[Haskell-cafe] newbie question about Functional dependencies conflict between instance declarations:.....
Hello,
I largely don't know what I'm doing or even trying to do, it is a voyage into
the unknownbutif I go...
{-# LANGUAGE MultiParamTypeClasses #-}
{-# LANGUAGE FunctionalDependencies #-}
{-# LANGUAGE FlexibleInstances #-}
{-# LANGUAGE UndecidableInstances #-}
class Foo x y | x - y, y - x
instance Foo Integer Integer
That seems to workand my head seems to say...your created some sort of
binary relation between 2 types...and made Integer,Integer a member of it...
Something like that anyway
Then I go
data Bar
instance Foo Bar x
Error!but I'm think I understand thisI can't claim that Bar,x is a
member of Foo and Integer,Integer is member of Foo and preserve my functional
dependencies, because Bar,Integer is now a member of Foo..
Bad programmer...
So how I naively go
class NotAnInteger a
instance (NotAnInteger x) = Foo Bar x
I haven't declared integer to be NotAnIntegerso (in a closed
world)this would seem to exclude the contradictionbut...
Functional dependencies conflict between instance declarations:
instance Foo Integer Integer -- Defined at liam1.lhs:7:12
instance NotAnInteger x = Foo Bar x -- Defined at liam1.lhs:13:12
So
i)I clearly don't understand something about the type
system.
ii) I don't know how to restrict type variables in instance
declarationsi.e. how do I use the notion of Foo across different
combinations of types, without them colliding.
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Re: [Haskell-cafe] newbie question about Functional dependencies conflict between instance declarations:.....
Ah
So it isn't a closed world
So how do I stop my instances clashing?
The x in
instance Foo Bar x
is never intended to be Integer.
Mark Nicholls | Lead broadcast corporate architect, Programmes Development
- Viacom International Media Networks
A: 17-29 Hawley Crescent London NW1 8TT | e:
nicholls.m...@vimn.commailto:m...@vimn.com T: +44 (0)203 580 2223
[Description: cid:image001.png@01CD488D.9204D030]
From: Tikhon Jelvis [mailto:tik...@jelv.is]
Sent: 05 July 2013 2:08 PM
To: Nicholls, Mark
Cc: haskell-cafe
Subject: Re: [Haskell-cafe] newbie question about Functional dependencies
conflict between instance declarations:.
You're running into the open worldassumption--anybody could come along and
make Integer part of your NotAnInteger class, and there's nothing you can do to
stop them. This is a design tradeoff for typeclasses: typeclass instances are
always global and are exported to all other modules you use. This means you
cannot ensure a type is *not* part of a typeclass. (Or, at the very least, you
can't convince GHC of this fact.)
For more information about this, take a look at the following StackOverflow
question: http://stackoverflow.com/questions/8728596/explicitly-import-instances
On Jul 5, 2013 8:47 AM, Nicholls, Mark
nicholls.m...@vimn.commailto:nicholls.m...@vimn.com wrote:
Hello,
I largely don't know what I'm doing or even trying to do, it is a voyage into
the unknownbutif I go...
{-# LANGUAGE MultiParamTypeClasses #-}
{-# LANGUAGE FunctionalDependencies #-}
{-# LANGUAGE FlexibleInstances #-}
{-# LANGUAGE UndecidableInstances #-}
class Foo x y | x - y, y - x
instance Foo Integer Integer
That seems to workand my head seems to say...your created some sort of
binary relation between 2 types...and made Integer,Integer a member of it...
Something like that anyway
Then I go
data Bar
instance Foo Bar x
Error!but I'm think I understand thisI can't claim that Bar,x is a
member of Foo and Integer,Integer is member of Foo and preserve my functional
dependencies, because Bar,Integer is now a member of Foo..
Bad programmer...
So how I naively go
class NotAnInteger a
instance (NotAnInteger x) = Foo Bar x
I haven't declared integer to be NotAnIntegerso (in a closed
world)this would seem to exclude the contradictionbut...
Functional dependencies conflict between instance declarations:
instance Foo Integer Integer -- Defined at liam1.lhs:7:12
instance NotAnInteger x = Foo Bar x -- Defined at liam1.lhs:13:12
So
i)I clearly don't understand something about the type
system.
ii) I don't know how to restrict type variables in instance
declarationsi.e. how do I use the notion of Foo across different
combinations of types, without them colliding.
CONFIDENTIALITY NOTICE
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(and other intellectual property rights). If you are not the intended recipient
please e-mail the sender and then delete the email and any attached files
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While MTV Networks Europe has taken steps to ensure that this email and any
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message and any attachments are virus free and do not affect your systems /
data.
Communicating by email is not 100% secure and carries risks such as delay, data
corruption, non-delivery, wrongful interception and unauthorised amendment. If
you communicate with us by e-mail, you acknowledge and assume these risks, and
you agree to take appropriate measures to minimise these risks when e-mailing
us.
MTV Networks International, MTV Networks UK Ireland, Greenhouse, Nickelodeon
Viacom Consumer Products, VBSi, Viacom Brand Solutions International, Be
Viacom, Viacom International Media Networks and VIMN and Comedy Central are all
trading names of MTV Networks Europe. MTV Networks Europe is a partnership
between MTV Networks Europe Inc. and Viacom Networks Europe Inc. Address for
service in Great Britain is 17-29 Hawley Crescent, London, NW1 8TT.
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Communicating by email is not 100
RE: [Haskell-cafe] type questions again....
Is my problem here, simply that the forall extension in GHC is misleading.that the forall in MkSwizzle :: (forall a. Ord a = [a] - [a]) - Swizzle is not the same beast as the forall in.. data Accum a = forall s. MkAccum s (a - s - s) (s - a) really data Accum a = exists s. MkAccum s (a - s - s) (s - a) would be much better syntax don't get me wrongI still don't especially understand...but if it's a misleading label...I can mentally substitute exists whenever I see a forall without a =. From: Luke Palmer [mailto:[EMAIL PROTECTED] Sent: Fri 11/01/2008 18:03 To: Nicholls, Mark Cc: haskell-cafe@haskell.org Subject: Re: [Haskell-cafe] type questions again On Jan 11, 2008 5:47 PM, Nicholls, Mark [EMAIL PROTECTED] wrote: If you wrap an existential type up in a constructor, not much changes: If you wrap a what?should this read existential or universal? Whoops, right, universal. newtype ID = ID (forall a. a - a) ID can hold any value of type forall a. a - a; i.e. it can hold any value which exhibits the property that it can give me a value of type a - a for any type a I choose. In this case the only things ID can hold are id and _|_, because id is the only function that has that type. Here's how I might use it: It's the only function you've defined the type of Id2 :: forall a. a - a Now it can hold id2? Well, that's not what I meant, but yes it can hold id2. What I meant was that, in this case, id2 = _|_ or id2 = id, there are no other possibilities. id' :: forall a. Num a = a - a id' = id -- it doesn't have to use the constraint if it doesn't want to it doesn't have to use the constraint if it doesn't want to ? If id was of type.. Id::forall a. Ord a = a - a Then I assume it would complain? Yes. but you need to use constructors to use them. I'll write them using GADTs, because I think they're a lot clearer that way: data NUM' where NUM' :: Num a = a - NUM' Look at the type of the constructor NUM'. It has a universal type, meaning whatever type a you pick (as long as it is a Num), you can create a NUM' value with it. yes and then it goes wrong... So the type contained inside the NUM' constructor ? is called existential (note that NUM' itself is just a regular ADT; NUM' is not existential). Why existentialsee below...I have a guess Okay, I was being handwavy here. Explaining this will allow me to clear this up. If you take the non-GADT usage of an existential type: data Foo = forall a. Num a = Foo a That is isomorphic to a type: data Foo = Foo (exists a. Num a = a) Except GHC doesn't support a keyword 'exists', and if it did, it would only be able to be used inside constructors like this (in order for inference to be decidable), so why bother? That's what I meant by the type inside the constructor, Foo is not existential, (exists a. Num a = a) is. So when you have: negNUM' :: NUM' - NUM' negNUM' (NUM' n) = NUM' (negate n) Here n has an existential type, specifically (exists a. Num a = a). Here the argument could have been constructed using any numeric type n, so we know very little about it. The only thing we know is that it is of some type a, and that type has a Num instance. I think one of the harrowing things about Haskell is the practice of overloading data constructors with type namesit confuses the hell out of me Yeah that took a little getting used to for me too. But how am I supposed to come up with enough names if I want to name them differently!? That would require too much creativity... :-) OK so this declaration says that forall x constructed using NUM' n...there *exists* a type T s.t. T is a member of type class NUM... (you probably meant type class Num here) which in term implies that that there exists the function negate... yes? Huh, I had never thought of it like that, but yes. I just realized that I think of programming in a way quite different than I think of logic. Maybe I should try to have my brain unify them. doubleNUM' :: NUM' - NUM' doubleNUM' (NUM' n) = NUleM' (n + n) We can add it to itself, but note: addNUM' :: NUM' - NUM' - NUM' addNUM' (NUM' a) (NUM' b) = NUM (a + b) -- Illegal! We can't add them to each other, because the first argument could have been constructed with, say, a Double and the other with a Rational. But do you see why we're allowed to add it to itself? We can add it to itself because + is of type a-a-a... Yep, so whatever type a n happens to have, it matches both arguments. How about this: data Variant where Variant :: a - Variant This is a type that can be constructed with any value whatsoever. Looks pretty powerful... but it isn't. Why not? Eeek. Because a could be of any type whatsover?...so how I actually do anything with it? Right. Luke
[Haskell-cafe] type questions again....
