如何鏈 - > IO（M B）功能

Question

有功能與類似簽名：

a -> IO (m b) 
b -> IO (m c) 
c -> IO (m d) 

如何把它們連在

a -> IO (m d) 

？

實際應用：假設有一組REST端點。每個返回值和下一個值都要求前面返回的值作爲參數。從終端獲取

因此函數是這樣的：

Value1 -> IO (Maybe Value2) 
Value2 -> IO (Maybe Value3) 
Value3 -> IO (Maybe Value4) 

Answer 1

有與類似的簽名功能：

a -> IO (m b) 
b -> IO (m c) 
c -> IO (m d) 

如何把它們連在

a -> IO (m d) 

屬屬l，你可能無法。例如，如果m是Const，我不確定問這甚至是有道理的。一般而言，您可能預期m爲Monad。但是，請注意，即使m是 a Monad，其與IO的組合可能不是（see this）。

Value1 -> IO (Maybe Value2) 
Value2 -> IO (Maybe Value3) 
Value3 -> IO (Maybe Value4) 

啊，現在你說！您在這裏尋找的抽象是MaybeT和Kleisli組合(>=>)。然後，例如，

import Control.Monad ((>=>)) 
import Control.Monad.Trans.Maybe (MaybeT(..)) 

rest1 :: Value1 -> IO (Maybe Value2) 
rest2 :: Value2 -> IO (Maybe Value3) 
rest3 :: Value3 -> IO (Maybe Value4) 

rest4 :: Value1 -> IO (Maybe Value4) 
rest4 x = runMaybeT ((MaybeT . rest1 >=> MaybeT . rest2 >=> MaybeT . rest3) x) 

這仍然看起來有點難看。要做的事情可能是重構您的rest1,rest2和rest3函數。正如已經在評論MaybeT IO a中指出的那樣，可以將其轉換爲IO (Maybe a)（其實這正是runMaybeT和MaybeT所做的）。

import Control.Monad ((>=>)) 
import Control.Monad.Trans.Maybe (MaybeT(..)) 

rest1 :: Value1 -> MaybeT IO Value2 
rest2 :: Value2 -> MaybeT IO Value3 
rest3 :: Value3 -> MaybeT IO Value4 

rest4 :: Value1 -> MaybeT IO Value4 
rest4 = rest1 >=> rest2 >=> rest3 

Answer 2

如果m具有Traversable一個實例（Maybe有一個）這裏是另一種選擇：

import   Control.Monad (join) 

f :: (Traversable m, Monad m) 
    => (a -> IO (m b)) 
    -> (b -> IO (m c)) 
    -> (c -> IO (m d)) 
    -> a 
    -> IO (m d) 
f fa fb fc a = fa a 
      >>= traverse fb 
      >>= fmap join . traverse fc . join