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Today's Topics:
1. Why does sequence (map print [1, 2, 3, 4, 5]) produce [(),
(), (), (), ()] at the end? (Olumide)
2. Re: Why does sequence (map print [1, 2, 3, 4, 5]) produce
[(), (), (), (), ()] at the end? (David McBride)
3. foldr on infinite list to decide prime number (Chul-Woong Yang)
4. Re: foldr on infinite list to decide prime number
(Chul-Woong Yang)
5. Re: foldr on infinite list to decide prime number
(Francesco Ariis)
6. Re: foldr on infinite list to decide prime number
(Chul-Woong Yang)
----------------------------------------------------------------------
Message: 1
Date: Mon, 1 Feb 2016 17:28:27 +0000
From: Olumide <[email protected]>
To: [email protected]
Subject: [Haskell-beginners] Why does sequence (map print [1, 2, 3, 4,
5]) produce [(), (), (), (), ()] at the end?
Message-ID: <[email protected]>
Content-Type: text/plain; charset=utf-8; format=flowed
Hello list,
The question says it all.
BTW, I'm studying LYH and I'm on the chapter on IO. The book offers and
explanation but its not very clear -- to me at least.
Thanks,
- Olumide
------------------------------
Message: 2
Date: Mon, 1 Feb 2016 12:39:50 -0500
From: David McBride <[email protected]>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <[email protected]>
Subject: Re: [Haskell-beginners] Why does sequence (map print [1, 2,
3, 4, 5]) produce [(), (), (), (), ()] at the end?
Message-ID:
<can+tr40r6kgwia8tdda2untckvvgd9rp-mcvpzzrsorkiys...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
map :: (a -> b) -> [a] -> [b]
print :: Show a => a -> IO ()
map print :: Show a => [a] -> [IO ()]
print takes a showable a and creates a procedure that prints it out (IO ()).
So therefore map print causes each element in the array to become a
procedure that prints out its element, thus the return value is [IO()].
Note that it has not actually printed them out. It merely has an array of
as yet unexecuted actions. To print them you'd go like sequence (map
print) [1,2,3], or better yet, sequence_ which returns () instead of [()]
On Mon, Feb 1, 2016 at 12:28 PM, Olumide <[email protected]> wrote:
> Hello list,
>
> The question says it all.
>
> BTW, I'm studying LYH and I'm on the chapter on IO. The book offers and
> explanation but its not very clear -- to me at least.
>
> Thanks,
>
> - Olumide
>
> _______________________________________________
> Beginners mailing list
> [email protected]
> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>
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Message: 3
Date: Tue, 2 Feb 2016 10:32:10 +0900
From: Chul-Woong Yang <[email protected]>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <[email protected]>
Subject: [Haskell-beginners] foldr on infinite list to decide prime
number
Message-ID:
<calmycjpezhgjjz_e1xkej3nxzgxihytqrv-u6dzh4e+syff...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
Hi, all.
While I know that foldr can be used on infinite list to generate infinite
list,
I'm having difficulty in understaind following code:
isPrime n = n > 1 && -- from haskell wiki foldr (\p r -> p*p > n || ((n
`rem` p) /= 0 && r)) True primes primes = 2 : filter isPrime [3,5..]
primes is a infinite list of prime numbers, and isPrime does foldr to get a
boolean value.
What causes foldr to terminate folding?
Any helps will be deeply appreciated.
Thank you.
Chul-Woong
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Message: 4
Date: Tue, 2 Feb 2016 10:36:03 +0900
From: Chul-Woong Yang <[email protected]>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <[email protected]>
Subject: Re: [Haskell-beginners] foldr on infinite list to decide
prime number
Message-ID:
<calmycjr-rt1ctx_+kksjztxbdlqrpfqz3q2+ag7423w63ld...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
I feel sorry for posting mis-formatted code.
I re-post the question.
--
Hi, all.
While I know that foldr can be used on infinite list to generate infinite
list,
I'm having difficulty in understaind following code:
isPrime n = n > 1 && -- from haskell wiki
foldr (\p r -> p*p > n || ((n `rem` p) /= 0 && r)) True primes
primes = 2 : filter isPrime [3,5..]
primes is a infinite list of prime numbers, and isPrime does foldr to get a
boolean value.
