gmane.comp.lang.haskell.cafe

Andrew Coppin | 12 Oct 19:08

Still stacking monad transformers

I am becoming extremely frustrated now. The task I want to perform is 
simple, yet I simply cannot make Haskell do what I want.

I've given up hope of ever getting my program to handle infinite result 
sets. That means I can make do with just ListT. So I have the following 
monad:

  type MyMonad x = StateT MyState (ListT Identity) x

Now I'm trying to run two computations, starting from _the same state_, 
and combine the two resulting lists. The trouble is, I am literally 
losing the will to live trying to comprehend the whinings of the type 
checker. The operation I'm trying to perform is perfectly simple; I 
don't understand why this has to be so damned *difficult*! >_<

Any suggestions?

I found that by using the brief and easily memorable construction 
"runIdentity $ runListT $ runStateT foo state" I can get at the result 
set for each action, and combine them. But nothing in hell seems to 
transform this from [((), MyState)] back into MyMonad ().

Duncan Coutts | 12 Oct 19:18

Re: Still stacking monad transformers

On Sun, 2008-10-12 at 18:08 +0100, Andrew Coppin wrote:
> I am becoming extremely frustrated now. The task I want to perform is 
> simple, yet I simply cannot make Haskell do what I want.
> 
> I've given up hope of ever getting my program to handle infinite result 
> sets. That means I can make do with just ListT. So I have the following 
> monad:
> 
>   type MyMonad x = StateT MyState (ListT Identity) x
> 
> Now I'm trying to run two computations, starting from _the same state_, 
> and combine the two resulting lists. The trouble is, I am literally 
> losing the will to live trying to comprehend the whinings of the type 
> checker. The operation I'm trying to perform is perfectly simple; I 
> don't understand why this has to be so damned *difficult*! >_<
> 
> Any suggestions?

Have you tried pure lazy functional programming without stacked monads?

I've never been convinced that stacked monads is a good way to write
ordinary code. Monad transformers are great for building your own custom
monads but they should be wrapped in a newtype and made abstract. One
shouldn't have to see the multiple layers. If ordinary code is full of
'lift' then it would seem to me that one is doing something wrong.

Duncan

Andrew Coppin | 12 Oct 19:43

Re: Still stacking monad transformers

Duncan Coutts wrote:
> Have you tried pure lazy functional programming without stacked monads?
>   

I want to process multiple results. The list monad seems a natural way 
to do this. (All of which goes horribly wrong when you try to add error 
processing...)

> I've never been convinced that stacked monads is a good way to write
> ordinary code. Monad transformers are great for building your own custom
> monads but they should be wrapped in a newtype and made abstract. One
> shouldn't have to see the multiple layers. If ordinary code is full of
> 'lift' then it would seem to me that one is doing something wrong.
>   

Given that what I'm attempting to do is extremely simple, yet I am 
having extreme difficulty doing it, yeah, I think we can safely conclude 
I'm doing something wrong.

Miguel Mitrofanov | 12 Oct 19:25

Re: Still stacking monad transformers

On 12 Oct 2008, at 21:08, Andrew Coppin wrote:

> I found that by using the brief and easily memorable construction  
> "runIdentity $ runListT $ runStateT foo state" I can get at the  
> result set for each action, and combine them. But nothing in hell  
> seems to transform this from [((), MyState)] back into MyMonad ().

Well, State monad (and StateT transformer) doesn't work with STATE,  
they work with STATE CHANGES. So, instead of [((), MyState], you  
should have something like (MyState -> [((), MyState)]). And that can  
be transformed to MyMonad () quite easily:

Prelude Control.Monad.State Control.Monad.List  
Control.Monad.Identity> :t \f -> StateT $ ListT . Identity . f
\f -> StateT $ ListT . Identity . f :: (s -> [(a, s)]) -> StateT s  
(ListT Identity) a

Reid Barton | 12 Oct 20:59

Re: Still stacking monad transformers

On Sun, Oct 12, 2008 at 06:08:22PM +0100, Andrew Coppin wrote:
> Now I'm trying to run two computations, starting from _the same state_,  
> and combine the two resulting lists. The trouble is, I am literally  
> losing the will to live trying to comprehend the whinings of the type  
> checker. The operation I'm trying to perform is perfectly simple; I  
> don't understand why this has to be so damned *difficult*! >_<

It's not difficult: the operation is called

mplus :: MyMonad a -> MyMonad a -> MyMonad a

and already exists (assuming the author of ListT has not forgotten to
write a MonadPlus instance).

