[Haskell-cafe] New version of itself

[Erik Hesselink]

do r1 <- m1
   r2 <- m2
   return (r1 && r2)

Erik

On 6 June 2016 at 11:11, martin <martin.drautzburg at web.de> wrote:
> Applicative does the trick. But using do-notation was my first attempts, but I could not figure it out. Can you see how
> to do this (other than using `ap`)?
>
>
> Am 06/06/2016 um 10:20 AM schrieb Erik Hesselink:
>> Since `Mealy a` is an instance of `Applicative` you can do:
>>
>>   (&&) <$> m1 <*> m2
>>
>> It's also a monad, so do notation would also work.
>>
>> Erik
>>
>> On 6 June 2016 at 10:00, martin <martin.drautzburg at web.de> wrote:
>>> Carl,
>>>
>>> thanks for pointing me towards mealy machines.
>>>
>>> Maybe you can help me with this one: I am trying to compose two mealy machines into one. Suppose I have
>>>
>>> m1 :: Mealy a Bool
>>> m2 :: Mealy a Bool
>>>
>>> now I want a combinded mealy machine which outputs True when both m1 and m2 return True and False otherwise. I can write
>>> this using runMealy. But this looks awkward, because it is a bare-bones implementation and doesn't make use of any of
>>> the typeclasses Mealy is a member of. I.e. the machines package is basically not used at all.
>