Can someone explain (in simple terms) what is meant by existential and universal types. Preferably illustrating it in terms of logic rather than lambda calculus. There's plenty of stuff out there on itbut most of it seems double dutch (no offense to the dutch intended). ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
RE: [Haskell-cafe] type questions again....
-Original Message- From: Luke Palmer [mailto:[EMAIL PROTECTED] Sent: 11 January 2008 17:11 To: Nicholls, Mark Cc: haskell-cafe@haskell.org Subject: Re: [Haskell-cafe] type questions again 2008/1/11 Nicholls, Mark [EMAIL PROTECTED]: Can someone explain (in simple terms) what is meant by existential and universal types. Preferably illustrating it in terms of logic rather than lambda calculus. Well, I don't know about logic. While they are certainly related to existential and universal types in logic, I don't really see a way to explain them in terms of that. Universal types are easy, you use them every day in Haskell. Take for example id: id :: a - a Or better illustrated (using ghc extension): id :: forall a. a - a That means that for any type a I pick, I can get a value of type a - a from id. Yepit's universal because forall types a. If you wrap an existential type up in a constructor, not much changes: If you wrap a what?should this read existential or universal? newtype ID = ID (forall a. a - a) ID can hold any value of type forall a. a - a; i.e. it can hold any value which exhibits the property that it can give me a value of type a - a for any type a I choose. In this case the only things ID can hold are id and _|_, because id is the only function that has that type. Here's how I might use it: It's the only function you've defined the type of Id2 :: forall a. a - a Now it can hold id2? applyID :: ID - (Int,String) - (Int,String) applyID (ID f) (i,s) = (f i, f s) Note how I can use f, which is a universal type, on both an Int and a String in the same place. Yep. You can also put typeclass constraints on universals. Take the universal type forall a. Num a = a - a. Among functions that have this type are: add1 :: forall a. Num a = a - a add1 x = x + 1 id' :: forall a. Num a = a - a id' = id -- it doesn't have to use the constraint if it doesn't want to it doesn't have to use the constraint if it doesn't want to ? If id was of type.. Id::forall a. Ord a = a - a Then I assume it would complain? Wrapping this up in a constructor: newtype NUM = NUM (forall a. Num a = a - a) We can create values: NUM add1 :: NUM NUM id :: NUM And use them: applyNUM :: NUM - (Int, Double) - (Int, Double) applyNUM (NUM f) (i,d) = (f i, f d) Yep. Existential types are dual, Dual? (like a dual basis rather than double?) but you need to use constructors to use them. I'll write them using GADTs, because I think they're a lot clearer that way: data NUM' where NUM' :: Num a = a - NUM' Look at the type of the constructor NUM'. It has a universal type, meaning whatever type a you pick (as long as it is a Num), you can create a NUM' value with it. yes and then it goes wrong... So the type contained inside the NUM' constructor ? is called existential (note that NUM' itself is just a regular ADT; NUM' is not existential). Why existentialsee below...I have a guess So when you have: negNUM' :: NUM' - NUM' negNUM' (NUM' n) = NUM' (negate n) Here the argument could have been constructed using any numeric type n, so we know very little about it. The only thing we know is that it is of some type a, and that type has a Num instance. I think one of the harrowing things about Haskell is the practice of overloading data constructors with type namesit confuses the hell out of me OK so this declaration says that forall x constructed using NUM' n...there *exists* a type T s.t. T is a member of type class NUM... which in term implies that that there exists the function negate... yes? It's existential...because the word exists occurred in the above translation. So we can perform operations to it which work for any Num type, such as negate, but not things that only work for particular Num types, such as div. Because the existence of the value implies the existence of a type in the typeclass? doubleNUM' :: NUM' - NUM' doubleNUM' (NUM' n) = NUleM' (n + n) We can add it to itself, but note: addNUM' :: NUM' - NUM' - NUM' addNUM' (NUM' a) (NUM' b) = NUM (a + b) -- Illegal! We can't add them to each other, because the first argument could have been constructed with, say, a Double and the other with a Rational. But do you see why we're allowed to add it to itself? We can add it to itself because + is of type a-a-a... I think How about this: data Variant where Variant :: a - Variant This is a type that can be constructed with any value whatsoever. Looks pretty powerful... but it isn't. Why not? Eeek. Because a could be of any type whatsover?...so how I actually do anything with it? Don't know complete guess really. Luke ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell
[Haskell-cafe] confusion about 'instance'....
Should be straight forwardsimplest example is...
class A a
data D = D1
instance A D
fine.D is declared to be a member of type class A
what about.
class A a
type T = (forall x.Num x=x)
instance A T
error!...
Illegal polymorphic or qualified type: forall x. (Num x) = x
In the instance declaration for `A T'
I am simply trying to state that all members of typeclass Num are of
typeclass A
Doesn't like it.
Does this mean that instance only operates on 'atomic' (for want of a
better word) types?
-Original Message-
From: Peter Verswyvelen [mailto:[EMAIL PROTECTED] On Behalf Of
Peter Verswyvelen
Sent: 03 January 2008 12:02
To: Nicholls, Mark
Cc: haskell-cafe@haskell.org
Subject: RE: [Haskell-cafe] Is there anyone out there who can translate
C# generics into Haskell?
Hi Mark,
foo1 :: Int - obj - String
Yep...I think that's what I'd dothough I would have done...
foo1 :: obj - Int - String
Does that matter?
Well, it's a good habit in Haskell to move the most important
parameter to
the end of the argument list. See e.g.
http://www.haskell.org/haskellwiki/Parameter_order.
