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Today's Topics:
1. Re: Need help understanding the tell function in the Monad
Writer example in LYAH (Olumide)
2. Re: Ambiguous type variable prevents the constraint `(Ord
t0)' from being solved. (Sylvain Henry)
3. Applicative for State (mike h)
4. Re: Applicative for State (Francesco Ariis)
5. Re: Applicative for State (mike h)
6. Re: Semigroup Instances (Theodore Lief Gannon)
----------------------------------------------------------------------
Message: 1
Date: Mon, 6 Feb 2017 14:50:25 +0000
From: Olumide <50...@web.de>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <beginners@haskell.org>
Subject: Re: [Haskell-beginners] Need help understanding the tell
function in the Monad Writer example in LYAH
Message-ID: <de5d8587-2942-e47c-ceb2-474161180...@web.de>
Content-Type: text/plain; charset=utf-8; format=flowed
I think I get it now. tell() is defined in Control.Monad.Writer as:
tell :: w -> m ()
tell w = writer ((),w)
*also* the result if the do notation is the last expression; and that's
why the result of the computation will be lost (or disregarded) if
tell() comes last.
- Olumide
On 03/02/2017 13:54, Francesco Ariis wrote:
> a Writer do block can be read as a series of function which all have
> a "hidden parameter". This parameter is the pile of log messages.
> So you could as well substitute `tell ...` with
>
> myTell :: String -> Writer [String] ()
> myTell s = writer ((), [s])
>
> and then in the do block
>
> -- ... receiving a list of log messages
> c <- myTell "something" -- adding mine to the list (and binding
> -- a variable)
> return (a*b) -- c is not being used!
> -- but the log message *is* there
>
> You can verify this yourself by adding `logNumber` statement in a do
> block and not using them in the last return statement. There too log
> will appear even if the bound variable is unused.
>
> multWithLog :: Writer [String] Int
> multWithLog = do
> a <- logNumber 3
> b <- logNumber 5 -- not used but logged
> -- equivalent to: logNumber 5 (without b <-)
> return (a)
>
>> Also, I don't understand the paragraph following the example:
>>
>> "It's important that return (a*b) is the last line, because the result of
>> the last line in a do expression is the result of the whole do expression.
>> Had we put tell as the last line, () would have been the result of this do
>> expression. We'd lose the result of the multiplication. However, the log
>> would be the same."
>
> `tell` is really not much different from `myTell`. Let's examine it again:
>
> myTell :: String -> Writer [String] ()
> myTell s = writer ((), [s])
>
> See the ()? It means it is *actually* returning something, a ().
> Remember that `return` isn't the same `return` as in some imperative
> languages: it only wraps a value in the monad we are using:
>
> return 5
> -- takes `5` and 'lifts' so it is usable inside the Writer
> -- monad: `(5, [])`
>
> Putting a `tell "something"` after a return statement would overwrite
> that result (and gives us back a () instead).
>
> Did this help?
> My tip for really getting a Monad in your brain is to reimplement it.
> It is a very useful exercise.
> Also learning *not* to use the `do notation` helps too, as having
> operators instead of magic makes things easier to understand.
>
> _______________________________________________
> Beginners mailing list
> Beginners@haskell.org
> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>
------------------------------
Message: 2
Date: Mon, 6 Feb 2017 16:09:14 +0100
From: Sylvain Henry <sylv...@haskus.fr>
To: beginners@haskell.org
Subject: Re: [Haskell-beginners] Ambiguous type variable prevents the
constraint `(Ord t0)' from being solved.
Message-ID: <52a6bdf3-95df-93c9-9140-0787ff05f...@haskus.fr>
Content-Type: text/plain; charset=utf-8; format=flowed
Try removing the lambda:
applyd f d x =
let apply_listd l d x = [..]
in let k = 5 -- hash x - todo
[...]
Regards,
Sylvain
On 28/01/2017 21:09, Ivan Kush wrote:
> I get this error (full message at the end of the mail). How could I correct
> my code?
>
>
> ===================
> Code:
> ===================
>
> module Intro where
>
> import Data.Bits -- for xor, .&.
>
> data Func a b
> = Empty
> | Leaf Int [(a, b)]
> | Branch Int Int (Func a b) (Func a b)
>
> applyd =
> let apply_listd l d x =
> case l of
> [] -> d x
> (a, b) : t ->
> let c = compare x a
> in if c == EQ then b
> else if c == GT then apply_listd t d x
> else d x
>
> in \f d x ->
> let k = 5 -- hash x - todo
> in let look t =
> case t of
> Leaf h l | h == k ->
> apply_listd l d x
> Branch p b l r | (k `xor` p) .&. (b - 1) == 0 -> --
> (Branch p b l r) | ((k xor p) .&. (b - 1)) == 0 ->
> look (if k .&. b == 0 then l else r)
> _ -> d x
> in look f
>
>
>
> ===================
> Error:
> ===================
>
> Intro.hs:37:25: error:
> * Ambiguous type variable `t0' arising from a use of `apply_listd'
> prevents the constraint `(Ord t0)' from being solved.
