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Today's Topics:
1. Confusion over currying, , lambda, and closures
(Lawrence Bottorff)
2. Re: Confusion over currying, , lambda, and closures
(Bob Ippolito)
3. Re: Confusion over currying, , lambda, and closures
(Francesco Ariis)
4. mconcat (PICCA Frederic-Emmanuel)
5. Re: Confusion over currying, , lambda, and closures
(Jay Sulzberger)
----------------------------------------------------------------------
Message: 1
Date: Wed, 23 Dec 2020 21:12:08 -0600
From: Lawrence Bottorff <borg...@gmail.com>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <beginners@haskell.org>
Subject: [Haskell-beginners] Confusion over currying, , lambda, and
closures
Message-ID:
<cafahfsvu5l21g8gavjjsezxzdtgfrjocogy7ws39uheh+kh...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
I have these three versions of addition
addA x y = x + y
addB x = \y -> x + y
addC = \x -> \y -> x + y
and all three add two arguments just fine
> addA 2 3
5
> addB 2 3
5
> addC 2 3
5
but I can't see how addB and addC are actually accomplishing this. addA is
currying, which I don't fully follow. addC I understand beta reduction-wise
(\x -> \y -> x + y) 2 3
(\y -> 2 + y) 3
(2 + 3)
5
but I don't understand addB and what steps are happening currying/beta
reduction-wise. Can someone break down the steps with addA and addB?
LB
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Message: 2
Date: Wed, 23 Dec 2020 20:02:17 -0800
From: Bob Ippolito <b...@redivi.com>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <beginners@haskell.org>
Subject: Re: [Haskell-beginners] Confusion over currying, , lambda,
and closures
Message-ID:
<cacwmpm8g+wptieyrrazt7sotbz3cko0q+yxmqmro8hnqwsk...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
The steps and semantics are the same, the only meaningful difference is
syntax.
These two definitions are indistinguishable:
f x = y
f = \x -> y
In basically the same way that these two expressions are:
(+) 1 2
1 + 2
In Haskell there are many cases where a more convenient but equivalent
syntax exists to express certain terms. This is referred to as syntax sugar.
On Wed, Dec 23, 2020 at 19:12 Lawrence Bottorff <borg...@gmail.com> wrote:
> I have these three versions of addition
>
> addA x y = x + y
> addB x = \y -> x + y
> addC = \x -> \y -> x + y
>
> and all three add two arguments just fine
>
> > addA 2 3
> 5
> > addB 2 3
> 5
> > addC 2 3
> 5
>
> but I can't see how addB and addC are actually accomplishing this. addA is
> currying, which I don't fully follow. addC I understand beta reduction-wise
>
> (\x -> \y -> x + y) 2 3
> (\y -> 2 + y) 3
> (2 + 3)
> 5
>
> but I don't understand addB and what steps are happening currying/beta
> reduction-wise. Can someone break down the steps with addA and addB?
>
> LB
>
>
> _______________________________________________
> Beginners mailing list
> Beginners@haskell.org
> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>
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Message: 3
Date: Thu, 24 Dec 2020 05:37:22 +0100
From: Francesco Ariis <fa...@ariis.it>
To: beginners@haskell.org
Subject: Re: [Haskell-beginners] Confusion over currying, , lambda,
and closures
Message-ID: <20201224043722.GB911@extensa>
Content-Type: text/plain; charset=utf-8
Il 23 dicembre 2020 alle 21:12 Lawrence Bottorff ha scritto:
> I have these three versions of addition
>
> addA x y = x + y
> addB x = \y -> x + y
> addC = \x -> \y -> x + y
>
> […]
>
> I can't see how addB and addC are actually accomplishing this. addA is
> currying, which I don't fully follow.
Parameters on the left-hand side of a function definition can always be
expressed as a lambda, by «plucking» them from the rightmost one
addA x y = x + y
addA x = \y -> x + y
addA = \x -> \y -> x + y -- those three are equal!
Intuitively you can say that first we are binding the patterns in the
function definition and then the one left in the lambda right-hand side
(if any). So:
add 3 5
addA x = \y -> x + y 3 5
\y -> 3 + y 5
3 + 5
I suspect what the book is trying to teach you is that you can partially
apply a function and pass it around merrily:
map (addA 3) [1, 2, 3]
So by writing `addA 3` we obtain:
addA :: Int -> Int -> Int
addA :: Int -> (Int -> Int) -- can also be written like this
-- since `(->)` is associates to
-- the right
add3 :: Int -> Int -- applying addA to 3
another function with one parameter less. Once everything is applied, you
will get your result!
------------------------------
Message: 4
Date: Thu, 24 Dec 2020 06:48:51 +0000
From: PICCA Frederic-Emmanuel
<frederic-emmanuel.pi...@synchrotron-soleil.fr>
To: "beginners@haskell.org" <beginners@haskell.org>
Subject: [Haskell-beginners] mconcat
Message-ID:
<a2a20ec3b8560d408356cac2fc148e5301bf027...@sun-dag3.synchrotron-soleil.fr>
Content-Type: text/plain; charset="us-ascii"
Hello,
I have a monoid and I would like to know if there is a way to do the
concatenation in parallele ?
the mappend method takes time for my type.
thanks for you help.
