Re: [Haskell-beginners] Lost Monad Signature

Thanks for the explanation. But why did Haskell pick up the Reader monad in particular? Is this part of the HM type inference rule? Regards, Qingbo Liu On Dec 8, 2017, 13:12 -0500, David McBride , wrote:

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 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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