Thanks Daniel, correcting the function call to calcCUVs did the trick. On Sat, May 28, 2011 at 3:11 PM, Daniel Fischer < daniel.is.fischer@googlemail.com> wrote:
On Saturday 28 May 2011 23:35:11, Neil Jensen wrote:
I've been attempting to refactor some working code and running into confusion about returning IO values.
The basic sequence is to query a database, calculate some values, and then store the results back in the database.
The function which does the querying of the db and calculating results has the following type signature: calcCUVs :: AccountId -> IO [((ISODateInt, ISODateInt), CUV)]
This function stores the results back into the database saveCUVs :: AccountId -> [((ISODateInt, ISODateInt), CUV)] -> IO () saveCUVs account cuvs = do r' <- mapM (\x -> storeCUV (snd $ fst x) account (snd x)) cuvs return ()
You're not using the result of mapM, so you should use mapM_ here, if the list is long or the results of storeCUV are large, it's also more efficient.
saveCUVs account cuvs = mapM_ (\x -> storeCUV (snd $ fst x) account (snd x)) cuvs
or, avoiding the snd and fst by taking the argument apart via pattern- matching,
saveCUVs account cuvs = mapM_ (\((_,date),cuv) -> storeCUV date account cuv) cuvs
I had a working variation of the below using 'do' notation, but for some reason when I moved to using bind, I'm getting messed up with return values.
processAccountCUVs :: AccountId -> ISODateInt -> ISODateInt -> IO () processAccountCUVs account prevMonthEnd monthEnd = -- do if (prevMonthEnd == 0 && monthEnd == 0) then calcCUVs account >>= (\cuvs -> saveCUVs account cuvs) >>= return ()
The second argument of (>>=) must be a function, here of type (a -> IO b), ... >> return () or ... >>= return would typecheck, but the latter doesn't make any sense, since 'action >>= return' is the same as plain 'action'. The latter is only of use to finally get the right result type, which is what you're doing here.
else calcCUVs account prevMonthEnd monthEnd >>=
No, that can't be. calcCUVs takes one argument and returns an IO [...], here you pass it three.
(\cuvs -> saveCUVs account cuvs) >>= return ()
The compiler gives the following error message:
Couldn't match expected type `IO ()' against inferred type `()' In the first argument of `return', namely `()' In the second argument of `(>>=)', namely `return ()' In the expression: calcCUVs account >>= (\ cuvs -> saveCUVs account cuvs) >>= return ()
I thought the last return () would correctly return us IO () as we are in the IO monad... what am I missing?
Thanks for any input you can provide. Neil