Frederic
------------------------------
Message: 5
Date: Thu, 24 Dec 2020 07:15:25 +0000 ()
From: Jay Sulzberger <j...@panix.com>
To: The Haskell-Beginners Mailing List <beginners@haskell.org>
Subject: Re: [Haskell-beginners] Confusion over currying, , lambda,
and closures
Message-ID: <pine.neb.4.64.2012240642420.4...@panix3.panix.com>
Content-Type: TEXT/PLAIN; charset=US-ASCII; format=flowed
On Wed, 23 Dec 2020, Lawrence Bottorff <borg...@gmail.com> wrote:
> I have these three versions of addition
>
> addA x y = x + y
> addB x = \y -> x + y
> addC = \x -> \y -> x + y
>
> and all three add two arguments just fine
>
>> addA 2 3
> 5
>> addB 2 3
> 5
>> addC 2 3
> 5
>
> but I can't see how addB and addC are actually accomplishing this. addA is
> currying, which I don't fully follow. addC I understand beta reduction-wise
>
> (\x -> \y -> x + y) 2 3
> (\y -> 2 + y) 3
> (2 + 3)
> 5
>
> but I don't understand addB and what steps are happening currying/beta
> reduction-wise. Can someone break down the steps with addA and addB?
>
> LB
Dear Lawrence Bottorf, your questions are delightful!
Heaven forwarding, I will respond more fully and clearly.
The questions you ask strike at the heart of the issue:
Why is Haskell so hard to learn?
I offer no decent answer here, but I give below a transcript of an
incoherent session with ghci on panix3.panix.com,
panix3% ghci --version
The Glorious Glasgow Haskell Compilation System, version 7.8.4
panix3% uname -a
NetBSD panix3.panix.com 9.0 NetBSD 9.0 (PANIX-XEN-USER) #1: Sun May 3
21:15:18 EDT 2020
r...@juggler.panix.com:/misc/obj64/misc/devel/netbsd/9.0/src/sys/arch/amd64/compile/PANIX-XEN-USER
amd64
panix3%
Below is the whole transcript of the session. Note two things:
1. I do not know ghci.
2. I failed to ask the ':t" questions in the right order, for a good
training video.
But I hope the transcript is helpful.
oo--JS.
bash-5.0$ ghci
GHCi, version 7.8.4: http://www.haskell.org/ghc/ :? for help
Loading package ghc-prim ... linking ... done.
Loading package integer-gmp ... linking ... done.
Loading package base ... linking ... done.
Prelude> let addA x y = x + y
Prelude> addA
<interactive>:6:1:
No instance for (Show (a0 -> a0 -> a0))
arising from a use of `print'
In a stmt of an interactive GHCi command: print it
Prelude> print addA
<interactive>:7:1:
No instance for (Show (a0 -> a0 -> a0))
arising from a use of `print'
In the expression: print addA
In an equation for `it': it = print addA
<interactive>:7:7:
No instance for (Num a0) arising from a use of `addA'
The type variable `a0' is ambiguous
Note: there are several potential instances:
instance Num Double -- Defined in `GHC.Float'
instance Num Float -- Defined in `GHC.Float'
instance Integral a => Num (GHC.Real.Ratio a)
-- Defined in `GHC.Real'
...plus three others
In the first argument of `print', namely `addA'
In the expression: print addA
In an equation for `it': it = print addA
Prelude> let addA x y = x + y
Prelude> print it
<interactive>:9:7:
Not in scope: `it'
Perhaps you meant `id' (imported from Prelude)
Prelude> addA
<interactive>:10:1:
No instance for (Show (a0 -> a0 -> a0))
arising from a use of `print'
In a stmt of an interactive GHCi command: print it
Prelude> addA 7
<interactive>:11:1:
No instance for (Show (a0 -> a0)) arising from a use of `print'
In a stmt of an interactive GHCi command: print it
Prelude> addA 7 5
12
Prelude> (addA 7)
<interactive>:13:1:
No instance for (Show (a0 -> a0)) arising from a use of `print'
In a stmt of an interactive GHCi command: print it
Prelude> :t (addA 7)
(addA 7) :: Num a => a -> a
Prelude> :t addA
addA :: Num a => a -> a -> a
Prelude> :t addA 7
addA 7 :: Num a => a -> a
Prelude> :t addA 7 5
addA 7 5 :: Num a => a
Prelude> let addB x = \y -> x + y
Prelude> :t addB
addB :: Num a => a -> a -> a
Prelude> :t addB 7
addB 7 :: Num a => a -> a
Prelude> :t (addB 7)
(addB 7) :: Num a => a -> a
Prelude> :t addB 7 5
addB 7 5 :: Num a => a
Prelude> let addC = \x -> \y -> x + y
Prelude> :t addC
addC :: Num a => a -> a -> a
Prelude> :t (addC)
(addC) :: Num a => a -> a -> a
Prelude> :t addC 7
addC 7 :: Num a => a -> a
Prelude> :t (addC 7)
(addC 7) :: Num a => a -> a
Prelude> :t addC 7 5
addC 7 5 :: Num a => a
Prelude> addC 7 5
12
Prelude> (addC 7 5)
12
Prelude> (exit)
<interactive>:31:2: Not in scope: `exit'
Prelude>
Leaving GHCi.
bash-5.0$
------------------------------
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