do              
        tell [ "- Done(2): a=" ++ show a ]
        abCombo' a 2 lst
    | otherwise = do
        let res = (abCombo' a 2 lst) <<<<<<<<<<<<<<<<<<<<<< line of interest, here            
        tell ["- Processing: a=" ++ show a]
        abCombo (a + 1) (flatten snd res)

Answer 1

要在do - 块中绑定操作的结果，您需要使用<-而不是let。

res <- abCombo' a 2 lst      -- res :: [Int]

这是因为使用let，你只需在动作本身上加上一个名字。

let res = abCombo' a 2 lst   -- res :: Writer [String] [Int]

如何从此haskell片段返回列表值

1 个答案: