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Today's Topics:
1. Using a monad function inside the monad transfomer variant
(Moritz Tacke)
2. Re: Using a monad function inside the monad transfomer
variant (Seph Shewell Brockway)
3. Parsing Terms in Brackets for Calculator (Leonhard Applis)
----------------------------------------------------------------------
Message: 1
Date: Fri, 1 Feb 2019 15:14:32 +0100
From: Moritz Tacke <moritz.ta...@gmail.com>
To: beginners@haskell.org
Subject: [Haskell-beginners] Using a monad function inside the monad
transfomer variant
Message-ID:
<CANZtOB1riZq_TG8U9AghcX=7bh3k55bplzi0lzd4nn9nivl...@mail.gmail.com>
Content-Type: text/plain; charset="UTF-8"
Hi,
I am running into difficulties regarding the use of monad
transformers. The situation is as follows: I implemented some
functions returning "RVar" results. Now, in a different part of the
program, I am using STUArrays. Therefore my idea was to create a monad
transformer stack which would have this type:
someFunction:: a -> b -> RVarT (ST s) (STUArray s Int Double)
I still want to use the functions that return RVars, so eg.
rvarDouble :: RVar Double
and then the definition of the transformer function would be:
someFunction a b =
do ...
the_double <- rvarDouble
....
This does not compile, complaining that;
• Couldn't match type ‘Data.Functor.Identity.Identity’ with ‘ST s’
Expected type: RVarT (ST s) Double
Actual type: RVar Double
How can I re-user the RVar function in the RVarT monad transformer?
Sincerely yours,
Moritz
------------------------------
Message: 2
Date: Fri, 1 Feb 2019 17:57:17 +0000
From: Seph Shewell Brockway <s...@codex.scot>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <beginners@haskell.org>
Subject: Re: [Haskell-beginners] Using a monad function inside the
monad transfomer variant
Message-ID: <20190201175717.acbhtmqs2wv23c5p@leviathan>
Content-Type: text/plain; charset=utf-8
On Fri, Feb 01, 2019 at 03:14:32PM +0100, Moritz Tacke wrote:
> Hi,
>
> I still want to use the functions that return RVars, so eg.
>
> rvarDouble :: RVar Double
>
> and then the definition of the transformer function would be:
>
> someFunction a b =
> do ...
> the_double <- rvarDouble
> ....
>
> This does not compile, complaining that;
>
> • Couldn't match type ‘Data.Functor.Identity.Identity’ with ‘ST s’
> Expected type: RVarT (ST s) Double
> Actual type: RVar Double
>
> How can I re-user the RVar function in the RVarT monad transformer?
Your declaration of rvarDouble needs to be polymorphic in the monad:
rvarDouble :: Monad m => RVarT m Double
The crucial observation is that RVar is actually a type synonym for
RVarT Identity, so the function can still be made to return a plain
RVar Double, but it can also return an RVarT (ST s) Double, satisfying
the type-checker in the example that you gave.
--
Seph Shewell Brockway, BSc MSc (Glas.)
------------------------------
Message: 3
Date: Sat, 02 Feb 2019 08:02:36 +0000
From: Leonhard Applis <leonhard.app...@protonmail.com>
To: "beginners@haskell.org" <beginners@haskell.org>
Subject: [Haskell-beginners] Parsing Terms in Brackets for Calculator
Message-ID:
<HTHwJeVSXkupBGrfRHdmRdsujghkho3j4A8VdAJUMJfSIkmsTWQOmmxxrkkCAL9G91vTkmPiQKyMvorN1Uo0N0TcEkDSDo6MiwdbMSba8UM=@protonmail.com>
Content-Type: text/plain; charset="utf-8"
Hello,
I'm currently doing my first steps in Haskell with a calculator and I'm stuck
at the parser.
I have a data Term which will build ... basically a tree of operations, and
works fine.
I need help for the function
termify :: [Either Operator Term] -> Term
It takes operators (such as +,**) and terms and output a new, bigger term and
is mostly trivial.
However, all attempts I've done for parsing brackets seem very ... crude and
not like Haskell at all.
The very first pattern match should check for the innermost brackets, and
return termify for everything in between.
I guess that I'm missing some really cool, haskelly solution here.
Best Regards
Leonhard
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