I am having trouble writing a function with type
Int -> IO [Int] or Int -> [IO Int]
I have the following code that is not working:
createEIList :: Int -> IO [Int]
createEIList len = do
cur <- createEI
(return cur):(createEIList (len-1))
Where createEI is createEI :: IO Int
What is the best way to do something like this?
To keep it close to what you originally had
createEIList :: Int -> IO [Int]
createEIList len = do
cur <- createEI
rest <- createEIList (len-1)
return (cur:rest)
This way, you don't try to append IO Int to your list, but rather return the whole list lifting your [Int] to an IO [Int]
You also probably want a base case
createEIList 0 = return []
createEIList n = ...
So that the recursion actually terminates.
Also worth noting that you can remove the explicit recursion with the combinator
replicateM :: Int -> IO a -> IO [a] -- Restricted for clarity
I'll leave it to you to figure out how to use this.
: have a left argument of type Int and a right argument of IO Int?
<- removes the IO and then you can just cons it with the new elementThe last line should be createEIList (len-1) >>= \a -> return (cur:a)
do-notation instead of lambda, since we're already in that.
createEIList 0 = return []?
createEIList (len-1) >>= return . (cur :)createEIList :: Int -> IO Int
createEIList len = createEI >>= return . replicate len
or
createEIList :: Int -> IO Int
createEIList len = createEI >>= replicateM len . return
better solution:
createEIList :: Int -> IO Int
createEIList len = replicateM len createEI
tersest:
createEIList :: Int -> IO Int
createEIList = flip replicateM createEI
applicative:
import Control.Applicative
createEIList len = replicate <$> pure len <*> createEI
createEIList' len = liftA2 replicate (pure len) createEI
