Haskell: No instance of (Num ()) - defining my own monadic loop - loops

Question

I am attempting this exercise in Haskell wiki: https://en.wikibooks.org/wiki/Haskell/Higher-order_functions
The following exercise combines what you have learned about higher
order functions, recursion and I/O. We are going to recreate what is
known in imperative languages as a for loop. Implement a function
for :: a -> (a -> Bool) -> (a -> a) -> (a -> IO ()) -> IO ()
for i p f job = -- ???
so far I have:
-- 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 works OK:
for 1 (\i->i==10) (\i->i+1) (print)
1
2
3
4
5
6
7
8
9
for' gets an error:
*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>
I can't figure out what's wrong.

Answer 1

Here's what you wrote:
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 ~ ()
Thus 1 :: (Num a) => a ~ (Num ()) => (). But there's no Num instance for the type ().
This is GHC's roundabout way of telling you, you need a function of type a -> IO a, not a -> IO ().
Hindley-Milner allows unification with a "narrower" type, but here it would be better if it weren't -- the error message would be much clearer.
In any case it is easy to conjure up your own value–returning print, like
myprint :: a -> IO a
myprint x = do { print x -- or, print x >> return x
; return x
}

Answer 2

sequenceIO return type is IO [a], but ioFn (as in the definition of for') is "print" which returns IO(). You need ioFn to return IO a, so that return of sequenceIO will be IO [a].
e.g. try "return" instead of "print" and then print the value. "return" is a function in haskell which simply converts the argument passed into monad, in this case IO monad.
main = do p <- for' 1 (\i->i==10) (\i->i+1) (return)
print(p)
This will print simply list of 1 to 9 numbers.
Or
p :: Int -> IO [Int]
p x = for' 1 (\i->i==x) (\i->i+1) (return)
main = do y <- p 10
print (y)