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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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 wrote:

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

_______________________________________________ Beginners mailing list Beginners@haskell.org mailto:Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners

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-- Rahul Muttineni _______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners

I did, sorry! On Tue, 21 Feb 2017 at 15:53 Olumide <50295@web.de> wrote:

On 21/02/2017 15:08, Benjamin Edwards wrote:

...

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

Did you mean to write (h w)?

- Olumide

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Thanks a bunch. Simply gorgeous answer. - Olumide On 21/02/17 14:32, Rahul Muttineni wrote:

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 mailto: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 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

_______________________________________________ Beginners mailing list Beginners@haskell.org mailto: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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-- Rahul Muttineni

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