OK but I was going to go onto
Interface IXA where A : IXA {}
and
Interface IXA,B where A : B {}
No, I would not know how to that in Haskell using type classes. It seems
Haskell does not allow cycles in type class definitions. But as I'm new,
this does not mean it's not possible. It's more important to know *what*
you
are trying to do, than to give a solution in a different language, since
OO
and FP are kind of orthogonal languages.
Where I cannot see a way to do the above in Haskell at allas
interfaces effectively operator on type classes not typeswhich
seems
inherently more powerful
Yeah, kind of makes sense. I liked interfaces in C# a lot, but when I
started doing everything with interfaces, I found the lack of support
for
mixins or default implementations problematic. This ended up in a
lot of
copy/paste or encapsulating the implementations into a static class with
plain functions, a mess.
But if these could be done in Haskell the see what could be made of
stuff likewhich is obviously problematic in C# it obviously
doesn't
workbut potentially does make 'sense'.
Interface IXA : A {}
Ah! That's one of the bigger restrictions in C# yes! C++ can do that;
ATL
uses it a lot, and I also like that approach. You can emulate mixins
with
that, and still stay in the single inheritance paradigm. In Haskell you
don't do that at all of course, since you avoid thinking about objects
and
inheritance in the first place.
OO is strange. They offer you the nice concept of inheritance, and then
the
guidelines tell you: don't use too many levels of inheritance...
Although
I've build huge projects using OO, it always felt a bit like unsafe
hacking.
I don't really have that feeling with Haskell, but that could also be
because I'm too new to the language ;-)
I'm looking at Haskell because of the formality of it's type
systembut I'm actually not convinced it is as powerful as an OO
onei.e. OO ones operatate principally (in Haskell speak) on type
classes not types
Maybe you are right, I don't know, my theoritical skills are not high
enough
to answer that. Haskell just feels better to me, although the lack of
a
friendly productive IDE and large standard framework remains a bit of a
burden.
Good luck,
Peter
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Nicholls, Mark
Sent: Wednesday, January 02, 2008 5:41 PM
To: haskell-cafe@haskell.org
Subject: [Haskell-cafe] Is there anyone out there who can translate C#
generics into Haskell?
I'm trying to translate some standard C# constucts into Haskell... some
of this seems easy
Specifically
1)
Interface IX
{
}
2)
Interface IXA
{
}
3)
Interface IXA
Where A : IY
{
}
4)
Interface IXA : IZ
Where A : IY
{
}
I can take a punt at the first 2but then it all falls apart
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RE: [Haskell-cafe] confusion about 'instance'....
Thanks for your response, I think you helped me on one of my previous
abberations.
Hmmmthis all slightly does my head inon one hand we have
typesthen type classes (which appear to be a relation defined on
types)then existential types...which now appear not to be treated
quite in the same way as 'normal' typesand in this instance the
syntax even seems to changedoes
instance Num a = A a
Mean the same thing as
instance A (forall a.Num a=a)
seems like a weird questions, as you're saying my version doesn't mean
anythingbut
does it mean that forall types 'a', if 'a' is a member of the class
Num, then 'a' is a member of class 'A'
and secondly in what way can this construct lead to undecidable
instances
What are the instances, and what about them is undecidableseems
pretty decidable to me?
What is the ramifications of turning this option on?
-Original Message-
From: Luke Palmer [mailto:[EMAIL PROTECTED]
Sent: 10 January 2008 13:14
To: Nicholls, Mark
Cc: haskell-cafe@haskell.org
Subject: Re: [Haskell-cafe] confusion about 'instance'
On Jan 10, 2008 1:03 PM, Nicholls, Mark [EMAIL PROTECTED] wrote:
Should be straight forwardsimplest example is...
class A a
data D = D1
instance A D
fine.D is declared to be a member of type class A
what about.
class A a
type T = (forall x.Num x=x)
instance A T
error!...
Illegal polymorphic or qualified type: forall x. (Num x) = x
In the instance declaration for `A T'
I am simply trying to state that all members of typeclass Num are of
typeclass A
Ahh, you want:
instance Num a = A a
Sorry to lead you on, but that actually is not legal (and
-fallow-undecidable-instances
will make it legal, but you don't want that, because instances of this
particular form
are very likely to lead to an infinite loop).
Adding supertypes like this is not possible in Haskell. I really want
it to be, but alas...
Luke
Doesn't like it.
Does this mean that instance only operates on 'atomic' (for want of a
better word) types?
-Original Message-
From: Peter Verswyvelen [mailto:[EMAIL PROTECTED] On Behalf Of
Peter Verswyvelen
Sent: 03 January 2008 12:02
To: Nicholls, Mark
Cc: haskell-cafe@haskell.org
Subject: RE: [Haskell-cafe] Is there anyone out there who can
translate
C# generics into Haskell?
Hi Mark,
foo1 :: Int - obj - String
Yep...I think that's what I'd dothough I would have done...
foo1 :: obj - Int - String
Does that matter?
Well, it's a good habit in Haskell to move the most important
parameter to
the end of the argument list. See e.g.
http://www.haskell.org/haskellwiki/Parameter_order.
OK but I was going to go onto
Interface IXA where A : IXA {}
and
Interface IXA,B where A : B {}
No, I would not know how to that in Haskell using type classes. It
seems
Haskell does not allow cycles in type class definitions. But as I'm
new,
this does not mean it's not possible. It's more important to know
*what*
you
are trying to do, than to give a solution in a different language,
since
OO
and FP are kind of orthogonal languages.
Where I cannot see a way to do the above in Haskell at allas
interfaces effectively operator on type classes not typeswhich
seems
inherently more powerful
Yeah, kind of makes sense. I liked interfaces in C# a lot, but when I
started doing everything with interfaces, I found the lack of support
for
mixins or default implementations problematic. This ended up in a
lot of
copy/paste or encapsulating the implementations into a static class
with
plain functions, a mess.
But if these could be done in Haskell the see what could be made of
stuff likewhich is obviously problematic in C# it obviously
doesn't
workbut potentially does make 'sense'.
Interface IXA : A {}
Ah! That's one of the bigger restrictions in C# yes! C++ can do that;
ATL
uses it a lot, and I also like that approach. You can emulate mixins
with
that, and still stay in the single inheritance paradigm. In Haskell
you
don't do that at all of course, since you avoid thinking about
objects
and
inheritance in the first place.
OO is strange. They offer you the nice concept of inheritance, and
then
the
guidelines tell you: don't use too many levels of inheritance...
Although
I've build huge projects using OO, it always felt a bit like unsafe
hacking.
I don't really have that feeling with Haskell, but that could also be
because I'm too new to the language ;-)
I'm looking at Haskell because of the formality of it's type
systembut I'm actually not convinced it is as powerful as an OO
onei.e. OO ones operatate principally (in Haskell speak) on
type
classes not types
Maybe you are right, I don't know, my theoritical skills are not high
enough
to answer that. Haskell just feels better to me, although the lack
of
a
friendly productive IDE and large standard framework remains a bit
RE: [Haskell-cafe] confusion about 'instance'....
class A a
type T = (forall x.Num x=x)
instance A T
type declares a synonym, like #define in C - but working only on
types.
So, essentially, you wrote
Yep that's fine..
instance A (forall x.Num x = x)
Yep
which is not very Haskelly.
Hmmm...
I am simply trying to state that all members of typeclass Num are of
typeclass A
You can't do that.
Because it wont let me...or because it makes no sense?