> Relevant bindings include
> l :: [(t0, t)] (bound at Intro.hs:36:28)
> t :: Func t0 t (bound at Intro.hs:34:21)
> look :: Func t0 t -> t (bound at Intro.hs:34:16)
> x :: t0 (bound at Intro.hs:32:15)
> d :: t0 -> t (bound at Intro.hs:32:13)
> f :: Func t0 t (bound at Intro.hs:32:11)
> (Some bindings suppressed; use -fmax-relevant-binds=N or
> -fno-max-relevant-binds)
> Probable fix: use a type annotation to specify what `t0' should be.
> These potential instances exist:
> instance Ord Ordering -- Defined in `GHC.Classes'
> instance Ord Integer
> -- Defined in `integer-gmp-1.0.0.1:GHC.Integer.Type'
> instance Ord a => Ord (Maybe a) -- Defined in `GHC.Base'
> ...plus 22 others
> ...plus five instances involving out-of-scope types
> (use -fprint-potential-instances to see them all)
> * In the expression: apply_listd l d x
> In a case alternative: Leaf h l | h == k -> apply_listd l d x
> In the expression:
> case t of {
> Leaf h l | h == k -> apply_listd l d x
> Branch p b l r
> | (k `xor` p) .&. (b - 1) == 0
> -> look (if k .&. b == 0 then l else r)
> _ -> d x }
>
>
> --
> Best wishes,
> Ivan Kush
> _______________________________________________
> Beginners mailing list
> Beginners@haskell.org
> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
------------------------------
Message: 3
Date: Mon, 6 Feb 2017 19:40:12 +0000
From: mike h <mike_k_hough...@yahoo.co.uk>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <beginners@haskell.org>
Subject: [Haskell-beginners] Applicative for State
Message-ID: <d49355af-31c2-4795-a610-7a547f9af...@yahoo.co.uk>
Content-Type: text/plain; charset=utf-8
Hi,
I have a State by another name, Stat, just to experiment and learn.
newtype Stat s a = Stat { runStat :: s -> (a, s) }
instance Functor (Stat s) where
fmap f (Stat g) = Stat $ \s -> (f a, s) where
(a, s) = g s
instance Applicative (Stat s) where
pure a = Stat $ \s -> (a, s)
(Stat f) <*> (Stat g) = Stat $ \s -> (a, s) where
(a, s) = undefined
I really can’t get what the <*> in the Applicative should be!
I just do see how I ‘get the f out of the Stat’ and then apply it.
I’d be really grateful if someone would explain what it should be and the
steps/reasoning needed to get there.
Many thanks
Mike
------------------------------
Message: 4
Date: Mon, 6 Feb 2017 21:15:29 +0100
From: Francesco Ariis <fa...@ariis.it>
To: beginners@haskell.org
Subject: Re: [Haskell-beginners] Applicative for State
Message-ID: <20170206201529.ga1...@casa.casa>
Content-Type: text/plain; charset=utf-8
On Mon, Feb 06, 2017 at 07:40:12PM +0000, mike h wrote:
> I have a State by another name, Stat, just to experiment and learn.
> [...]
> I really can’t get what the <*> in the Applicative should be!
> I just do see how I ‘get the f out of the Stat’ and then apply it.
>
> I’d be really grateful if someone would explain what it should be and
> the steps/reasoning needed to get there.
Hello Mike,
when writing an instance, you always have to keep in mind: (a) the
signature of the function you are writing and (b) what the instance
is designed to do.
In our case, (<*>) is:
(<*>) :: Applicative f => f (a -> b) -> f a -> f b
-- which we could 'rewrite' as
(<*>) :: Stat s (a -> b) -> Stat s a -> Stat s b
so we grab the results, one being a function and the other a value,
and apply the first to the second.
(b) is "pass the state around in the background". Good, let's put this
in action:
(Stat f) <*> (Stat g) = Stat $ \s ->
let (h, s') = f s -- h is a function :: a -> b
(a, s'') = g s' -- state passing
b = h a in -- the application
(b, s'') -- we're not returning just the tuple, we're returning
-- even the bit before the 'let' statement
And that is that. Was this clear?
