Haskell 新手我是怎么得到 IO ( IO String )

Question

所以我想学习一些 Haskell（无聊的夜班），我编写了一个程序，通过计算任何给定时间段的班次，可以简化我的假期计划.

import System.Environment
import Data.List
import Data.List.Split
import Data.Time
import Data.Time.Calendar.WeekDate (toWeekDate)


-- merge xs and ys lists alternating value from each [ xs(0),ys(0),(xs(1),ys(1)..xs(n),ys(n) ]
merge :: [a] -> [a] -> [a]
merge xs [] = xs
merge [] ys = ys
merge (x:xs) (y:ys) = x : y : merge xs ys     

-- get part of list from index 'start' to 'stop'
slice :: Int -> Int -> [a] -> [a]
slice start_from stop_at xs = fst $ splitAt (stop_at - start_from) (snd $ splitAt start_from xs)

timeFormat = "%d-%m-%Y"
timeFormatOut :: Day -> String
timeFormatOut = formatTime defaultTimeLocale "%d-%m-%Y" 
-- parses Strings to DateTime Day1
parseMyDate :: String -> Day
parseMyDate = parseTimeOrError True defaultTimeLocale timeFormat

-- 8 week shift rotation
shiftRotation = cycle ["NAT","NAT","NAT","NAT","NAT","NAT","NAT","-","-","-","-","-","-","-","DAG","DAG","DAG","DAG","-","AFT","AFT","-","-","DAG","DAG","DAG","-","-","DAG","DAG","DAG","-","DAG","DAG","DAG","DAG","DAG","-","DAG","DAG","-","-","AFT","AFT","AFT","AFT","AFT","-","-","DAG(r)","DAG(r)","DAG(r)","DAG(r)","DAG(r)","(r)","(r)"]

hs_findshift :: String -> String -> String -> IO String
hs_findshift anchor start end = do
    let dayZero = parseMyDate anchor
    let startDate = parseMyDate start
    let endDate = parseMyDate end
    let startPos = fromIntegral (diffDays startDate dayZero)
    let endPos = fromIntegral (diffDays endDate dayZero) + 1
    let period = slice startPos endPos shiftRotation
    let dates = map (timeFormatOut) [startDate..endDate]
    let listStr = (concat(intersperse "," (merge dates period)))
    putStrLn listStr

这很好用。现在我会尝试将它导出到 C# 应用程序，这样我就可以制作一个漂亮的界面。我遇到了一些麻烦。我加了

module Shiftlib where

import Foreign.C.String
import Foreign.C.Types

到顶部。在导入的正下方，我添加了一个块来将输入和输出转换为 C 类型。

foreign export ccall findshift :: CString -> CString -> CString -> IO CString

findshift :: CString -> CString -> CString -> IO CString
findshift a s e = do
    anchor <- peekCString a
    start <- peekCString s
    end <- peekCString e
    result <- hs_findshift anchor start end
    return $ newCString result

现在无法编译。 "return $ newCString result" return 似乎是 "foreign" 调用不接受的 IO ( IO CString )。

:l shiftlib
[1 of 1] Compiling Shiftlib         ( shiftlib.hs, interpreted )
shiftlib.hs:53:1: error:
    * Illegal foreign declaration: requires unregisterised, llvm (-fllvm) or native code generation (-fasm)
    * When checking declaration:
        foreign export ccall "findshift" findshift
          :: CString -> CString -> CString -> IO CString
   |
53 | foreign export ccall findshift :: CString -> CString -> CString -> IO CString
   | ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

shiftlib.hs:61:5: error:
    * Couldn't match type `IO CString' with `GHC.Ptr.Ptr CChar'
      Expected type: IO CString
        Actual type: IO (IO CString)
    * In a stmt of a 'do' block: return $ newCString result
      In the expression:
        do anchor <- peekCString a
           start <- peekCString s
           end <- peekCString e
           result <- hs_findshift anchor start end
           ....
      In an equation for `findshift':
          findshift a s e
            = do anchor <- peekCString a
                 start <- peekCString s
                 end <- peekCString e
                 ....
   |
61 |     return $ newCString result
   |     ^^^^^^^^^^^^^^^^^^^^^^^^^^
Failed, no modules loaded.

我似乎无法绕过它。如何从我的小模块中 return CString？

Answer 1

                newCString :: String -> IO String
                    result :: String
         newCString result ::           IO String
                    return ::                   a -> IO a
return $ newCString result ::                        IO (IO String)

当你已经拥有你想要的东西时就停下来，newCString result。

Answer 2

当您在 IO monad 中工作时，在 foo :: .. -> .. -> IO T 类型的函数中 do 块中的最后一个条目必须具有类型 IO T.

return 存在，因此如果您只有 T，您可以将其包装在 IO 中。本质上，T 类型的表达式求值为 T 类型的值，没有任何副作用（例如打印消息或改变 IORefs）。 return 将该值转换为 IO T 类型的 IO 计算。类型 IO T 适用于可能在 returning T 之前有副作用的表达式。 return 创建一个没有副作用的计算，作为 IO T.

的一个简单子案例

newCString result 已经是 IO CString 类型了，所以如果你使用 return 你会把它包装太多次，就像 IO (IO CString) 类型的东西一样。这是一个不 return 字符串的计算，而是运行的另一个计算（最终会产生一个字符串）。

这里显而易见的解决方案是删除 return 以避免多余的包装。

另一个不推荐的解决方案是

do ...
   ...
   string <- newCString result
   return string

这会起作用，因为 string 的类型是 String，而不是 IO CString。不推荐这样做，因为它是多余的。事实上，我们甚至可以添加更多的冗余

do ...
   ...
   string <- newCString result
   string2 <- return string
   string3 <- return string2
   return string3

效果相同。但这没有任何意义，而且会影响可读性。

你的代码是完美的。也许您还想考虑使用 "applicative style".

的一些替代方案

findshift :: CString -> CString -> CString -> IO CString
findshift a s e = do
    result <- hs_findshift <$> peekCString a <*> peekCString s <*> peekCString e
    newCString result

甚至

findshift :: CString -> CString -> CString -> IO CString
findshift a s e =
    hs_findshift <$> peekCString a <*> peekCString s <*> peekCString e >>= newCString

（不过我不喜欢最后一个。）

Haskell 新手我是怎么得到 IO ( IO String )

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Haskell 新手 我是怎么得到 IO ( IO String )

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Haskell 新手我是怎么得到 IO ( IO String )