[Haskell-cafe] How to define a Monad instance
holmisen at gmail.com
Sun Jul 29 11:43:49 CEST 2012
As for understanding monads, you can try to define the State monad
. Not sure if it's the best example but it's intuitive in that it
let's you thread a state "behind the scenes".
Not related to your question -- in your example if you want to
translate characters but do not plan to change the length of the
input, you don't need Maybe. Your 'table' can then be defined as:
table :: Char -> Char
table 'a' = 'b'
table 'A' = 'B'
table x = x
Then your 'replaceAll' is simply 'map':
replaceAll = map
2012/7/28 Steffen Schuldenzucker <sschuldenzucker at uni-bonn.de>:
> 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 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
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