------------------------------
Message: 5
Date: Mon, 6 Feb 2017 23:35:31 +0000
From: mike h <mike_k_hough...@yahoo.co.uk>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <beginners@haskell.org>
Subject: Re: [Haskell-beginners] Applicative for State
Message-ID: <cf95cc11-8cf3-4266-b080-9666b327c...@yahoo.co.uk>
Content-Type: text/plain; charset=utf-8
Thanks again Francesco.
Part of my problem was confusing the data and type constructors.
With your solution and my renaming of the data constructors it all became much
clearer!
:)
Mike
> On 6 Feb 2017, at 20:15, Francesco Ariis <fa...@ariis.it> wrote:
>
> On Mon, Feb 06, 2017 at 07:40:12PM +0000, mike h wrote:
>> I have a State by another name, Stat, just to experiment and learn.
>> [...]
>> I really can’t get what the <*> in the Applicative should be!
>> I just do see how I ‘get the f out of the Stat’ and then apply it.
>>
>> I’d be really grateful if someone would explain what it should be and
>> the steps/reasoning needed to get there.
>
> Hello Mike,
>
> when writing an instance, you always have to keep in mind: (a) the
> signature of the function you are writing and (b) what the instance
> is designed to do.
>
> In our case, (<*>) is:
>
> (<*>) :: Applicative f => f (a -> b) -> f a -> f b
>
> -- which we could 'rewrite' as
> (<*>) :: Stat s (a -> b) -> Stat s a -> Stat s b
>
> so we grab the results, one being a function and the other a value,
> and apply the first to the second.
>
> (b) is "pass the state around in the background". Good, let's put this
> in action:
>
> (Stat f) <*> (Stat g) = Stat $ \s ->
> let (h, s') = f s -- h is a function :: a -> b
> (a, s'') = g s' -- state passing
> b = h a in -- the application
> (b, s'') -- we're not returning just the tuple, we're returning
> -- even the bit before the 'let' statement
>
> And that is that. Was this clear?
> _______________________________________________
> Beginners mailing list
> Beginners@haskell.org
> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
------------------------------
Message: 6
Date: Mon, 6 Feb 2017 15:56:33 -0800
From: Theodore Lief Gannon <tan...@gmail.com>
To: Atrudyjane <atrudyj...@protonmail.com>, The Haskell-Beginners
Mailing List - Discussion of primarily beginner-level topics related
to Haskell <beginners@haskell.org>
Subject: Re: [Haskell-beginners] Semigroup Instances
Message-ID:
<CAJoPsuAFrmcVm-t_m4fVcs75R2G=xqvafoqc_nvsdnl3oyb...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
Gmail put you in spam.
If you haven't figured this out since you asked -- it's a matter of
confusing (IMO bad) variable names. Check the data definition:
data Validation a b
= Failure a
| Success b
deriving (Eq, Show)
Failures are always type a, and successes are always type b. The type
variables used in the first line correspond to these. But in the
definitions of (<>), they are just local values. The instance could be
rewritten like so:
instance Semigroup a => Semigroup (Validation a b) where
Success x <> Success y = Success x
Failure x <> Success y = Success y
Success x <> Failure y = Success x
Failure x <> Failure y = Failure (x <> y)
On Thu, Jan 26, 2017 at 1:55 PM, Atrudyjane <atrudyj...@protonmail.com>
wrote:
> I'm currently studying semigroups and trying to figure out how to
> determine which type variables need a semigroup instance. Here are a couple
> of examples from Evan Cameron's github (https://github.com/leshow/
> haskell-programming-book/blob/master/src/Ch15ex.hs):
> (1)
> data Validation a b
> = Failure a
> | Success b
> deriving (Eq, Show)
>
> instance Semigroup a => Semigroup (Validation a b) where
> Success a <> Success b = Success a
> Failure a <> Success b = Success b
> Success a <> Failure b = Success a
> Failure a <> Failure b = Failure (a <> b)
>
> * Why doesn't 'b' need an instance of semigroup?
> (2)
> newtype AccumulateRight a b = AccumulateRight (Validation a b) deriving (
> Eq, Show)
>
> instance Semigroup b => Semigroup (AccumulateRight a b) where
> AccumulateRight (Success a) <>AccumulateRight (Failure b) =AccumulateRight
> (Success a)
> AccumulateRight (Failure a) <>AccumulateRight (Success b) =AccumulateRight
> (Success b)
> AccumulateRight (Failure a) <>AccumulateRight (Failure b) =AccumulateRight
> (Failure a)
>
> AccumulateRight (Success a) <> AccumulateRight (Success b) =
> AccumulateRight (Success (a <> b))
>
> * Why doesn't 'a' need an instance of semigroup?
>
>
> Thank you,
> Andrea
>
>
> Sent with ProtonMail <https://protonmail.com> Secure Email.
>
>
> _______________________________________________
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