But, if there would not be any other instances of A,
then you don't need it at all, you can just use Num class. And if
there
are some, for example
Ok but there may beI'm just trying to get my head around Haskells
type system
data D = D
instance A D
Yep.
D is a member of A
then it can happen (well, it's unlikely, but possible) that you or
some
other developer working on your code would declare
instance Num D where ...
D is a member of Num
(and I'm assuming that we've gotAll Nums are also members of
Awhich is fine...so far).
So...Num x implies A x
So...D is a member of A
Fine.
After that, you would have two instances of A for the type D, one
defined
explicitly and one derived from the Num instance.
I would have 2 declarations that D is a member of Aboth consistent.
That's not a problem for
this empty class, but if class A is not empty, say
class A x where a :: x
then the compiler would be unable to decide which instance (and which
a)
to choose. So allowing that leads to non-obvious bugs.
This slightly bamboozles me.
I only have 1 type.
If I say my name is mark twice, it doesn't mean I belong to set of
objects called Mark twice
which makes me think that type classes are not simple relations on types
after allthey appear to be relations on declarations of types being
members of a class.
So in my example...there exists two instances of me claiming my name is
Mark.
However, if you trust yourself enough, you can do what you want to in
this
way:
{-# OPTIONS_GHC -fglasgow-exts -fallow-undecidable-instances #-}
class A a
instance Num a = A a
Hmmm...OK...
It all seems a little odd
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RE: Re[2]: [Haskell-cafe] confusion about 'instance'....
-Original Message- From: Bulat Ziganshin [mailto:[EMAIL PROTECTED] Sent: 10 January 2008 13:36 To: Nicholls, Mark Cc: Luke Palmer; haskell-cafe@haskell.org Subject: Re[2]: [Haskell-cafe] confusion about 'instance' Hello Mark, Thursday, January 10, 2008, 4:25:20 PM, you wrote: instance Num a = A a Mean the same thing as instance A (forall a.Num a=a) programmers going from OOP world always forget that classes in Haskell doesn't the same as classes in C++. *implementation* of this instance require to pass dictionary of Num class along with type. now imagine the following code: My confusion is not between OO classes and Haskell classes, but exactly are the members of a Haskell type class...I'd naively believed them to be types (like it says on the packet!)...but now I'm not so sure. f :: A a = a - a f cannot use your instance because it doesn't receive Num dictionary of type `a`. it is unlike OOP situation where every object carries the generic VMT which includes methods for every class/interface that object supports as usual, i suggest you to study http://haskell.org/haskellwiki/OOP_vs_type_classes first and especially two papers mentioned in References there I have donelearning is not an atomic operationi.e. I can only believe what I understand...academic papers are especially beyond me at this point. I can translate OO into mathematical logic pretty easily, I was trying to do the same thing (informally of course) with Haskellbut things are not quite what they appearnot because of some OO hang up (which I probably have many)...but because of what type class actually means. So you may be right, I think I need to understand more about the sematics of Haskell...I was hoping to stay (initially) ignorant. I will try the postscript doc and see if it makes any sense. -- Best regards, Bulatmailto:[EMAIL PROTECTED] ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
RE: [Haskell-cafe] confusion about 'instance'....
Thanks for your response, I think you helped me on one of my previous abberations. Hmmmthis all slightly does my head inon one hand we have typesthen type classes (which appear to be a relation defined on types)then existential types...which now appear not to be treated quite in the same way as 'normal' typesand in this instance the syntax even seems to changedoes instance Num a = A a Mean the same thing as instance A (forall a.Num a=a) Uh... that second one is pretty much nonsensical to me. I could imagine it meaning the type (forall a.Num a = a) itself is an instance of A, but not specializations of it (like Int). But without an identity in the type system, the meaning of that would be convoluted. It's kind of off topic, but just for the interested, here are two similar, legal constructions: ok Existential: newtype Numeric = forall a. Num a = Numeric a My compiler doesn't like this A newtype constructor cannot have an existential context, Universal: newtype Numeric' = Numeric' (forall a. Num a = a) Not so sure I understand the difference here. Both of which are easily declared to be instances of Num. They're not what you want though, because Haskell doesn't support what you want :-(. Anyway, if you have a value of type Numeric, you know it contains some value of a Num type, but you don't know what type exactly (and you can never find out). If you have a value of type Numeric', then you can produce a value of any Num type you please (i.e. the value is built out of only operations in the Num class, nothing more specific). But that was a digression; ignore at your leisure (now that you've already read it :-). Makes little sense to meNumeric looks reasonable...I think... Numeric'...seems weirdand I'm not sure I understood the explanation. and secondly in what way can this construct lead to undecidable instances Okay, read: instance A a = B b (where a and be might be more complex expressions) not as b is an instance of B whenever a is an instance of A, but rather as b is an instance of B, and using it as such adds a constraint of A a. Let's look at a slightly more complex (and contrived) example: class Foo a where foo :: a - a instance (Foo [a]) = Foo a where foo x = head $ foo [x] Then when checking the type of the expression foo (0::Int), we'd have to check if Foo Int, Foo [Int], Foo [[Int]], Foo [[[Int]]], ad infinitum. Ooo blimeythat sort of makes sense. What are the instances, and what about them is undecidableseems pretty decidable to me? What is the ramifications of turning this option on? Theoretically, compile time fails to guarantee to ever finish. Practically, ghc will give you a very-difficult-to-reason-about message when constraint checking stack overflows. Luke ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
RE: [Haskell-cafe] confusion about 'instance'....
-Original Message- From: Jules Bean [mailto:[EMAIL PROTECTED] Sent: 10 January 2008 14:22 To: Nicholls, Mark Cc: Bulat Ziganshin; haskell-cafe@haskell.org Subject: Re: [Haskell-cafe] confusion about 'instance' Nicholls, Mark wrote: My confusion is not between OO classes and Haskell classes, but exactly are the members of a Haskell type class...I'd naively believed them to be types (like it says on the packet!)...but now I'm not so sure. Which packet? The packet labelled type classyou're from .co.ukyou should understand my English idioms. :-) Classes are not types. Yep. Classes are groups of types. Sets of types. Classifications of types. I had them down as an n-ary relation on typessomeone's said something somewhere that's made me question that...but I think I misinterpreted themso I may default back to n-ary relation. For any type, you can ask the quesiton is this type a member of this class, or not? yep Without wishing to split hairs too finely, I find it a useful intuition not to consider the class context part of the type somehow. So, when you see this: (Num a, Eq b) = a - b - a Rather than thinking of that whole thing as a type, it helps to think of the part on the right of the = as the 'actual type' and the part on the left of the = as some extra constraints on the type. Hmmm...I'm not sure that helpsit may just make me more confused. So you might say this has the type a - b - a, providing that a is a Num and b is an Eq. Jules ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
RE: Re[2]: [Haskell-cafe] confusion about 'instance'....
Someone said something about having 2 instances of the type in the typeclass.maybe I misinterpreted it. -Original Message- From: Luke Palmer [mailto:[EMAIL PROTECTED] Sent: 10 January 2008 14:12 To: Nicholls, Mark Cc: Bulat Ziganshin; haskell-cafe@haskell.org Subject: Re: Re[2]: [Haskell-cafe] confusion about 'instance' On Jan 10, 2008 2:04 PM, Nicholls, Mark [EMAIL PROTECTED] wrote: I can translate OO into mathematical logic pretty easily, I was trying to do the same thing (informally of course) with Haskellbut things are not quite what they appearnot because of some OO hang up (which I probably have many)...but because of what type class actually means. But you can think of a type class as a set of types! The problem is that if we allow certain kinds of instances (such as the Foo instance I gave earlier) then the set is allowed to be non-recursive (only recursively enumerable), so determining whether a particular type is a member of it would be undecidable. Luke ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
RE: Re[6]: [Haskell-cafe] Is there anyone out there who can translate C# generics into Haskell?
