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You can reach the person managing the list at beginners-ow...@haskell.org When replying, please edit your Subject line so it is more specific than "Re: Contents of Beginners digest..." Today's Topics: 1. Re: Grappling with State Monad get (Olumide) ---------------------------------------------------------------------- Message: 1 Date: Sat, 5 Aug 2017 00:12:17 +0100 From: Olumide <50...@web.de> To: The Haskell-Beginners Mailing List - Discussion of primarily beginner-level topics related to Haskell <beginners@haskell.org> Subject: Re: [Haskell-beginners] Grappling with State Monad get Message-ID: <c42730bc-120c-e1e2-9926-38e28ff9d...@web.de> Content-Type: text/plain; charset=utf-8; format=flowed Thanks Theodore. I'm still pondering your answer and I've tried to write desugar the do-notation like so: stackyStack :: State Stack () stackyStack = get >>= (\stackNow -> ... ) I can _sort of_ see how the type of stackyStack 'imposes' a type on the polymorphic function get, even though I'm not sure that I can precisely explain *exactly* how it happens. Here is my attempt: Comparing the definition of bind (>>=) :: m a -> (a -> m b) -> m b with the desugared form of stackyStack, it seems to me that a is stackNow is an [Int], and m is State [Int]. Is this correct? On a not-too-unrelated note, why is the State (with upper case S) used in the following instance of the state monad instance Monad (State s) where return x = State $ \s -> (x,s) (State h) >>= f = State $ \s ... bearing in mind that Control.Monad.State does not expose its value constructor. Thanks, - Olumide On 02/08/17 03:26, Theodore Lief Gannon wrote: > Actually, get is constrained immediately to be State Stack Stack. If > later statements tried to use it as something else, it would be a type > error. > > The types in a do block can only vary in their final parameter. The type > of stackyStack is State Stack (), so in that do block all the statements > must be forall a. State Stack a. > > Now, check the type of get in GHCi: > get :: State a a > > We know that the first type parameter here is Stack, and the second > position is the same type, thus State Stack Stack. > > > > On Tue, Aug 1, 2017 at 5:13 PM, Olumide <50...@web.de > <mailto:50...@web.de>> wrote: > > Of course 's' is not a type variable as its lowercase. > > Therefore, get is a monad 'constructed' by the state function, correct? > > So, in the do notation the lambda is extracted and used as per the > definition of bind. The context of which we speak therefore must > derive from the next expression(?) in the do notation, which is > somewhat confusing to determine in the example > > stackyStack :: State Stack () > stackyStack = do > stackNow <- get > if stackNow == [1,2,3] > then put [8,3,1] > else put [9,2,1] > > Regards, > > - Olumide > > On 02/08/17 00:35, Theodore Lief Gannon wrote: > > 's' here is not a type variable, it's an actual variable. \s -> > (s, s) defines a lambda function which takes any value, and > returns a tuple with that value in both positions. > > So yes, get is point-free, but the missing argument is right > there in-line. And yes, its type is simply implied from context. > > > On Aug 1, 2017 4:17 PM, "Olumide" <50...@web.de > <mailto:50...@web.de> <mailto:50...@web.de > <mailto:50...@web.de>>> wrote: > > Ahoy Haskellers, > > In the section "Getting and Setting State" > (http://learnyouahaskell.com/for-a-few-monads-more#state > <http://learnyouahaskell.com/for-a-few-monads-more#state> > <http://learnyouahaskell.com/for-a-few-monads-more#state > <http://learnyouahaskell.com/for-a-few-monads-more#state>>) in LYH > get is defined as > > get = state $ \s -> (s, s) > > How does does get determine the type s, is considering that > it has > no argument as per the definition given above? Or is the > definition > written in some sort of point-free notation where an > argument has > been dropped? > > I find the line stackNow <- get in the the function(?) > stackyStack > confusing for the same reason. I guess my difficulty is > that state $ > \s ->(s , s) has a generic type (s) whereas the stackyStack > has a > concrete type. Is the type of s determined from the type of the > stateful computation/do notation? > > Regards, > > - Olumide > _______________________________________________ > Beginners mailing list > Beginners@haskell.org <mailto:Beginners@haskell.org> > <mailto:Beginners@haskell.org <mailto:Beginners@haskell.org>> > http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners > <http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners> > <http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners > <http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners>> > > > > > _______________________________________________ > Beginners mailing list > Beginners@haskell.org <mailto:Beginners@haskell.org> > http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners > <http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners> > > _______________________________________________ > Beginners mailing list > Beginners@haskell.org <mailto:Beginners@haskell.org> > http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners > <http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners> > > > > > _______________________________________________ > Beginners mailing list > Beginners@haskell.org > http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners > ------------------------------ Subject: Digest Footer _______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners ------------------------------ End of Beginners Digest, Vol 110, Issue 10 ******************************************