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Today's Topics:
1. Re: Need help understanding the tell function in the Monad
Writer example in LYAH (Francesco Ariis)
2. Bit confused... (mike h)
3. Re: Bit confused... (Francesco Ariis)
4. Re: Bit confused... (mike h)
----------------------------------------------------------------------
Message: 1
Date: Fri, 3 Feb 2017 14:54:20 +0100
From: Francesco Ariis <fa...@ariis.it>
To: beginners@haskell.org
Subject: Re: [Haskell-beginners] Need help understanding the tell
function in the Monad Writer example in LYAH
Message-ID: <20170203135420.ga23...@casa.casa>
Content-Type: text/plain; charset=us-ascii
On Fri, Feb 03, 2017 at 11:31:32AM +0000, Olumide wrote:
> Hello List,
>
> Would someone kindly explain how the tell function interacts with the Writer
> Monad as show below:
> (Example taken from chapter 14 of LYAH
> http://learnyouahaskell.com/for-a-few-monads-more#reader )
>
> multWithLog :: Writer [String] Int
> multWithLog = do
> a <- logNumber 3
> b <- logNumber 5
> tell ["Gonna multiply these two"]
> return (a*b)
>
> Result:
> ghci> runWriter multWithLog
> (15,["Got number: 3","Got number: 5","Gonna multiply these two"])
>
> I know that tell function binds to an argument that is discarded but I don't
> know how its argument "Gonna multiply these two" is concatenated with the
> other the Writer created by logNumber 3 and logNumber 5.
>
> 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."
>
>
> Regards,
>
> - Olumide
Hello Olumide,
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.
------------------------------
Message: 2
Date: Fri, 3 Feb 2017 21:47:04 +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] Bit confused...
Message-ID: <abfba73e-0586-48b4-8d0c-709c8dc1f...@yahoo.co.uk>
Content-Type: text/plain; charset=utf-8
I have
----------
import qualified Data.Map as M
type Link a = (a, Int)
data MChain a = Map a [Link a] deriving (Show)
-------------------
and want to make a Monoid of MChain. So I have
-------------------
instance Monoid (MChain a) where
mempty = M.empty
mappend = undefined
-------------------
this won’t compile and I need M.empty to be Map a [Link a]
The error is
Couldn't match expected type ‘MChain a1’
with actual type ‘M.Map k0 a0’
• In the expression: M.empty :: MChain a
In an equation for ‘mempty’: mempty = M.empty :: MChain a
In the instance declaration for ‘Monoid (MChain a)’
This part confuses me "Couldn't match expected type ‘MChain a1 with actual type
‘M.Map k0 a0’ "
How should I proceed.
Many thanks
Mike
------------------------------
Message: 3
Date: Fri, 3 Feb 2017 22:59:48 +0100
From: Francesco Ariis <fa...@ariis.it>
To: beginners@haskell.org
Subject: Re: [Haskell-beginners] Bit confused...
Message-ID: <20170203215948.ga11...@casa.casa>
Content-Type: text/plain; charset=utf-8
On Fri, Feb 03, 2017 at 09:47:04PM +0000, mike h wrote:
> I have
>
> ----------
> import qualified Data.Map as M
>
> type Link a = (a, Int)
> data MChain a = Map a [Link a] deriving (Show)
>
> -------------------
>
> and want to make a Monoid of MChain. So I have
>
> -------------------
> instance Monoid (MChain a) where
> mempty = M.empty
> mappend = undefined
> -------------------
>
> this won’t compile and I need M.empty to be Map a [Link a]
Hello Mike, I think the error lies in the confusion between
`type` and `data` declaration.
type Something = Int
but
data Something = SomeConstructor Int
So I bet you wanted to write
data MChain a = MChain (M.Map a [Link a]) deriving (Show)
`M.empty` returns a Map.
λ> :t M.empty
M.empty :: M.Map k aj
Hence this will work:
instance Monoid (MChain a) where
mempty = MChain M.empty
mappend = undefined
Does this help?
------------------------------
Message: 4
Date: Sat, 4 Feb 2017 08:13:09 +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] Bit confused...
Message-ID: <6f289e37-3f47-42f7-ad8d-a8c153c36...@yahoo.co.uk>
Content-Type: text/plain; charset="utf-8"
Yes, that’s exactly what I meant to write!!! :)
Thank you Francesco.
Mike
> On 3 Feb 2017, at 21:59, Francesco Ariis <fa...@ariis.it> wrote:
>
> On Fri, Feb 03, 2017 at 09:47:04PM +0000, mike h wrote:
>> I have
>>
>> ----------
>> import qualified Data.Map as M
>>
>> type Link a = (a, Int)
>> data MChain a = Map a [Link a] deriving (Show)
>>
>> -------------------
>>
>> and want to make a Monoid of MChain. So I have
>>
>> -------------------
>> instance Monoid (MChain a) where
>> mempty = M.empty
>> mappend = undefined
>> -------------------
>>
>> this won’t compile and I need M.empty to be Map a [Link a]
>
> Hello Mike, I think the error lies in the confusion between
> `type` and `data` declaration.
>
> type Something = Int
>
> but
>
> data Something = SomeConstructor Int
>
> So I bet you wanted to write
>
> data MChain a = MChain (M.Map a [Link a]) deriving (Show)
>
> `M.empty` returns a Map.
>
> λ> :t M.empty
> M.empty :: M.Map k aj
>
> Hence this will work:
>
> instance Monoid (MChain a) where
> mempty = MChain M.empty
> mappend = undefined
>
> Does this help?
>
>
>
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