Question

一个玩具示例，但仍然令人沮丧：

numberMapper:: IO ()
numberMapper = do codes <- forM [1 .. 4] (\num ->
                   do putStrLn $ "Enter a code for " ++ show num
                       code <- getLine
                       return code)
                   let numberCodes = zip [1 .. 4] codes
                   in forM numberCodes (\(num,code) ->
                   putStrLn $ "Got code " ++ show code ++ " for " ++ show num)

ghci告诉我，我有一个Parse error in pattern: putStrLn，我无法弄清楚为什么它无法解析。

Answer 1

校正：

numberMapper:: IO ()
numberMapper = do
    codes <- forM [1 .. 4] $ \num -> do
        putStrLn $ "Enter a code for " ++ show num
        getLine
    let numberCodes = zip [1 .. 4] codes
    forM_ numberCodes $ \(num,code) ->
        putStrLn $ "Got code " ++ show code ++ " for " ++ show num

修复： do块内的行应排成一行。

-- wrong
a = do codes <- something
        let numberCodes = zip [1..4] codes

-- right
a = do codes <- something
       let numberCodes = zip [1..4] codes

修复2：在let区块内使用do时，请勿使用in。

-- wrong
func = do
    let x = 17
    in print x

-- right
func = do
    let x = 17
    print x

修复3：使用forM_（返回()，a.k.a。void）而不是forM（返回列表）。

codes <- forM [1..4] func...  -- returns a list
forM_ numberCodes $ ...       -- discards list, returns ()

所以forM_可以（差不多）这样写：

forM_ xs f = do forM xs f
                return ()

小改变：您不需要return：

do func1
   x <- func2
   return x

您可以将其更改为等效的

do func1
   func2 -- value of func2 is returned

Answer 2

你在do-blocks中过度缩进行。此外，in - let中的do语句不需要numberMapper:: IO () numberMapper = do codes <- forM [1 .. 4] (\num -> do putStrLn $ "Enter a code for " ++ show num code <- getLine return code) let numberCodes = zip [1 .. 4] codes forM numberCodes (\(num,code) -> putStrLn $ "Got code " ++ show code ++ " for " ++ show num)。

这对我有用：

numberMapper:: IO ()
numberMapper = do codes <- forM [1 .. 4] $ \num ->
                   do putStrLn $ "Enter a code for " ++ show num
                      code <- getLine
                      return code
                  let numberCodes = zip [1 .. 4] codes
                  forM numberCodes $ \(num,code) ->
                    putStrLn $ "Got code " ++ show code ++ " for " ++ show num

你也可以像这样构建它：

do

（允许您避开括号;或者，将\num ->放在{{1}}的末尾并排列后续语句

Haskell在一个陌生的地方抛出一个解析错误

2 个答案: