[Haskell-cafe] How to define a Monad instance
ryani.spam at gmail.com
Tue Jul 31 07:39:53 CEST 2012
A couple typos:
instance Monad Replacer1 where
instance Monad (Replacer1 k) where
instance Monad Replacer2 k where
instance Monad (Replacer2 k) where
I haven't tested any of this code, so you may have to fix some minor type
On Mon, Jul 30, 2012 at 10:38 PM, Ryan Ingram <ryani.spam at gmail.com> wrote:
> To take this a step further, if what you really want is the syntax sugar
> for do-notation (and I understand that, I love sweet, sweet syntactical
> sugar), you are probably implementing a Writer monad over some monoid.
> Here's two data structures that can encode this type;
> data Replacer1 k a = Replacer1 (k -> Maybe k) a
> data Replacer2 k a = Replacer2 [(k,k)] a
> instance Monad Replacer1 where
> return x = Replacer1 (\_ -> Nothing) x
> Replacer1 ka a >>= f = result where
> Replacer1 kb b = f a
> result = Replacer1 (\x -> ka x `mplus` kb x) b
> (!>) :: Eq k => k -> k -> Replacer1 k ()
> x !> y = Replacer1 (\k -> if k == x then Just y else Nothing) ()
> replace1 :: Replacer1 k () -> [k] -> [k] -- look ma, no Eq requirement!
> replace1 (Replacer k ()) = map (\x -> fromMaybe x $ k x) -- from Data.Maybe
> table1 :: Replacer1 Char ()
> table1 = do
> 'a' !> 'b'
> 'A' !> 'B'
> test = replace1 table1 "All I want"
> -- Exercise: what changes if we switch ka and kb in the result of (>>=)?
> When does it matter?
> -- Exercises for you to implement:
> instance Monad Replacer2 k where
> replacer :: Eq k => Replacer2 k -> [k] -> [k]
> ($>) :: k -> k -> Replacer2 k
> -- Exercise: Lets make use of the fact that we're a monad!
> -- What if the operator !> had a different type?
> -- (!>) :: Eq k => k -> k -> Replacer k Integer
> -- which returns the count of replacements done.
> -- table3 = do
> -- count <- 'a' !> 'b'
> -- when (count > 3) ('A' !> 'B')
> -- return ()
> -- Do any of the data structures I've given work? Why or why not?
> -- Can you come up with a way to implement this?
> -- ryan
> On Sat, Jul 28, 2012 at 10:07 AM, Steffen Schuldenzucker <
> sschuldenzucker at uni-bonn.de> wrote:
>> 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
>> Haskell-Cafe mailing list
>> Haskell-Cafe at haskell.org
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