You may be right...but learning is not an atomic thingwherever I start I will get strange things happening. -Original Message- From: Bulat Ziganshin [mailto:[EMAIL PROTECTED] Sent: 03 January 2008 18:59 To: Nicholls, Mark Cc: Bulat Ziganshin; haskell-cafe@haskell.org Subject: Re[6]: [Haskell-cafe] Is there anyone out there who can translate C# generics into Haskell? Hello Mark, Thursday, January 3, 2008, 6:40:13 PM, you wrote: it would be hard to understand overlap without knowing both systems. you will believe that you understand it, but things will go strange ways :) I do not necessarily disagree But if I can identify the overlapthen I have leant the overlap...on the cheap. -Original Message- From: Bulat Ziganshin [mailto:[EMAIL PROTECTED] Sent: 03 January 2008 14:39 To: Nicholls, Mark Cc: Bulat Ziganshin; haskell-cafe@haskell.org Subject: Re[4]: [Haskell-cafe] Is there anyone out there who can translate C# generics into Haskell? Hello Mark, Thursday, January 3, 2008, 2:13:08 PM, you wrote: of course *some* overlap exists but in order to understand it you should know exact shape of both methods when i tried to develop complex library without understanding t.c. implementation, i constantly goes into the troubles - things that i (using my OOP experience) considered as possible, was really impossible in Haskell so i'm really wonder why you don't want to learn the topic thoroughly I loosely do understandbut very looselybut I'm not, as yet, convinced it is completely relevant. The implementation may differ, but that does not mean that there is no overlapI am not expecting one model to be a superset of the other, but I am expecting some sort of overlap between 'interface' implementation and type class instance declaration. -Original Message- From: Bulat Ziganshin [mailto:[EMAIL PROTECTED] Sent: 03 January 2008 10:54 To: Nicholls, Mark Cc: Bulat Ziganshin; haskell-cafe@haskell.org Subject: Re[2]: [Haskell-cafe] Is there anyone out there who can translate C# generics into Haskell? Hello Mark, Thursday, January 3, 2008, 1:22:26 PM, you wrote: because they have different models. i recommend you to start from learning this model, otherwise you will don't understand how Haskell really works and erroneously apply your OOP knowledge to Haskell data structures. shortly said, there are 3 ways to polymorphism: 1) C++ templates - type-specific code generated at compile time 2) OOP classes - every object carries VMT which allows to select type-specific operation 3) type classes - dictionary of type-specific operations is given as additional hidden argument to each function Haskell uses t.c. and its abilities are dictated by this implementation. there is no simple and direct mapping between features provided by OOP and t.c. Can you give me a summary of why it's meaningless.both would seem to describe/construct values/objectsthey may not be equivalent, but I would expect some considerable overlap. -Original Message- From: Bulat Ziganshin [mailto:[EMAIL PROTECTED] Sent: 02 January 2008 20:29 To: Nicholls, Mark Cc: haskell-cafe@haskell.org Subject: Re: [Haskell-cafe] Is there anyone out there who can translate C# generics into Haskell? Hello Mark, Wednesday, January 2, 2008, 7:40:31 PM, you wrote: I'm trying to translate some standard C# constucts into Haskell... some it's meaningless. read http://haskell.org/haskellwiki/OOP_vs_type_classes and especially papers mentioned in the References -- Best regards, Bulatmailto:[EMAIL PROTECTED] ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
RE: [Haskell-cafe] Is there anyone out there who can translate C# generics into Haskell?
I was thinking more along type classesand then I was going to throw
some spanners in the works
From: Ryan Ingram [mailto:[EMAIL PROTECTED]
Sent: 02 January 2008 17:41
To: Nicholls, Mark
Cc: haskell-cafe@haskell.org
Subject: Re: [Haskell-cafe] Is there anyone out there who can translate
C# generics into Haskell?
Of course it depends what's inside the braces, and what you want to do
with it, but I'd be inclined to do something like this:
1) data IX a = IX { constructor :: Int - a, ... }
2) data IX a b = IX { constructor :: Int - b, func :: a - b, ... }
3) data IX a b = IX { iy :: IY a, ... }
4) data IX a b = IX { iz :: IZ b, iy :: IY a, ... }
Can you specify more clearly what the goal of the conversion is? If you
want OO style behavior the thing that is most important is existential
quantification.
-- ryan
On 1/2/08, Nicholls, Mark [EMAIL PROTECTED] wrote:
I'm trying to translate some standard C# constucts into Haskell... some
of this seems easy
Specifically
1)
Interface IX
{
}
2)
Interface IXA
{
}
3)
Interface IXA
Where A : IY
{
}
4)
Interface IXA : IZ
Where A : IY
{
}
I can take a punt at the first 2but then it all falls apart
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RE: [Haskell-cafe] Is there anyone out there who can translate C# generics into Haskell?
[snip]
-- C#: interface IX1 { String foo1(int); }
class IX1 obj where
foo1 :: Int - obj - String
Yep...I think that's what I'd dothough I would have done...
foo1 :: obj - Int - String
Does that matter?
-- C#: interface IX2A { String foo2(A); }
class IX2 obj a where
foo2 :: a - obj - String
Ok same here
--C#: interface IX3A where A : IY { String foo3(A); }
class IY a where {- ... -}
class IY a = IX3 obj a where
foo3 :: a - obj - String
Yep I think again I would have guessed at that
--C#: interface IX4A : IZ where A : IY
class IZ a where {- ... -}
class (IY a, IZ obj) = IX4 obj a where
foo4 :: a - obj - String
H...this would have been one of my guessesbut let me have a
go...
This assumes your objects are immutable, otherwise you would have to
return (obj,String) instead of just String and then you most likely
want to
use the state monad and do notation to make functional programming
look
more like imperative programming.
This is finemy oop is largely immutable.
You really have to drop the OO way of thinking,
which I find the hardest :)
Haskell's type classes are more powerful in some sense than C#
interfaces;
for example, in C# you can't attach an interface to any class (take for
example the Point struct), it has to be your own class, while in
Haskell,
you can implement an instance of a type class for any datatype!
OK but I was going to go onto
Interface IXA
where A : IXA
{
}
And
Interface IXA,B
where A : B
{
}
Where I cannot see a way to do the above in Haskell at allas
interfaces effectively operator on type classes not typeswhich seems
inherently more powerful
But if these could be done in Haskell the see what could be made of
stuff likewhich is obviously problematic in C# it obviously doesn't
workbut potentially does make 'sense'.
Interface IXA : A
{
}
Hope this helps a bit. As I'm new myself to Haskell, so take this with
a
grain of salt.
It does...I will have a go with your sample answers.
Once you bite the bullet, I found Haskell a lot more fun than C#,
although
C# of course comes with a gigantic .NET framework and tools...
I'm looking at Haskell because of the formality of it's type
systembut I'm actually not convinced it is as powerful as an OO
onei.e. OO ones operatate principally (in Haskell speak) on type
classes not types
Peter
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Nicholls, Mark
Sent: Wednesday, January 02, 2008 5:41 PM
To: haskell-cafe@haskell.org
Subject: [Haskell-cafe] Is there anyone out there who can translate C#
generics into Haskell?
