CODEWITHSUNDEEP
haskell
Rog Matthews
Rog Matthews

Reputation: 3247

Error in the code due to type mismatch

In this code there is an error sue to mismatch.

   import System.IO
   loop :: Int -> [Int] -> IO ()
   loop 0 ls = return ls
   loop n ls = do  newNumber <- readLn
            loop (n-1) (newNumber:ls)


   main = do 
   putStrLn "Please enter the number"
   number <- readLn :: IO Int
   putStrLn $ "The num is:" ++ show number
   xs <- loop number []
   print xs

The error message is:

treep.hs:4:20:
Couldn't match expected type `()' with actual type `[Int]'
In the first argument of `return', namely `ls'
In the expression: return ls
In an equation for `loop': loop 0 ls = return ls

how can i remove the error

Upvotes: 0

Views: 391

Answers (2)

Andre
Andre

Reputation: 1607

From your comment, your loop function should return IO [Int], like so:

loop :: Int -> IO [Int]

Then the first case is:

loop 0 = return []

The second one should be similar to what you wrote (without passing in ls which is your result, not an input).

BTW, the error can be explained this way:

loop :: Int -> [Int] -> IO ()     -- line 1
loop 0 ls = return ls             -- line 2

In line 1, you tell the compiler that the loop function takes two arguments, an Int and a [Int]. In line 2, you use 0 (which is an Int, thus it's fine), and ls. So, ls is of type [Int]. The signature for return is return :: a -> m a, the return type depends on the Monad m (in your case IO) and on the input type which is [Int]. Therefore, the type of return ls is IO [Int], but you told the compiler on line 1 that you it should be `IO ()'. (Exactly what the compiler told you by the error message. ;-) )

Upvotes: 2

hammar
hammar

Reputation: 139930

ls has type [Int], so return ls has type IO [Int]. However, your type signature says that the function should return IO (). It looks like you simply got the type signature wrong. This should work:

loop :: Int -> [Int] -> IO [Int]

Upvotes: 2

Related Questions