[Haskell-cafe] How to define a Monad instance
sschuldenzucker at uni-bonn.de
Sat Jul 28 19:07:14 CEST 2012
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'
> 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'
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
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
> 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)
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