如何解决有关IO的Haskell练习

Question

I have been learning Haskell for few months on my own now by doing some exercises I found on some lecturers site, and I stumbled upon this one and cannot figure how to solve it.

There are two tasks that are kinda "connected". I solved the first one with ease, but don't really understand how to do the other one.

Here's the first one:

Exercise 2. (Recall IO programming in Haskell and the do notation)

Write a recursive function

 sumInts :: Integer -> IO Integer 

that repeatedly reads integer numbers from IO until the number 0 is given. At that point, the function should return the sum of all the entered numbers plus the original (default) value, which is given as a function parameter.

I solved this one this way:

getInt :: IO Integer          
getInt = do  
  s <- getLine
  return (read s)


sumInts :: Integer -> IO Integer
sumInts input = do
  x<-getInt
  if x==0 then return input else (x+) <$> sumInts input

It was pretty easy to do. Here's the other one, the one I cannot understand:

Exercise 3. Generalize the previous IO interaction into a higher order function

 whileIO :: IO a -> (a -> Bool) -> (a -> a -> a) -> a -> IO a 

which, for the given reading IO action, termination condition, folding function, and the original value, returns the required IO action. Check that for some values of getIO, condF, foldF, we can redefine sumInts as

 sumInts = whileIO getIO condF foldF 

Would love some help with this one. :)

Answer 1

Hint:

try to generalize your code, making pieces of it as parameters instead of hardcoding them. For instance, replace the getInt with a more general getIO parameter.

sumInts :: IO Integer -> Integer -> IO Integer
sumInts getIO input = do
    --  ^^^^^
  x<-getIO -- <------
  if x==0 then return input else (x+) <$> sumInts input

Then replace x==0 with a generic predicate.

Then replace (x+) using a generic folding function.

And so on.

At the very end you'll get the wanted whileIO , and you can also give it the more general type suggested by the exercise.