[Haskell-cafe] Funny State monad dependency

2008-04-16 Thread 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)


___
Haskell-Cafe mailing list
Haskell-Cafe@haskell.org
http://www.haskell.org/mailman/listinfo/haskell-cafe


Re: [Haskell-cafe] Funny State monad dependency

2008-04-16 Thread Miguel Mitrofanov
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


Re: [Haskell-cafe] Funny State monad dependency

2008-04-16 Thread Miguel Mitrofanov

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

2008-04-16 Thread Roberto Zunino

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

2008-04-16 Thread Daniel Fischer
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

2008-04-16 Thread Hans Aberg

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

2008-04-16 Thread Hans Aberg

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