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 <50295@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" <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
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