The thing that you might also be missing is that function application binds
tightest. Hopefully the parenthesis that Rahul has added help you out
there. If not:
\w -> f (h w) w
f will be applied to the result of (h r) which yields another function,
which is then applied to r
that is
\w ->
let x = h w
g = f x
in g w
would yield exactly the same result. I apologise for the indentation, I
need a better mail client.
Ben
On Tue, 21 Feb 2017 at 14:34 Rahul Muttineni
Hi Olumide,
Let the types help you out.
The Monad typeclass (omitting the superclass constraints):
class Monad m where return :: a -> m a (>>=) :: m a -> (a -> m b) -> m b
Write out the specialised type signatures for (->) r:
{-# LANGUAGE InstanceSigs #-} -- This extension allows you to specify the type signatures in instance declarations
instance Monad ((->) r) where return :: a -> (r -> a) (>>=) :: (r -> a) -> (a -> (r -> b)) -> (r -> b)
Now we look at how to make some definition of return that type checks. We're given an a and we want to return a function that takes an r and returns an a. Well the only way you can really do this is ignoring the r and returning the value you were given in all cases! Because 'a' can be *anything*, you really don't have much else you can do! Hence:
return :: a -> (r -> a) return a = \_ -> a
Now let's take a look at (>>=). Since this is a bit complicated, let's work backwards from the result type. We want a function that gives us a b given an r and given two functions with types (r -> a) and (a -> (r -> b)). To get a b, we need to use the second function. To use the second function, we must have an a, which we can get from the first function!
(>>=) :: (r -> a) -> (a -> (r -> b)) -> (r -> b) (>>=) f g = \r -> (g (f r)) r
Hope that helps! Rahul
On Tue, Feb 21, 2017 at 5:04 PM, Olumide <50295@web.de> wrote:
On 21/02/2017 10:25, Benjamin Edwards wrote:
What is it that you are having difficulty with? Is it "why" this is a good definition? Is it that you don't understand how it works?
I simply can't grok f (h w) w.
- Olumide
On Tue, 21 Feb 2017 at 10:15 Olumide <50295@web.de mailto:50295@web.de> wrote:
Hello List,
I am having enormous difficulty understanding the definition of the bind operator of ((->) r) as show below and would appreciate help i this regard.
instance Monad ((->) r) where return x = \_ -> x h >>= f = \w -> f (h w) w
Thanks,
- Olumide
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