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" <50295@web.de mailto:50295@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) 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 http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
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