[Haskell-beginners] return IO () vs. ()

[Neil Jensen]

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 at 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
>
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