Question

我正在Haskell Wiki：https://en.wikibooks.org/wiki/Haskell/Higher-order_functions

中尝试此练习

以下练习结合了您所学的高级知识订单功能，递归和I / O。我们将重现什么是在命令式语言中称为 for 循环。实现功能
for :: a -> (a -> Bool) -> (a -> a) -> (a -> IO ()) -> IO () 
for i p f job = -- ???

到目前为止，我有：

-- for : init value, end condition function, increment function, IO function, 
--       returns IO action

generate :: a -> (a->Bool) -> (a->a) -> [a]
generate s cnd incr = if (cnd s) then [] else [s] ++ generate (incr s) cnd incr

printToList  = do
               u <- print 1
               v <- print 2
               return [u,v]

ioToASingle :: [IO a] -> IO [a]
ioToASingle (x:xs) = do
                 x' <- x
                 return [x']

sequenceIO :: [IO a] -> IO [a]
sequenceIO [] = return []
sequenceIO (x:xs) = do
                     x' <- x
                     xs' <- sequenceIO xs
                     return ([x'] ++ xs')

for::a->(a->Bool)->(a->a)->(a->IO())->IO()
for s cnd incr ioFn = sequence_ (map (ioFn) (generate s cnd incr))

for'::a->(a->Bool)->(a->a)->(a->IO a)->IO [a]
for' s cnd incr ioFn = sequenceIO (map (ioFn) (generate s cnd incr))

for可以正常运行：

for 1 (\i->i==10) (\i->i+1) (print)
1
2
3
4
5
6
7
8
9

for'出现错误：

*Main> for' 1 (\i->i==10) (\i->i+1) (print)

<interactive>:323:6: error:
    • No instance for (Num ()) arising from the literal ‘1’
    • In the first argument of ‘for'’, namely ‘1’
      In the expression: for' 1 (\ i -> i == 10) (\ i -> i + 1) (print)
      In an equation for ‘it’:
          it = for' 1 (\ i -> i == 10) (\ i -> i + 1) (print)
*Main>

我不知道怎么了。

Answer 1

这是你写的：

    for' :: a -> (a -> Bool) -> (a -> a ) -> (a -> IO a ) -> IO [a]
    for'    s    cnd            incr         ioFn         =  sequenceIO (map (ioFn) ...

*Main> for' 1    (\i->i==10)    (\i->i+1)    (print     )      
            a                                (a -> IO ())
                                             ------------
            ()                                      a ~ ()

因此1 :: (Num a) => a ~ (Num ()) => ()。但是没有Num类型的()实例。

这是GHC告诉您的回旋方式，您需要类型为a -> IO a而不是a -> IO ()的函数。

Hindley-Milner允许使用“更窄”的类型进行统一，但是如果不是这样，那就更好了-错误消息会更清晰。

在任何情况下，很容易变出自己的值-返回 print ，如

myprint :: a -> IO a
myprint x = do { print x      -- or,  print x >> return x
               ; return x 
               }

Answer 2

sequenceIO返回类型为IO [a]，但ioFn（如for'的定义）为“ print”，返回IO（）。您需要ioFn返回IO a，以便sequenceIO的返回将是IO [a]。

例如请尝试“返回”而不是“打印”，然后打印该值。 “返回”是haskell中的一个函数，它仅将传递的参数转换为monad，在这种情况下为IO monad。

main = do p <- for' 1 (\i->i==10) (\i->i+1) (return) 
         print(p)

这将仅打印1到9个数字的列表。

或

p :: Int -> IO [Int]
p x =  for' 1 (\i->i==x) (\i->i+1) (return)

main = do y <- p 10
         print (y)

Haskell：没有（Num（））的实例-定义我自己的单子循环

2 个答案: