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Today's Topics:
1. Re: Trying to prove Applicative is superclass of Functor,
etc (D?niel Arat?)
----------------------------------------------------------------------
Message: 1
Date: Fri, 29 Apr 2016 12:51:43 +0200
From: D?niel Arat? <exitcons...@gmail.com>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <beginners@haskell.org>
Subject: Re: [Haskell-beginners] Trying to prove Applicative is
superclass of Functor, etc
Message-ID:
<CAHvKd2JnRYx2JB16Ka08Yk+M7fon_Z+KAnz_3WFKAJ=cakc...@mail.gmail.com>
Content-Type: text/plain; charset=UTF-8
For a drunk person you're getting these equivalences surprisingly right. :)
>> Ah true, I heard about this D?niel. But then it would be generalized to
> all classes, or just those three ones?
It would make sense for it to work for any class hierarchy, but
honestly I just don't know and I've got an old GHC on my laptop so I
can't check. Maybe someone who actually knows will come along and
enlighten us. ?\_(?)_/?
>> Anyway, trying the same with Applicative and its *sub*class Monad:
>>> pure = return
That's right.
>>> mf <*> ma =
>> let mg = mf >>= \f -> return (return . f)
>> in mg >>= \g -> (ma >>= g)
>> Is there an easier solution? It's easy enough, but not as trivial as for
> Applicative => Functor.
I've got (an alpha-equivalent of) the following on a scrap piece of paper:
mf <*> mx = mf >>= \f -> mx >>= \x -> return . f $ x
It reads nice and fluent.
>> Why do we write constraints like that:
>>> Functor f => (a -> b) -> f a -> f b
>> or:
>>> Functor f => Applicative f
That's a very good question. I agree that it would make a bit more
sense if the arrows were reversed.
I think it helps to read
"Functor f => Applicative f where ..."
as
"if you've got a Functor then you can have an Applicative if you just
implement these functions"
as opposed to
"if f is a Functor that implies that f is an Applicative" (wrong).
Daniel
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