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:

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

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