Understanding the function monad ((->) r)

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Understanding the function monad ((->) r)

 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 [hidden email] http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
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Re: Understanding the function monad ((->) r)

 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?BenOn Tue, 21 Feb 2017 at 10:15 Olumide <[hidden email]> 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 [hidden email] http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners _______________________________________________ Beginners mailing list [hidden email] http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
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Re: Understanding the function monad ((->) r)

 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 <[hidden email] > > 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 >     [hidden email] >     http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners> > > > _______________________________________________ > Beginners mailing list > [hidden email] > http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners> _______________________________________________ Beginners mailing list [hidden email] http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
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Re: Understanding the function monad ((->) r)

 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 bWrite out the specialised type signatures for (->) r:{-# LANGUAGE InstanceSigs #-}-- This extension allows you to specify the type signatures in instance declarationsinstance 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 = \_ -> aNow 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)) rHope that helps!Rahul  On Tue, Feb 21, 2017 at 5:04 PM, Olumide 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 <[hidden email] > 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     [hidden email]     http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners _______________________________________________ Beginners mailing list [hidden email] http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners _______________________________________________ Beginners mailing list [hidden email] http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners -- Rahul Muttineni _______________________________________________ Beginners mailing list [hidden email] http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
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Re: Understanding the function monad ((->) r)

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Re: Understanding the function monad ((->) r)

 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 _______________________________________________ Beginners mailing list [hidden email] http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
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Re: Understanding the function monad ((->) r)

 I did, sorry!On Tue, 21 Feb 2017 at 15:53 Olumide <[hidden email]> 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 _______________________________________________ Beginners mailing list [hidden email] http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners _______________________________________________ Beginners mailing list [hidden email] http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners