[Haskell-beginners] Lost Monad Signature

[David McBride]

If you type :i Monad in ghci, you will see this instance

instance Monad ((->) r)

What this means is a function where the first argument is of type 'r'... is
a monad.  You can in fact use do notation / return on a tuple to manipulate
its second argument monadically.

So let's look at what that does to the type signature of join when 'm' is
((->) b)

join :: Monad m => m (m a) -> m a

 -- m = ((->) b)

join :: ((->) b ((->) b a)) -> (((->) b a))

Now we just have to move the arrows from prefix to infix.  Let's do it step
by step.

join :: ((->) b (b -> a)) -> (b -> a)
join :: (b -> (b -> a)) -> (b -> a)

x -> (y -> z) is equivalent to x -> y -> z

join :: (b -> b -> a) -> (b -> a)
join :: (b -> b -> a) -> b -> a

So now when you put an operator into it that takes two arguments

(,) :: a -> b -> (a,b)

You get the type you saw.

join (,) :: b -> (b, b)





On Fri, Dec 8, 2017 at 10:37 AM, Quentin Liu <quentin.liu.0415 at gmail.com>
wrote:

> Hi,
>
> The function `join` flattens a double-layered monad into one layer and its
> type signature is
>
> join :: (Monad m) => m (m a) -> m a
>
> But when the first argument supplied is `(,)`, the type signature becomes
>
> join (,) :: b -> (b, b)
>
> in ghci. The monad constraint is lost when supplied the first argument. So
> my question is why the type constraint is lost and what monad is supplied
> here.
>
> Regards,
> Qingbo Liu
>
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