Or maybe what you really want is mapM instead of foldM? In this case, your function f2 will look like this: f2 :: [(String, String)] -> IO [Maybe String] f2 xs = mapM f1 xs And this will first map (=apply) your function f1 over all elements in the given input list xs and then it will sequence this list, thus creating a resulting list of type IO [Maybe String] David McBride writes:

You have a foldM function which takes a " a -> b -> m a " as a first argument. But you are passing it f1 which has a type (a, b) -> m a.

Unfortunately I'm not sure what you were trying to do. The function you have just doesn't fit. :(

Maybe you intended to have f1 :: String -> (String, String) -> IO String, and then go

:t foldM f1 "" x foldM f1 "" [("a","b"),("c","d")] :: IO String

?

On Mon, Dec 30, 2013 at 6:42 PM, Miro Karpis wrote:

...

Hi, please can you help me with following?

I have a following function: f1 :: (String, String) -> IO (Maybe String)

which I would like to apply to a list of tuples.I have defined the function like this, but that gives me error (below):

f2 :: [(String, String)] -> IO (Maybe String) f2 x = foldM f1 0 x

error: Couldn't match expected type `IO (Maybe String)' with actual type `[b0] -> m0 (String, String)' In the return type of a call of `foldM'

thanks, m.

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