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Today's Topics:
1. Re: lazyness and tail recursion (Ovidiu Deac)
2. cabal check complains about missing Setup.hs (Ovidiu Deac)
3. Re: cabal check complains about missing Setup.hs (Daniel Fischer)
4. Re: cabal check complains about missing Setup.hs (Ovidiu Deac)
5. question on types (Jake Penton)
6. Re: question on types (Brandon Allbery)
7. Re: question on types (Julian Porter)
----------------------------------------------------------------------
Message: 1
Date: Thu, 28 Jul 2011 13:27:44 +0300
From: Ovidiu Deac <ovidiud...@gmail.com>
Subject: Re: [Haskell-beginners] lazyness and tail recursion
To: Will Ness <will_...@yahoo.com>
Cc: beginners@haskell.org
Message-ID:
<CAKVsE7viO23K_Jn4=jspwantl2c4jfg1ev10qy1km171ucg...@mail.gmail.com>
Content-Type: text/plain; charset=ISO-8859-1
If I got it right, the idea is to keep the number of operations
between the recursive call and the end of the function constant.
On Wed, Jul 27, 2011 at 1:38 AM, Will Ness <will_...@yahoo.com> wrote:
> Ovidiu Deac <ovidiudeac <at> gmail.com> writes:
>
>>
>> I've read the paragraph below from Real World Haskell
>>
> (http://book.realworldhaskell.org/read/efficient-file-processing-regular-expressions-and-file-name-matching.html
>> section "An important aside: writing lazy functions")
>>
>> I can get a feeling why lazyness compensates for the lack of tail
>> recursion but I don't really understand why it would work in general.
>>
>> My understanding is that that a function which returns a list with
>> non-tail recursion would only work this way with certain usage
>> patterns. In this case the caller of the funtion will have to use the
>> string in a stream-like fashion.
>
> The concept of tail recursion modulo cons is similar. Wikipedia has an
> article.
> Basically it's about keeping one execution frame above the tail-recursive one.
> Lazily accessing a list from the top is a lot like adding elements to it one
> by
> one at the bottom. Prefixing a recursive call with an operation of fixed
> number
> of operations is OK, because they will get consumed and the recursive case
> called, in a fixed number of steps - the instance between current point and
> next
> recursive invocation point won't grow, only get increased at times by a set
> amount and then get consumed.
>
> The problem with non-tail recursion in lazy setting is that the distance grows
> between the current point and the next invocation, and so the amount of "what
> to
> do next?" information grows all the time. In imperative setting it gets
> flipped
> into "what to do after the invocation" but it still grows. In tail-recursion
> (even modulo cons, or (++) with fixed prefix) it's bounded, finite, constant.
> Looking at Scheme example code in WP "continuation-passing style" article
> might
> help. There we see as in translations of tail-recursive functions the
> constructed continuation does not grow in unlimited fashion, instead only
> maybe
> getting prefixed by a fixed amount of operations at times.
>
> Does it make any sense? :)
>
>
> _______________________________________________
> Beginners mailing list
> Beginners@haskell.org
> http://www.haskell.org/mailman/listinfo/beginners
>
------------------------------
Message: 2
Date: Thu, 28 Jul 2011 15:17:04 +0300
From: Ovidiu Deac <ovidiud...@gmail.com>
Subject: [Haskell-beginners] cabal check complains about missing
Setup.hs
To: beginners <beginners@haskell.org>
Message-ID:
<CAKVsE7uR0AX2DV+ugqEMc=uzgmzsej0zrmst83t6nevao2j...@mail.gmail.com>
Content-Type: text/plain; charset=ISO-8859-1
When I run cabal check I get the following message:
$ cabal check
The following errors will cause portability problems on other environments:
* The package is missing a Setup.hs or Setup.lhs script.
Hackage would reject this package.
...but except this the package builds fine so far and the executable runs fine.
So what does this exactly mean?
ovidiu
------------------------------
Message: 3
Date: Thu, 28 Jul 2011 14:47:43 +0200
From: Daniel Fischer <daniel.is.fisc...@googlemail.com>
Subject: Re: [Haskell-beginners] cabal check complains about missing
Setup.hs
To: beginners@haskell.org
Message-ID: <201107281447.44048.daniel.is.fisc...@googlemail.com>
Content-Type: Text/Plain; charset="iso-8859-1"
On Thursday 28 July 2011, 14:17:04, Ovidiu Deac wrote:
> When I run cabal check I get the following message:
>
> $ cabal check
> The following errors will cause portability problems on other
> environments: * The package is missing a Setup.hs or Setup.lhs script.
>
> Hackage would reject this package.
>
>
> ...but except this the package builds fine so far and the executable
> runs fine.
>
> So what does this exactly mean?
