Haskell维基有以下问题:
https://en.wikibooks.org/wiki/Haskell/Higher-order_functions
for :: a -> (a -> Bool) -> (a -> a) -> (a -> IO ()) -> IO ()
for i p f job = -- ???
我能够提出以下实现:
generate :: a -> (a->Bool) -> (a->a) -> [a]
generate s cnd incr = if (cnd s) then [] else [s] ++ generate (incr s) cnd incr
-- collapse :: [IO ()] -> IO ()
-- collapse (x:xs) = x ++ collapse xs
-- does not work ^^^^^^
for::a->(a->Bool)->(a->a)->(a->IO())->IO()
for s cnd incr ioFn = map (ioFn) (generate s cnd incr)
当然,map (ioFn) (generate s cnd incr)
的结果是[IO ()]
.我不确定如何将其转换为IO ()
我需要类似foldl
的东西,但可以使用[IO ()]
而不是[a]
.
您要查找的函数是:
sequence_ :: (Foldable t, Monad m) =>t (m a) ->m ()
但是我们实际上可以 替换map
,这样我们就不需要额外的功能。您可以在此处使用mapM_ :: Monad m => (a -> m b) -> [a] -> m ()
而不是map
,因此:
for :: a -> (a -> Bool) -> (a -> a) -> (a -> IO ()) -> IO()
for s cnd incr ioFn =mapM_ioFn (generate s cnd incr)
因此,这将在generate s cnd incr
的所有元素上应用函数ioFun
,并最终返回单位()
。