Regards,
Reid Barton

Andrew Coppin | 13 Oct 19:28

Re: Still stacking monad transformers

Reid Barton wrote:
> It's not difficult: the operation is called
>
> mplus :: MyMonad a -> MyMonad a -> MyMonad a
>
> and already exists (assuming the author of ListT has not forgotten to
> write a MonadPlus instance).
>   

I see... I was under the impression that "mplus" is just any arbitrary 
binary operation over a given monad. How do you know what it does for a 
specific monad?

Anyway, utilising this trick, I now have my function working quite well. 
Implementing negation is the only hard part; I need to unwind everything 
down to the list level, and check whether the list is empty, and do 
something different depending on whether it is or it isn't:

  foo = do
    ...
    let x = run_some_moadic_action
    if null x
      then ...
      else ...

This obviously fails since "x" isn't a list, it's a StateT MyState 
(ListT (ErrorT MyError Ideneity)) x. I can't see a "nice" way to handle 
this. I found a way that works, but it's quite ugly...

Jonathan Cast | 13 Oct 19:33

Re: Still stacking monad transformers

On Mon, 2008-10-13 at 18:28 +0100, Andrew Coppin wrote:
> Reid Barton wrote:
> > It's not difficult: the operation is called
> >
> > mplus :: MyMonad a -> MyMonad a -> MyMonad a
> >
> > and already exists (assuming the author of ListT has not forgotten to
> > write a MonadPlus instance).
> >   
> 
> I see... I was under the impression that "mplus" is just any arbitrary 
> binary operation over a given monad. How do you know what it does for a 
> specific monad?

Process of elimination.  Sometimes, this doesn't narrow things down to a
single operation, but it gives you a good idea of what you're supposed
to expect.

Firstly, mplus and mzero form a (natural) monoid, put together.  That
rules out a number of binary operations right there.

Secondly, mzero has a null law with (>>=):

  mzero >>= f = mzero

So, if you have

  a `mplus` b

and a calls mzero at some point (not inside another call to mplus ---

(Continue reading)

Andrew Coppin | 13 Oct 19:58

Re: Still stacking monad transformers

Jonathan Cast wrote:
>> I see... I was under the impression that "mplus" is just any arbitrary 
>> binary operation over a given monad. How do you know what it does for a 
>> specific monad?
>>     
>
> Process of elimination.  Sometimes, this doesn't narrow things down to a
> single operation, but it gives you a good idea of what you're supposed
> to expect.
>
> Firstly, mplus and mzero form a (natural) monoid, put together.  That
> rules out a number of binary operations right there.
>
> Secondly, mzero has a null law with (>>=):
>
>   mzero >>= f = mzero
>
> So, if you have
>
>   a `mplus` b
>
> and a calls mzero at some point (not inside another call to mplus ---
> nice and informal, that description :), then you know b will be executed
> instead.  (Maybe b will be executed *anyway*.  I didn't say anything
> about that).
>
> So mplus and mzero are basically suitable for three kinds of things:
>
> * Exception handling
> * Back-tracking

(Continue reading)

Jonathan Cast | 13 Oct 20:02

Re: Still stacking monad transformers

On Mon, 2008-10-13 at 18:58 +0100, Andrew Coppin wrote:
> Jonathan Cast wrote:
> >> I see... I was under the impression that "mplus" is just any arbitrary 
> >> binary operation over a given monad. How do you know what it does for a 
> >> specific monad?
> >>     
> >
> > Process of elimination.  Sometimes, this doesn't narrow things down to a
> > single operation, but it gives you a good idea of what you're supposed
> > to expect.
> >
> > Firstly, mplus and mzero form a (natural) monoid, put together.  That
> > rules out a number of binary operations right there.
> >
> > Secondly, mzero has a null law with (>>=):
> >
> >   mzero >>= f = mzero
> >
> > So, if you have
> >
> >   a `mplus` b
> >
> > and a calls mzero at some point (not inside another call to mplus ---
> > nice and informal, that description :), then you know b will be executed
> > instead.  (Maybe b will be executed *anyway*.  I didn't say anything
> > about that).
> >
> > So mplus and mzero are basically suitable for three kinds of things:
> >
> > * Exception handling

(Continue reading)

Arnar Birgisson | 13 Oct 20:17

Re: Still stacking monad transformers

Hi Andrew,

On Mon, Oct 13, 2008 at 19:58, Andrew Coppin
<andrewcoppin <at> btinternet.com> wrote:
> Right. OK. So... isn't there a class somewhere called MonadChoice or
> similar, which defines (<|>)?

