[Haskell-beginners] Monad problem

[Marcin Mrotek]

Imants Cekusins,

Well using monad transformers like that is a fairly common design
pattern in Haskell, so I think you should get used to it ;) But you
don't really need to understand monad transformers in depth just to
use the ig library, it appears that all actions defined there have a
type like

    :: (MonadBaseControl IO m, MonadResource m) => _ some arguments _
-> InstagramT m _result_

The smallest monad that satisfies both (MonadBaseControl IO m) and
(MonadResource m) is just (ResourceT IO), so the code suggested by Bob
Ippolito

        token <- runResourceT . runInstagramT credentials manager $
                            getUserAccessTokenURL2 redirectUrl code

works, because "getUserAccessTokenURL2 redirectUrl code" has type
"InstagramT (ResourceT IO) OAuthToken". It is one of Haskell's
features that the actual concrete type of an expression is inferred
according to the context in which it appears, and thus may vary.

* The whole thing starts with the type "MonadBaseControl IO m,
MonadResource m => InstagramT m OAuthToken"
* runInstagramT strips the InstagramT layer, leaving "MonadBaseControl
IO m, MonadResource m =>m OAuthToken"
* runResourceT has type "MonadBaseControl IO m => ResourceT m a -> m
a", so Haskell infers that the "m OAuthToken" above is actually
"MonadBaseControl IO m => ResourceT m OAuthToken". The satisfaction of
the MonadResource constraint is provided by the ResouceT layer itself,
which is stripped by runResourceT then.
* at the end, Haskell is left with "MonadBaseControl IO m => m
OAuthToken", which it tries to unify with "IO something", because
you're using it in the do block of main, that must have the type IO ()
in the end. As "IO OAuthToken" does satisfy this constraint,
everything is fine.

The library uses type classes instead of this concrete type, because
some users might want to supply a more complicated monad transformer
stack than that, but of course you don't actually have to do this.

Best regards,
Marcin Mrotek