What causes foldr to terminate folding?
Any helps will be deeply appreciated.
Thank you.
2016-02-02 10:32 GMT+09:00 Chul-Woong Yang <[email protected]>:
> Hi, all.
>
> While I know that foldr can be used on infinite list to generate infinite
> list,
> I'm having difficulty in understaind following code:
>
> isPrime n = n > 1 && -- from haskell wiki foldr (\p r -> p*p > n || ((n
> `rem` p) /= 0 && r)) True primes primes = 2 : filter isPrime [3,5..]
>
> primes is a infinite list of prime numbers, and isPrime does foldr to get
> a boolean value.
> What causes foldr to terminate folding?
>
> Any helps will be deeply appreciated.
>
> Thank you.
>
> Chul-Woong
>
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Message: 5
Date: Tue, 2 Feb 2016 03:01:23 +0100
From: Francesco Ariis <[email protected]>
To: [email protected]
Subject: Re: [Haskell-beginners] foldr on infinite list to decide
prime number
Message-ID: <[email protected]>
Content-Type: text/plain; charset=us-ascii
On Tue, Feb 02, 2016 at 10:32:10AM +0900, Chul-Woong Yang wrote:
> Hi, all.
>
> While I know that foldr can be used on infinite list to generate infinite
> list,
> I'm having difficulty in understaind following code:
>
> isPrime n = n > 1 && -- from haskell wiki foldr (\p r -> p*p > n || ((n
> `rem` p) /= 0 && r)) True primes primes = 2 : filter isPrime [3,5..]
>
> primes is a infinite list of prime numbers, and isPrime does foldr to get a
> boolean value.
> What causes foldr to terminate folding?
foldr _immediately_ calls the passed function, hence /it can short
circuit/, that isn't the case for foldl.
I wrote an article to explain it [1]. It was drafted in a time when
foldr and friends were monomorphic (i.e. they only worked with lists),
but it should illustrate the point nicely.
Current polymorphic implementation of foldr is:
foldr :: (a -> b -> b) -> b -> t a -> b
foldr f z t = appEndo (foldMap (Endo #. f) t) z
and I must admit I have problems explaining why it terminates
early (as it does).
[1] http://ariis.it/static/articles/haskell-laziness/page.html (more
complex cases section)
------------------------------
Message: 6
Date: Tue, 2 Feb 2016 11:12:35 +0900
From: Chul-Woong Yang <[email protected]>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <[email protected]>
Subject: Re: [Haskell-beginners] foldr on infinite list to decide
prime number
Message-ID:
<CALmycjowjev+dS3vt-fxCz06=mhy342yzaifoe33ectn_em...@mail.gmail.com>
Content-Type: text/plain; charset=UTF-8
Aha! I see.
Thank you for pointing the short-circuiting of (||),
and for nice article.
Regards,
2016-02-02 11:01 GMT+09:00 Francesco Ariis <[email protected]>:
> On Tue, Feb 02, 2016 at 10:32:10AM +0900, Chul-Woong Yang wrote:
>> Hi, all.
>>
>> While I know that foldr can be used on infinite list to generate infinite
>> list,
>> I'm having difficulty in understaind following code:
>>
>> isPrime n = n > 1 && -- from haskell wiki foldr (\p r -> p*p > n || ((n
>> `rem` p) /= 0 && r)) True primes primes = 2 : filter isPrime [3,5..]
>>
>> primes is a infinite list of prime numbers, and isPrime does foldr to get a
>> boolean value.
>> What causes foldr to terminate folding?
>
> foldr _immediately_ calls the passed function, hence /it can short
> circuit/, that isn't the case for foldl.
>
> I wrote an article to explain it [1]. It was drafted in a time when
> foldr and friends were monomorphic (i.e. they only worked with lists),
> but it should illustrate the point nicely.
>
> Current polymorphic implementation of foldr is:
>
> foldr :: (a -> b -> b) -> b -> t a -> b
> foldr f z t = appEndo (foldMap (Endo #. f) t) z
>
> and I must admit I have problems explaining why it terminates
> early (as it does).
>
> [1] http://ariis.it/static/articles/haskell-laziness/page.html (more
> complex cases section)
> _______________________________________________
> Beginners mailing list
> [email protected]
> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
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