[Haskell-cafe] How to define a Monad instance

[Steffen Schuldenzucker]

On 07/28/2012 03:35 PM, Thiago Negri wrote:
 > [...]
> As Monads are used for sequencing, first thing I did was to define the
> following data type:
>
> data TableDefinition a = Match a a (TableDefinition a) | Restart

So TableDefinition a is like [(a, a)].

> [...]
 >
> So, to create a replacement table:
>
> table' :: TableDefinition Char
> table' =
>          Match 'a' 'b'
>          (Match 'A' 'B'
>           Restart)
>
> It look like a Monad (for me), as I can sequence any number of
> replacement values:
>
> table'' :: TableDefinition Char
> table'' = Match 'a' 'c'
>           (Match 'c' 'a'
>           (Match 'b' 'e'
>           (Match 'e' 'b'
>            Restart)))

Yes, but monads aren't just about sequencing. I like to see a monad as a 
generalized computation (e.g. nondeterministic, involving IO, involving 
state etc). Therefore, you should ask yourself if TableDefinition can be 
seen as some kind of abstract "computation". In particular, can you 
"execute" a computation and "extract" its result? as in

   do
     r <- Match 'a' 'c' Restart
     if r == 'y' then Restart else Match 2 3 (Match 3 4 Restart)

Doesn't immediately make sense to me. In particular think about the 
different possible result types of a TableDefinition computation.

If all you want is sequencing, you might be looking for a Monoid 
instance instead, corresponding to the Monoid instance of [b], where 
b=(a,a) here.

>  [...]
 >
> I'd like to define the same data structure as:
>
> newTable :: TableDefinition Char
> newTable = do
>          'a' :>  'b'
>          'A' :>  'B'
>
> But I can't figure a way to define a Monad instance for that. :(

The desugaring of the example looks like this:

   ('a' :> 'b') >> ('A' :> 'B')

Only (>>) is used, but not (>>=) (i.e. results are always discarded). If 
this is the only case that makes sense, you're probably looking for a 
Monoid instead (see above)

-- Steffen