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
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
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
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
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 St
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
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 th
