CODEWITHSUNDEEP
haskellio
Iulius Curt
Iulius Curt

Reputation: 5114

Output Integer to stdout in Haskell

I have a simple function like:

nth :: Integer -> Integer

And I try to print it's result as follows:

main = do
    n <- getLine
    result <- nth (read n :: Integer)
    print result

The following error is generated:

Couldn't match expected type `IO t0' with actual type `Integer'
In the return type of a call of `nth'
In a stmt of a 'do' expression:
    result <- nth (read n :: Integer)

Also tried with putStrLn and a lot of other combinations with no luck.
I can't figure it out and I would need some help, being that I don't fully understand how stuff works around these IOs.

Upvotes: 21

Views: 38730

Answers (2)

rotskoff
rotskoff

Reputation: 704

The do syntax unwraps something within a monad. Everything on the right hand side of the arrow must live within the IO monad, otherwise the types don't check. An IO Integer would be fine in your program. do is syntactic sugar for the more explicit function which would be written as follows:

Recall that (>>=) :: m a -> (a -> m b) -> m b

main = getLine >>= (\x ->
       nth >>= (\y ->
       print y))

But nth is not a monadic value, so it doesn't make sense to apply the function (>>=), which requires something with the type IO a.

Upvotes: 7

Thomas M. DuBuisson
Thomas M. DuBuisson

Reputation: 64750

nth is a function, not an IO action:

main = do
  n <- getLine
  let result = nth (read n :: Integer)
  print result

Upvotes: 20

Related Questions