I'm trying to translate some standard C# constucts into Haskell... some
of this seems easy
Specifically
1)
Interface IX
{
}
2)
Interface IXA
{
}
3)
Interface IXA
Where A : IY
{
}
4)
Interface IXA : IZ
Where A : IY
{
}
I can take a punt at the first 2but then it all falls apart
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RE: [Haskell-cafe] Is there anyone out there who can translate C# generics into Haskell?
Can you give me a summary of why it's meaningless.both would seem to describe/construct values/objectsthey may not be equivalent, but I would expect some considerable overlap. -Original Message- From: Bulat Ziganshin [mailto:[EMAIL PROTECTED] Sent: 02 January 2008 20:29 To: Nicholls, Mark Cc: haskell-cafe@haskell.org Subject: Re: [Haskell-cafe] Is there anyone out there who can translate C# generics into Haskell? Hello Mark, Wednesday, January 2, 2008, 7:40:31 PM, you wrote: I'm trying to translate some standard C# constucts into Haskell... some it's meaningless. read http://haskell.org/haskellwiki/OOP_vs_type_classes and especially papers mentioned in the References -- Best regards, Bulatmailto:[EMAIL PROTECTED] ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
RE: Re[2]: [Haskell-cafe] Is there anyone out there who can translate C# generics into Haskell?
I loosely do understandbut very looselybut I'm not, as yet, convinced it is completely relevant. The implementation may differ, but that does not mean that there is no overlapI am not expecting one model to be a superset of the other, but I am expecting some sort of overlap between 'interface' implementation and type class instance declaration. -Original Message- From: Bulat Ziganshin [mailto:[EMAIL PROTECTED] Sent: 03 January 2008 10:54 To: Nicholls, Mark Cc: Bulat Ziganshin; haskell-cafe@haskell.org Subject: Re[2]: [Haskell-cafe] Is there anyone out there who can translate C# generics into Haskell? Hello Mark, Thursday, January 3, 2008, 1:22:26 PM, you wrote: because they have different models. i recommend you to start from learning this model, otherwise you will don't understand how Haskell really works and erroneously apply your OOP knowledge to Haskell data structures. shortly said, there are 3 ways to polymorphism: 1) C++ templates - type-specific code generated at compile time 2) OOP classes - every object carries VMT which allows to select type-specific operation 3) type classes - dictionary of type-specific operations is given as additional hidden argument to each function Haskell uses t.c. and its abilities are dictated by this implementation. there is no simple and direct mapping between features provided by OOP and t.c. Can you give me a summary of why it's meaningless.both would seem to describe/construct values/objectsthey may not be equivalent, but I would expect some considerable overlap. -Original Message- From: Bulat Ziganshin [mailto:[EMAIL PROTECTED] Sent: 02 January 2008 20:29 To: Nicholls, Mark Cc: haskell-cafe@haskell.org Subject: Re: [Haskell-cafe] Is there anyone out there who can translate C# generics into Haskell? Hello Mark, Wednesday, January 2, 2008, 7:40:31 PM, you wrote: I'm trying to translate some standard C# constucts into Haskell... some it's meaningless. read http://haskell.org/haskellwiki/OOP_vs_type_classes and especially papers mentioned in the References -- Best regards, Bulatmailto:[EMAIL PROTECTED] ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
RE: [Haskell-cafe] Is there anyone out there who can translate C# generics into Haskell?
Ahh ok I see what is meant by the parameter order
-Original Message-
From: Peter Verswyvelen [mailto:[EMAIL PROTECTED] On Behalf Of
Peter Verswyvelen
Sent: 03 January 2008 12:02
To: Nicholls, Mark
Cc: haskell-cafe@haskell.org
Subject: RE: [Haskell-cafe] Is there anyone out there who can translate
C# generics into Haskell?
Hi Mark,
foo1 :: Int - obj - String
Yep...I think that's what I'd dothough I would have done...
foo1 :: obj - Int - String
Does that matter?
Well, it's a good habit in Haskell to move the most important
parameter to
the end of the argument list. See e.g.
http://www.haskell.org/haskellwiki/Parameter_order.
OK but I was going to go onto
Interface IXA where A : IXA {}
and
Interface IXA,B where A : B {}
No, I would not know how to that in Haskell using type classes. It seems
Haskell does not allow cycles in type class definitions. But as I'm new,
this does not mean it's not possible. It's more important to know *what*
you
are trying to do, than to give a solution in a different language, since
OO
and FP are kind of orthogonal languages.
Where I cannot see a way to do the above in Haskell at allas
interfaces effectively operator on type classes not typeswhich
seems
inherently more powerful
Yeah, kind of makes sense. I liked interfaces in C# a lot, but when I
started doing everything with interfaces, I found the lack of support
for
mixins or default implementations problematic. This ended up in a
lot of
copy/paste or encapsulating the implementations into a static class with
plain functions, a mess.
But if these could be done in Haskell the see what could be made of
stuff likewhich is obviously problematic in C# it obviously
doesn't
workbut potentially does make 'sense'.
Interface IXA : A {}
Ah! That's one of the bigger restrictions in C# yes! C++ can do that;
ATL
uses it a lot, and I also like that approach. You can emulate mixins
with
that, and still stay in the single inheritance paradigm. In Haskell you
don't do that at all of course, since you avoid thinking about objects
and
inheritance in the first place.
OO is strange. They offer you the nice concept of inheritance, and then
the
guidelines tell you: don't use too many levels of inheritance...
Although
I've build huge projects using OO, it always felt a bit like unsafe
hacking.
I don't really have that feeling with Haskell, but that could also be
because I'm too new to the language ;-)
I'm looking at Haskell because of the formality of it's type
systembut I'm actually not convinced it is as powerful as an OO
onei.e. OO ones operatate principally (in Haskell speak) on type
classes not types
Maybe you are right, I don't know, my theoritical skills are not high
enough
to answer that. Haskell just feels better to me, although the lack of
a
friendly productive IDE and large standard framework remains a bit of a
burden.
Good luck,
Peter
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Nicholls, Mark
Sent: Wednesday, January 02, 2008 5:41 PM
To: haskell-cafe@haskell.org
Subject: [Haskell-cafe] Is there anyone out there who can translate C#
generics into Haskell?
I'm trying to translate some standard C# constucts into Haskell... some
of this seems easy
Specifically
1)
Interface IX
{
}
2)
Interface IXA
{
}
3)
Interface IXA
Where A : IY
{
}
4)
Interface IXA : IZ
Where A : IY
{
}
I can take a punt at the first 2but then it all falls apart
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RE: Re[4]: [Haskell-cafe] Is there anyone out there who can translate C# generics into Haskell?