Just to pitch in a helpful tip, Hoogle is excellent for these kind of
questions (which come up very often):

http://www.haskell.org/hoogle/?q=%3C|%3E

cheers,
Arnar

Dino Morelli | 13 Oct 20:15

Re: Still stacking monad transformers

On Mon, 13 Oct 2008, Andrew Coppin wrote:

> Reid Barton wrote:
>> It's not difficult: the operation is called
>> 
>> mplus :: MyMonad a -> MyMonad a -> MyMonad a
>> 
>> and already exists (assuming the author of ListT has not forgotten to
>> write a MonadPlus instance).
>> 
>
> I see... I was under the impression that "mplus" is just any arbitrary 
> binary operation over a given monad. How do you know what it does for a 
> specific monad?
>

I had imagined the definition of mplus to be similar in spirit to what
bind is for a specific monad.  i.e. it's part of that monad's strategy
for achieving what it does. As for knowing what it does, trial and error,
reading API docs and source. :)

Something similar to this discussion had come up recently for me, list
monad's MonadPlus implementation.

We found ourselves doing something like this to model 'use default value
if empty' for strings:

    let foo = case str of
                [] -> default
                s  -> s

(Continue reading)

David Menendez | 12 Oct 22:38

Re: Still stacking monad transformers

On Sun, Oct 12, 2008 at 1:08 PM, Andrew Coppin
<andrewcoppin <at> btinternet.com> wrote:
> I am becoming extremely frustrated now. The task I want to perform is
> simple, yet I simply cannot make Haskell do what I want.
>
> I've given up hope of ever getting my program to handle infinite result sets.

Did you miss this message?

<http://article.gmane.org/gmane.comp.lang.haskell.cafe/45952/>

--

-- 
Dave Menendez <dave <at> zednenem.com>
<http://www.eyrie.org/~zednenem/>

wren ng thornton | 12 Oct 22:54

Re: Still stacking monad transformers

David Menendez wrote:
> On Sun, Oct 12, 2008 at 1:08 PM, Andrew Coppin
> <andrewcoppin <at> btinternet.com> wrote:
>> I am becoming extremely frustrated now. The task I want to perform is
>> simple, yet I simply cannot make Haskell do what I want.
>>
>> I've given up hope of ever getting my program to handle infinite result sets.
> 
> Did you miss this message?
> 
> <http://article.gmane.org/gmane.comp.lang.haskell.cafe/45952/>

And if you don't like that one, there's also LogicT 
<http://hackage.haskell.org/cgi-bin/hackage-scripts/package/logict>. The 
function you're looking for is called Control.Monad.Logic.interleave. I 
know LogicT and fair disjunction were brought up earlier, though I seem 
to have mislaid the post.

In case you don't like the efficient Logic or LogicT implementations of 
MonadLogic, defining your own only requires that you can define msplit 
:: m a -> m (Maybe (a, m a)) which pulls the first content out of your 
monad without forcing the rest of it.

--

-- 
Live well,
~wren

Tillmann Rendel | 13 Oct 22:13

Re: Still stacking monad transformers

Dino Morelli wrote:
> I was wishing I could do this:
>
>    let foo = str `or-if-empty` default
>
>
> If it was a Maybe, this works with mplus:
>
>    (Just "foo") `mplus` (Just "bar") == Just "foo"
>    Nothing      `mplus` (Just "bar") == Just "bar"
>
>
> But not so much for list, mplus just ain't defined that way, instead
> doing concatination:
>
>    "foo" `mplus` "bar" == "foobar"
>    ""    `mplus` "bar" == "bar"

The difference between Maybe-style MonadPlus and List-style MonadPlus is
discussed on

   http://www.haskell.org/haskellwiki/MonadPlus_reform_proposal.

With that proposal, you could use morelse with both Maybe and lists.
Personally, I would like to call these beasts <|> (for Maybe-style
mplus) and <+> (for List-style mplus). One could even have

   a <| b = a <|> pure b
   a <+ b = a <+> pure b
   a +> b = pure a <+> b

(Continue reading)

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