It means that, while cabal can build packages with the `simple' build type
without a Setup.[l]hs, other ways to build and install a package, in
particular the old
$ runhaskell ./Setup.hs configure
$ runhaskell ./Setup.hs build
$ runhaskell ./Setup.hs install
need a Setup.[l]hs. Most of the time the simple
module Main (main) where
import Distribution.Simple
main :: IO ()
main = defaultMain
works. If it doesn't, probably even cabal won't be able to build the
package without a Setup script.
Cheers,
Daniel
------------------------------
Message: 4
Date: Thu, 28 Jul 2011 16:07:14 +0300
From: Ovidiu Deac <ovidiud...@gmail.com>
Subject: Re: [Haskell-beginners] cabal check complains about missing
Setup.hs
To: Daniel Fischer <daniel.is.fisc...@googlemail.com>, beginners
<beginners@haskell.org>
Message-ID:
<CAKVsE7sPPMgeXq73dxMG2WC2=tYEdR-dstPmYbGoXve=fnu...@mail.gmail.com>
Content-Type: text/plain; charset=ISO-8859-1
Thanks for the explanation.
And if I want to add the package in Hackage then I should provide a Setup.hs
On Thu, Jul 28, 2011 at 3:47 PM, Daniel Fischer
<daniel.is.fisc...@googlemail.com> wrote:
> On Thursday 28 July 2011, 14:17:04, Ovidiu Deac wrote:
>> When I run cabal check I get the following message:
>>
>> $ cabal check
>> The following errors will cause portability problems on other
>> environments: * The package is missing a Setup.hs or Setup.lhs script.
>>
>> Hackage would reject this package.
>>
>>
>> ...but except this the package builds fine so far and the executable
>> runs fine.
>>
>> So what does this exactly mean?
>
> It means that, while cabal can build packages with the `simple' build type
> without a Setup.[l]hs, other ways to build and install a package, in
> particular the old
>
> $ runhaskell ./Setup.hs configure
> $ runhaskell ./Setup.hs build
> $ runhaskell ./Setup.hs install
>
> need a Setup.[l]hs. Most of the time the simple
>
> module Main (main) where
>
> import Distribution.Simple
>
> main :: IO ()
> main = defaultMain
>
> works. If it doesn't, probably even cabal won't be able to build the
> package without a Setup script.
>
> Cheers,
> Daniel
>
------------------------------
Message: 5
Date: Thu, 28 Jul 2011 19:42:22 -0400
From: Jake Penton <d...@arqux.com>
Subject: [Haskell-beginners] question on types
To: Haskell Beginners <beginners@haskell.org>
Message-ID: <b8652fc3-b605-4b22-b36d-8b87fde04...@arqux.com>
Content-Type: text/plain; charset="us-ascii"
Yikes.
I have been doing a fair bit of productive programming in Haskell, thinking
that I am making a bit of progress. Then I hit something that is apparently
*really simple* that I do not understand at all. How discouraging.
Here is the code that makes me realize I don't understand types or type
inference very well at all yet:
f :: a
f = 1
When I try to load the above, ghci gives me:
No instance for (Num a)
arising from the literal `2'
In the expression: 2
In an equation for `f': f = 2
Failed, modules loaded: none.
Ok, so then I try:
g:: (Num a) => a
g = 2
This compiles.
Why? I mean, why is the first example (defining f) wrong, but the second
example (defining g) ok?
A slight variation on this is:
h:: a
h = 'a'
to which ghci replies:
Couldn't match type `a' with `Char'
`a' is a rigid type variable bound by
the type signature for c :: a
at /Users/David/Project/EoP/ch04/weak.hs:114:1
In the expression: 'a'
In an equation for `c': c = 'a'
This last example is probably the most basic one which I need to understand.
But, why is the problem apparently a different one than in the definition of
"f" above?
Of course, I cannot think of a reason to actually define things as shown above
under ordinary circumstances. The code above is just boiled down to the
simplest case I could find to illustrate my confusion.
I guess I interpret "f::a" to mean "f is some (any) type a". So why can't it be
whatever "1" is, which I suppose is Integer. What is the type system looking
for? And why does the constraint (Num a) make things ok?
Please point me in the direction of any reading I should do to clear up my
confusion.
TIA.