I do not necessarily disagree But if I can identify the overlapthen I have leant the overlap...on the cheap. -Original Message- From: Bulat Ziganshin [mailto:[EMAIL PROTECTED] Sent: 03 January 2008 14:39 To: Nicholls, Mark Cc: Bulat Ziganshin; haskell-cafe@haskell.org Subject: Re[4]: [Haskell-cafe] Is there anyone out there who can translate C# generics into Haskell? Hello Mark, Thursday, January 3, 2008, 2:13:08 PM, you wrote: of course *some* overlap exists but in order to understand it you should know exact shape of both methods when i tried to develop complex library without understanding t.c. implementation, i constantly goes into the troubles - things that i (using my OOP experience) considered as possible, was really impossible in Haskell so i'm really wonder why you don't want to learn the topic thoroughly I loosely do understandbut very looselybut I'm not, as yet, convinced it is completely relevant. The implementation may differ, but that does not mean that there is no overlapI am not expecting one model to be a superset of the other, but I am expecting some sort of overlap between 'interface' implementation and type class instance declaration. -Original Message- From: Bulat Ziganshin [mailto:[EMAIL PROTECTED] Sent: 03 January 2008 10:54 To: Nicholls, Mark Cc: Bulat Ziganshin; haskell-cafe@haskell.org Subject: Re[2]: [Haskell-cafe] Is there anyone out there who can translate C# generics into Haskell? Hello Mark, Thursday, January 3, 2008, 1:22:26 PM, you wrote: because they have different models. i recommend you to start from learning this model, otherwise you will don't understand how Haskell really works and erroneously apply your OOP knowledge to Haskell data structures. shortly said, there are 3 ways to polymorphism: 1) C++ templates - type-specific code generated at compile time 2) OOP classes - every object carries VMT which allows to select type-specific operation 3) type classes - dictionary of type-specific operations is given as additional hidden argument to each function Haskell uses t.c. and its abilities are dictated by this implementation. there is no simple and direct mapping between features provided by OOP and t.c. Can you give me a summary of why it's meaningless.both would seem to describe/construct values/objectsthey may not be equivalent, but I would expect some considerable overlap. -Original Message- From: Bulat Ziganshin [mailto:[EMAIL PROTECTED] Sent: 02 January 2008 20:29 To: Nicholls, Mark Cc: haskell-cafe@haskell.org Subject: Re: [Haskell-cafe] Is there anyone out there who can translate C# generics into Haskell? Hello Mark, Wednesday, January 2, 2008, 7:40:31 PM, you wrote: I'm trying to translate some standard C# constucts into Haskell... some it's meaningless. read http://haskell.org/haskellwiki/OOP_vs_type_classes and especially papers mentioned in the References -- Best regards, Bulatmailto:[EMAIL PROTECTED] ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] Is there anyone out there who can translate C# generics into Haskell?
I'm trying to translate some standard C# constucts into Haskell... some
of this seems easy
Specifically
1)
Interface IX
{
}
2)
Interface IXA
{
}
3)
Interface IXA
Where A : IY
{
}
4)
Interface IXA : IZ
Where A : IY
{
}
I can take a punt at the first 2but then it all falls apart
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[Haskell-cafe] what does @ mean?.....
Hello, I wonder if someone could answer the following... The short question is what does @ mean in mulNat a b | a = b = mulNat' a b b | otherwise = mulNat' b a a where mulNat' x@(S a) y orig | x == one = y | otherwise = mulNat' a (addNat orig y) orig The long version, explaining what everything means is here's a definition of multiplication on natural numbers I'm reading on a blog data Nat = Z | S Nat deriving Show one :: Nat one = (S Z) mulNat :: Nat - Nat - Nat mulNat _ Z = Z mulNat Z _ = Z mulNat a b | a = b = mulNat' a b b | otherwise = mulNat' b a a where mulNat' x@(S a) y orig | x == one = y | otherwise = mulNat' a (addNat orig y) orig Haskell programmers seem to have a very irritating habit of trying to be overly concise...which makes learnign the language extremely hard...this example is actually relatively verbosebut anyway... Z looks like Zero...S is the successor function...Nat are the Natural numbers. mulNat _ Z = Z mulNat Z _ = Z translates to... x * 0 = 0fine... 0 * x = 0fine.. mulNat a b | a = b = mulNat' a b b | otherwise = mulNat' b a a where mulNat' x@(S a) y orig | x == one = y | otherwise = mulNat' a (addNat orig y) orig is a bit more problematic... lets take a as 3 and b as 5... so now we have mulNat' 3 5 5 but what does the x@(S a) mean? in mulNat' x@(S a) y orig From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Nicholls, Mark Sent: 21 December 2007 17:47 To: David Menendez Cc: Jules Bean; haskell-cafe@haskell.org Subject: RE: [Haskell-cafe] nice simple problem for someone struggling Let me resend the code...as it stands module Main where data SquareType numberType = Num numberType = SquareConstructor numberType class ShapeInterface shape where area :: Num numberType = shape-numberType data ShapeType = forall a. ShapeInterface a = ShapeType a instance (Num a) = ShapeInterface (SquareType a) where area (SquareConstructor side) = side * side and the errors are for the instance declaration... [1 of 1] Compiling Main ( Main.hs, C:\Documents and Settings\nichom\Haskell\Shapes2\out/Main.o ) Main.hs:71:36: Couldn't match expected type `numberType' against inferred type `a' `numberType' is a rigid type variable bound by the type signature for `area' at Main.hs:38:15 `a' is a rigid type variable bound by the instance declaration at Main.hs:70:14 In the expression: side * side In the definition of `area': area (SquareConstructor side) = side * side I'm becoming lost in errors I don't comprehend What bamboozles me is it seemed such a minor enhancement. From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of David Menendez Sent: 21 December 2007 17:05 To: Nicholls, Mark Cc: Jules Bean; haskell-cafe@haskell.org Subject: Re: [Haskell-cafe] nice simple problem for someone struggling On Dec 21, 2007 11:50 AM, Nicholls, Mark [EMAIL PROTECTED] wrote: Now I have module Main where data SquareType numberType = Num numberType = SquareConstructor numberType This is a valid declaration, but I don't think it does what you want it to. The constraint on numberType applies only to the data constructor. That is, given an unknown value of type SquareType a for some a, we do not have enough information to infer Num a. For your code, you want something like: instance (Num a) = ShapeInterface (SquareType a) where area (SquareConstructor side) = side * side -- Dave Menendez [EMAIL PROTECTED] http://www.eyrie.org/~zednenem/ ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
RE: [Haskell-cafe] what does @ mean?.....
