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Today's Topics:
1. How does forever works ? (Olivier Revollat)
2. Re: How does forever works ? (Francesco Ariis)
3. Re: How does forever works ? (Olivier Revollat)
4. Storing the time difference between two Monotonic time
results (Awsaf Rahman)
----------------------------------------------------------------------
Message: 1
Date: Mon, 9 Jul 2018 13:31:07 +0200
From: Olivier Revollat <[email protected]>
To: [email protected]
Subject: [Haskell-beginners] How does forever works ?
Message-ID:
<ca+nxgrunlkv+ev9ymmrpmjvtmt+tyo35qsx-5k9djzukwtk...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
>From what I understand, this code
main = forever $ do
putStrLn "OK !"
is equivalent to this one :
main = do
putStrLn "OK !"
main
In the second case, it's a simple recursion, so far so good ... but when I
look at the implementation of "forever" i can't wrap my head around :
http://hackage.haskell.org/package/base-4.11.1.0/docs/src/Control.Monad.html#forever
forever
<http://hackage.haskell.org/package/base-4.11.1.0/docs/src/Control.Monad.html#forever>
a
<http://hackage.haskell.org/package/base-4.11.1.0/docs/src/Control.Monad.html#local-6989586621679327090>
= let a'
<http://hackage.haskell.org/package/base-4.11.1.0/docs/src/Control.Monad.html#local-6989586621679327091>
= a
<http://hackage.haskell.org/package/base-4.11.1.0/docs/src/Control.Monad.html#local-6989586621679327090>
*>
<http://hackage.haskell.org/package/base-4.11.1.0/docs/src/GHC.Base.html#%2A%3E>
a'
<http://hackage.haskell.org/package/base-4.11.1.0/docs/src/Control.Monad.html#local-6989586621679327091>
in a'
<http://hackage.haskell.org/package/base-4.11.1.0/docs/src/Control.Monad.html#local-6989586621679327091>
How does this works ? How does this make an infinite loop ?
I understand that *> discard his right argument but it doesn't help me
understand how forever implement an infinite loop ...
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Message: 2
Date: Mon, 9 Jul 2018 13:54:29 +0200
From: Francesco Ariis <[email protected]>
To: [email protected]
Subject: Re: [Haskell-beginners] How does forever works ?
Message-ID: <[email protected]>
Content-Type: text/plain; charset=us-ascii
Hello Olivier,
On Mon, Jul 09, 2018 at 01:31:07PM +0200, Olivier Revollat wrote:
> In the second case, it's a simple recursion, so far so good ... but when I
> look at the implementation of "forever" i can't wrap my head around :
>
> http://hackage.haskell.org/package/base-4.11.1.0/docs/src/Control.Monad.html#forever
forever a = let a' = a *> a' in a'
`*>` is the same as `>>` (if you have ever met `>>`), with the difference
that the latter works on monads only. In do notation, we could write
forever :: Monad m => m a -> m b
forever a = let a' = do a
a' in
do a'
Which should be easier to get: we `do` a'; a' is nothing but a and a',
so we do a and then a', which is nothing but a and a', etc.
Does this explanation feel right?
As soon as you can, familiarise yourself what the do notation desugars to:
in my opinion plain operators are clearer
-F
------------------------------
Message: 3
Date: Mon, 9 Jul 2018 13:58:56 +0200
From: Olivier Revollat <[email protected]>
To: [email protected]
Subject: Re: [Haskell-beginners] How does forever works ?
Message-ID:
<CA+nXgrU1CeaOw6T=xbCZCsD7u-VcBaFaoYUk=mxq0sarm1c...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
Hey I think it's a bit clearer now .... need to meditate on this :) Thanks !
Le lun. 9 juil. 2018 à 13:55, Francesco Ariis <[email protected]> a écrit :
> Hello Olivier,
>
> On Mon, Jul 09, 2018 at 01:31:07PM +0200, Olivier Revollat wrote:
> > In the second case, it's a simple recursion, so far so good ... but when
> I
> > look at the implementation of "forever" i can't wrap my head around :
> >
> >
> http://hackage.haskell.org/package/base-4.11.1.0/docs/src/Control.Monad.html#forever
>
> forever a = let a' = a *> a' in a'
>
> `*>` is the same as `>>` (if you have ever met `>>`), with the difference
> that the latter works on monads only. In do notation, we could write
>
> forever :: Monad m => m a -> m b
> forever a = let a' = do a
> a' in
> do a'
>
> Which should be easier to get: we `do` a'; a' is nothing but a and a',
> so we do a and then a', which is nothing but a and a', etc.
>
> Does this explanation feel right?
>
> As soon as you can, familiarise yourself what the do notation desugars to:
> in my opinion plain operators are clearer
> -F
> _______________________________________________
> Beginners mailing list
> [email protected]
> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>
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Message: 4
Date: Mon, 9 Jul 2018 14:30:04 +0200
From: Awsaf Rahman <[email protected]>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <[email protected]>
Subject: [Haskell-beginners] Storing the time difference between two
Monotonic time results
Message-ID:
<CAOH+Qtfj+qhxrZgsr2vhad2vmC1yu1A=mdgzypqw127tesv...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
Hello,
I am trying to time a function I have written in haskell using the clock
package in the following way:
*start <- getTime Monotonic*
*evaluate(something)*
*end <- getTime Monotonic*
*fprint (timeSpecs % "\n") start end*
Now what I want is to store the time difference between *start* and *end. *Is
there a way I can do that?
Regards
Awsaf
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