- Jake -
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Message: 6
Date: Thu, 28 Jul 2011 20:23:11 -0400
From: Brandon Allbery <allber...@gmail.com>
Subject: Re: [Haskell-beginners] question on types
To: Jake Penton <d...@arqux.com>
Cc: Haskell Beginners <beginners@haskell.org>
Message-ID:
<cakfcl4voo_xpj9hzmr5ifiaogd-yqixfy3muortwsrmteph...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
On Thu, Jul 28, 2011 at 19:42, Jake Penton <d...@arqux.com> wrote:
> h:: a
> h = 'a'
>
> to which ghci replies:
>
> Couldn't match type `a' with `Char'
> `a' is a rigid type variable bound by
> the type signature for c :: a
> at /Users/David/Project/EoP/ch04/weak.hs:114:1
> In the expression: 'a'
> In an equation for `c': c = 'a'
>
> This last example is probably the most basic one which I need to
> understand. But, why is the problem apparently a different one than in the
> definition of "f" above?
>
The other one was complicated by polymorphism: a numeric literal is
compiled into a call to fromIntegral, whose result type is Num a => a.
The problem is that, when you say something's type is "a", you are not
saying "pick an appropriate type for me"; you are saying "whoever invokes
this can ask for any type it wants" (equivalently: "I promise to be able to
produce *any possible* type"). But then you give the value as Num a => a in
the first example and Char in the second example, neither of which is "any
possible type".
An explicit type is a complete contract; having contracted to produce an "a"
(any type), you can't then offer only a Char or a Num a => a. You have to
satisfy the contract which says "any type", otherwise you're doing the type
checking equivalent of a bait-and-switch.
You can't express "pick a type for me" in a type signature; types are
concrete, and a type variable in a signature is a concrete "anything",
meaning the caller can request whatever it wants and you must produce it.
The type must be *completely* described by the signature; what it says is
what you're committed to, and you can't then offer something else. If you
need a partial type signature, there are some tricks you can use which let
you force types in the implementation without specifying a concrete
signature (see http://okmij.org/ftp/Haskell/types.html#partial-sigs).
--
brandon s allbery allber...@gmail.com
wandering unix systems administrator (available) (412) 475-9364 vm/sms
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Message: 7
Date: Fri, 29 Jul 2011 01:23:51 +0100
From: Julian Porter <julian.por...@porternet.org>
Subject: Re: [Haskell-beginners] question on types
To: Jake Penton <d...@arqux.com>
Cc: Haskell Beginners <beginners@haskell.org>
Message-ID: <0d3e3169-259b-47eb-8ff6-76a09748a...@porternet.org>
Content-Type: text/plain; charset="us-ascii"
The point is that in
f::a
you are saying that f is of an undefined type a. But in
f=1
you're saying that a must be a type that supports the value 1, and not all do
(eg strings, booleans).
Therefore you have to tell the compiler that you are constraining a to allow it
to take the value 1, hence the constraint Num a, which does just that.
The error message is telling you just this: it can't find a constraint that
tells it that a is in the class Num, so it is complaining because what you have
written could fail if you used f in a context where it was inferred to have a
non numerical type. Constraints allow the type-inference engine to spot such
errors, saving you from terrible grief in debugging.
Sent from my iPhone
On 29 Jul 2011, at 00:42, Jake Penton <d...@arqux.com> wrote:
> Yikes.
>
> I have been doing a fair bit of productive programming in Haskell, thinking
> that I am making a bit of progress. Then I hit something that is apparently
> *really simple* that I do not understand at all. How discouraging.
>
> Here is the code that makes me realize I don't understand types or type
> inference very well at all yet:
>
> f :: a
> f = 1
>
> When I try to load the above, ghci gives me:
>
> No instance for (Num a)
> arising from the literal `2'
> In the expression: 2
> In an equation for `f': f = 2
> Failed, modules loaded: none.
>
>
> Ok, so then I try:
>
> g:: (Num a) => a
> g = 2
>
> This compiles.
>
> Why? I mean, why is the first example (defining f) wrong, but the second
> example (defining g) ok?
>
> A slight variation on this is:
>
> h:: a
> h = 'a'
>
> to which ghci replies:
>
> Couldn't match type `a' with `Char'
> `a' is a rigid type variable bound by
> the type signature for c :: a
> at /Users/David/Project/EoP/ch04/weak.hs:114:1
> In the expression: 'a'
> In an equation for `c': c = 'a'
>
> This last example is probably the most basic one which I need to understand.
> But, why is the problem apparently a different one than in the definition of
> "f" above?
>
> Of course, I cannot think of a reason to actually define things as shown
> above under ordinary circumstances. The code above is just boiled down to the
> simplest case I could find to illustrate my confusion.
>
> I guess I interpret "f::a" to mean "f is some (any) type a". So why can't it
> be whatever "1" is, which I suppose is Integer. What is the type system
> looking for? And why does the constraint (Num a) make things ok?
>
> Please point me in the direction of any reading I should do to clear up my
> confusion.
>
> TIA.
>
> - Jake -
> _______________________________________________
> Beginners mailing list
> Beginners@haskell.org
> http://www.haskell.org/mailman/listinfo/beginners
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