Re: [Haskell-cafe] Funny State monad dependency
On 16 Apr 2008, at 15:14, Miguel Mitrofanov wrote: Before somebody noticed: I'm wrong. It's not List monad, but also a "(->) x" monad, also defined in Control.Monad. Therefore, "return y" is just "const y". Therefore, x >>= (return y) = x >>= (const y) = x >> y Right. It is an interesting monad, but it may cause unexpected effect, in view of its implicit name. Hans ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Funny State monad dependency
On 16 Apr 2008, at 15:22, Daniel Fischer wrote: The point is the instance Monad ((->) a) where return x = const x f >>= g = \x -> g (f x) x which is defined in Control.Monad.Instances... Thank you. I suspected there was an instance somewhere, and I wanted to know where it is defined. (try in GHCI: Prelude> let f x y = x >>= (return y) Prelude> :t f f :: (Monad ((->) a), Monad m) => m a -> m b -> m b ). It works in Hugs too. If I don't import Control.Monad.State, then f :: (Monad a, Monad ((->) b)) => a b -> a c -> a c This is imported into Control.Monad.State and hence the instance is visible. By the type of (>>=), (return y) must have type (a -> m b), on the other hand, if y has type c, then (return y) has type (m' c) for some monad m'. Unifying m' c and a -> m b gives then m' === ((->) a) and c === m b. Now according to the instance, return y === const y, so f is the same as g x y = x >>= (const y). Good to know the details. Thanks. Hans ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Funny State monad dependency
Am Mittwoch, 16. April 2008 14:56 schrieb Hans Aberg: > When I load the State module in Hugs, then I can define the function > f below, but I do not immediately see exactly what function "return" > returns. Explanation welcome. > > For example: >> f [2..4] [6..9] > >[6,7,8,9,6,7,8,9,6,7,8,9] > That is, it just repeats the second argument as many times as the > length of the second argument. > >Hans Aberg > > > import Control.Monad.State > > f :: Monad a => a b -> a c -> a c > f x y = x >>= (return y) > > The point is the instance Monad ((->) a) where return x = const x f >>= g = \x -> g (f x) x which is defined in Control.Monad.Instances (try in GHCI: Prelude> let f x y = x >>= (return y) Prelude> :t f f :: (Monad ((->) a), Monad m) => m a -> m b -> m b ). This is imported into Control.Monad.State and hence the instance is visible. By the type of (>>=), (return y) must have type (a -> m b), on the other hand, if y has type c, then (return y) has type (m' c) for some monad m'. Unifying m' c and a -> m b gives then m' === ((->) a) and c === m b. Now according to the instance, return y === const y, so f is the same as g x y = x >>= (const y). ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Funny State monad dependency
Miguel Mitrofanov wrote: It has nothing to do with State; it actually works in List monad. "return y" is just another way of writing "[y]". Actually, it seems that in this case return is from the ((->) a) monad, i.e. return=const. f x y = x >>= return y = x >>= const y = (concat . map) (const y) x = concat (map (const y) x) Zun. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Funny State monad dependency
Before somebody noticed: I'm wrong. It's not List monad, but also a "(->) x" monad, also defined in Control.Monad. Therefore, "return y" is just "const y". Therefore, x >>= (return y) = x >>= (const y) = x >> y On 16 Apr 2008, at 17:04, Miguel Mitrofanov wrote: It has nothing to do with State; it actually works in List monad. "return y" is just another way of writing "[y]". You don't need to import Control.Monad.State for this to work; you only need Control.Monad (which is imported by the former). On 16 Apr 2008, at 16:56, Hans Aberg wrote: When I load the State module in Hugs, then I can define the function f below, but I do not immediately see exactly what function "return" returns. Explanation welcome. For example: > f [2..4] [6..9] [6,7,8,9,6,7,8,9,6,7,8,9] That is, it just repeats the second argument as many times as the length of the second argument. Hans Aberg import Control.Monad.State f :: Monad a => a b -> a c -> a c f x y = x >>= (return y) ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Funny State monad dependency
It has nothing to do with State; it actually works in List monad. "return y" is just another way of writing "[y]". You don't need to import Control.Monad.State for this to work; you only need Control.Monad (which is imported by the former). On 16 Apr 2008, at 16:56, Hans Aberg wrote: When I load the State module in Hugs, then I can define the function f below, but I do not immediately see exactly what function "return" returns. Explanation welcome. For example: > f [2..4] [6..9] [6,7,8,9,6,7,8,9,6,7,8,9] That is, it just repeats the second argument as many times as the length of the second argument. Hans Aberg import Control.Monad.State f :: Monad a => a b -> a c -> a c f x y = x >>= (return y) ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] Funny State monad dependency
When I load the State module in Hugs, then I can define the function f below, but I do not immediately see exactly what function "return" returns. Explanation welcome. For example: > f [2..4] [6..9] [6,7,8,9,6,7,8,9,6,7,8,9] That is, it just repeats the second argument as many times as the length of the second argument. Hans Aberg import Control.Monad.State f :: Monad a => a b -> a c -> a c f x y = x >>= (return y) ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
