The dreaded monomorphism restriction strikes again, namely during the binding of let helper = f 2. Adding an explicit type signature let helper :: Show a => (Int -> a) -> Int -> IO a helper = f 2 or -XNoMonomorphismRestriction does the trick. On 4/20/2016 1:24 AM, James M wrote:

Below I have a contrived example. Please do not take this to be real world code.

f :: (Show a) => Int -> (Int -> a) -> Int -> IO a f i g x = do print i return $ g (x + i)

foo :: Bool -> IO (Either [Int] [String]) foo b = do let helper = f 2 if b then Left <$> sequence (fmap (helper negate) [0,1]) else Right <$> sequence (fmap (helper show) [0,1])

The above will fail stating

Example.hs:14:35:

Couldn't match type ‘Int’ with ‘[Char]’

Expected type: Int -> IO String

Actual type: Int -> IO Int

In the first argument of ‘fmap’, namely ‘(helper show)’

In the first argument of ‘sequence’, namely

‘(fmap (helper show) [0, 1])’

Example.hs:14:42:

Couldn't match type ‘[Char]’ with ‘Int’

Expected type: Int -> Int

Actual type: Int -> String

In the first argument of ‘helper’, namely ‘show’

In the first argument of ‘fmap’, namely ‘(helper show)’

However, if I simply change f's type signature to not have the typeclass constraint, the type checker is happy: f :: Int -> (Int -> a) -> Int -> IO a

Another possibility to remove the problem is to remove the let statement and instead put the entire expression in.

foo :: Bool -> IO (Either [Int] [String]) foo b = do if b then Left <$> sequence (fmap (f 2 negate) [0,1]) else Right <$> sequence (fmap (f 2 show) [0,1])

This seems to be a bug in the typechecker, and maybe it is a well known issue.

Can someone please confirm that this is a bug and whether or not it is known?

James

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