So in the example given... mulNat a b | a = b = mulNat' a b b | otherwise = mulNat' b a a where mulNat' x@(S a) y orig | x == one = y | otherwise = mulNat' a (addNat orig y) orig Is equivalent to mulNat a b | a = b = mulNat' a b b | otherwise = mulNat' b a a where mulNat' (S a) y orig | (S a) == one = y | otherwise = mulNat' a (addNat orig y) orig ? -Original Message- From: Alfonso Acosta [mailto:[EMAIL PROTECTED] Sent: 28 December 2007 11:20 To: Nicholls, Mark Cc: haskell-cafe@haskell.org Subject: Re: [Haskell-cafe] what does @ mean?. @ works as an aliasing primitive for the arguments of a function f x@(Just y) = ... using x in the body of f is equivalent to use Just y. Perhaps in this case is not really useful, but in some other cases it saves the effort and space of retyping really long expressions. And what is even more important, in case an error is made when choosing the pattern, you only have to correct it in one place. On Dec 28, 2007 12:05 PM, Nicholls, Mark [EMAIL PROTECTED] wrote: Hello, I wonder if someone could answer the following... The short question is what does @ mean in mulNat a b | a = b = mulNat' a b b | otherwise = mulNat' b a a where mulNat' x@(S a) y orig | x == one = y | otherwise = mulNat' a (addNat orig y) orig The long version, explaining what everything means is here's a definition of multiplication on natural numbers I'm reading on a blog data Nat = Z | S Nat deriving Show one :: Nat one = (S Z) mulNat :: Nat - Nat - Nat mulNat _ Z = Z mulNat Z _ = Z mulNat a b | a = b = mulNat' a b b | otherwise = mulNat' b a a where mulNat' x@(S a) y orig | x == one = y | otherwise = mulNat' a (addNat orig y) orig Haskell programmers seem to have a very irritating habit of trying to be overly concise...which makes learnign the language extremely hard...this example is actually relatively verbosebut anyway... Z looks like Zero...S is the successor function...Nat are the Natural numbers. mulNat _ Z = Z mulNat Z _ = Z translates to... x * 0 = 0fine... 0 * x = 0fine.. mulNat a b | a = b = mulNat' a b b | otherwise = mulNat' b a a where mulNat' x@(S a) y orig | x == one = y | otherwise = mulNat' a (addNat orig y) orig is a bit more problematic... lets take a as 3 and b as 5... so now we have mulNat' 3 5 5 but what does the x@(S a) mean? in mulNat' x@(S a) y orig From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Nicholls, Mark Sent: 21 December 2007 17:47 To: David Menendez Cc: Jules Bean; haskell-cafe@haskell.org Subject: RE: [Haskell-cafe] nice simple problem for someone struggling Let me resend the code...as it stands module Main where data SquareType numberType = Num numberType = SquareConstructor numberType class ShapeInterface shape where area :: Num numberType = shape-numberType data ShapeType = forall a. ShapeInterface a = ShapeType a instance (Num a) = ShapeInterface (SquareType a) where area (SquareConstructor side) = side * side and the errors are for the instance declaration... [1 of 1] Compiling Main ( Main.hs, C:\Documents and Settings\nichom\Haskell\Shapes2\out/Main.o ) Main.hs:71:36: Couldn't match expected type `numberType' against inferred type `a' `numberType' is a rigid type variable bound by the type signature for `area' at Main.hs:38:15 `a' is a rigid type variable bound by the instance declaration at Main.hs:70:14 In the expression: side * side In the definition of `area': area (SquareConstructor side) = side * side I'm becoming lost in errors I don't comprehend What bamboozles me is it seemed such a minor enhancement. From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of David Menendez Sent: 21 December 2007 17:05 To: Nicholls, Mark Cc: Jules Bean; haskell-cafe@haskell.org Subject: Re: [Haskell-cafe] nice simple problem for someone struggling On Dec 21, 2007 11:50 AM, Nicholls, Mark [EMAIL PROTECTED] wrote: Now I have module Main where data SquareType numberType = Num numberType = SquareConstructor numberType This is a valid declaration, but I don't think it does what you want it to. The constraint on numberType applies only to the data constructor. That is, given an unknown value of type SquareType a for some a, we do not have enough information to infer Num
RE: [Haskell-cafe] what does @ mean?.....
Lovelythank you very muchanother small step forward. -Original Message- From: Chaddaï Fouché [mailto:[EMAIL PROTECTED] Sent: 28 December 2007 11:29 To: Nicholls, Mark Cc: Alfonso Acosta; haskell-cafe@haskell.org Subject: Re: [Haskell-cafe] what does @ mean?. 2007/12/28, Nicholls, Mark [EMAIL PROTECTED]: So in the example given... mulNat a b | a = b = mulNat' a b b | otherwise = mulNat' b a a where mulNat' x@(S a) y orig | x == one = y | otherwise = mulNat' a (addNat orig y) orig Is equivalent to mulNat a b | a = b = mulNat' a b b | otherwise = mulNat' b a a where mulNat' (S a) y orig | (S a) == one = y | otherwise = mulNat' a (addNat orig y) orig ? Yes, but in the second version, it has to reconstruct (S a) before comparing it to one where in the first it could do the comparison directly. In this cas there may be some optimisation involved that negate this difference but in many case it can do a real performance difference. The as-pattern (@ means as) is both practical and performant in most cases. -- Jedaï ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] translating some C# abstractions into Haskell....
Lets say I've got
Interface IFooX,Y
Where X : IBar
Where Y : IBar
{
}
Would seem to translate roughly to
class (IBar x, IBar y) = IFoo foo x y
? (or does it?)
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Nicholls, Mark
Sent: 28 December 2007 11:30
To: Chaddaï Fouché
Cc: haskell-cafe@haskell.org
Subject: RE: [Haskell-cafe] what does @ mean?.
Lovelythank you very muchanother small step forward.
-Original Message-
From: Chaddaï Fouché [mailto:[EMAIL PROTECTED]
Sent: 28 December 2007 11:29
To: Nicholls, Mark
Cc: Alfonso Acosta; haskell-cafe@haskell.org
Subject: Re: [Haskell-cafe] what does @ mean?.
2007/12/28, Nicholls, Mark [EMAIL PROTECTED]:
So in the example given...
mulNat a b
| a = b = mulNat' a b b
| otherwise = mulNat' b a a
where
mulNat' x@(S a) y orig
| x == one = y
| otherwise = mulNat' a (addNat orig y) orig
Is equivalent to
mulNat a b
| a = b = mulNat' a b b
| otherwise = mulNat' b a a
where
mulNat' (S a) y orig
| (S a) == one = y
| otherwise = mulNat' a (addNat orig y) orig
?
Yes, but in the second version, it has to reconstruct (S a) before
comparing it to one where in the first it could do the comparison
directly. In this cas there may be some optimisation involved that
negate this difference but in many case it can do a real performance
difference.
The as-pattern (@ means as) is both practical and performant in most cases.
--
Jedaï
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[Haskell-cafe] nice simple problem for someone struggling....
I'm just trying to pick up the basicsand I've managed to write this code...which remarkably works.. module Main where data SquareType = SquareConstructor Int class ShapeInterface shape where area :: shape-Int data ShapeType = forall a. ShapeInterface a = ShapeType a instance ShapeInterface SquareType where area (SquareConstructor sideLength) = sideLength * sideLength main = do putStrLn (show (area (SquareConstructor 4))) name - getLine putStrLn But my next iteration was to try to parametise SquareType So something like data SquareType a = Num a = SquareConstructor a but of course doing this breaks everything...sepecifically the instance declaration `SquareType' is not applied to enough type arguments Expected kind `*', but `SquareType' has kind `* - *' In the instance declaration for `ShapeInterface SquareType' And I can't seem to get it to work. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
RE: [Haskell-cafe] nice simple problem for someone struggling....
ReallyI'm sure I tried that...(as it seemed obvious) ... and it failedbut I'll have another go -Original Message- From: Jules Bean [mailto:[EMAIL PROTECTED] Sent: 21 December 2007 15:33 To: Nicholls, Mark Cc: haskell-cafe@haskell.org Subject: Re: [Haskell-cafe] nice simple problem for someone struggling Nicholls, Mark wrote: *instance* ShapeInterface SquareType *where* area (SquareConstructor sideLength) = sideLength * sideLength *data* SquareType a = Num a = SquareConstructor a Now you have changed your type from SquareType to SquareType a, you need to change the instance to: instance ShapeInterface (SquareType a) where... Jules ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
RE: [Haskell-cafe] nice simple problem for someone struggling....
Now I have module Main where data SquareType numberType = Num numberType = SquareConstructor numberType data RectangleType = RectangleConstructor Int Int class ShapeInterface shape where area :: shape-Int data ShapeType = forall a. ShapeInterface a = ShapeType a instance ShapeInterface (SquareType numberType) where area (SquareConstructor sideLength) = sideLength * sideLength main = do putStrLn (show (area (SquareConstructor 4))) name - getLine putStrLn but get the errors In the expression: sideLength * sideLength In the definition of `area': area (SquareConstructor sideLength) = sideLength * sideLength In the definition for method `area' And Couldn't match expected type `Int' against inferred type `numberType' `numberType' is a rigid type variable bound by But to be fairI almost understand the errorswhich is not bad for me.surely class ShapeInterface shape where area :: shape-Int now looks dubiousI want it to be something like class ShapeInterface shape where area :: Num numberType = shape-Int ? but my instance declaration still complains with the errors above and I now get an error in the class declaration `numberType1' is a rigid type variable bound by It's slightly doing my head inand reminds me of trying to learn C++ oncenot a pleasant experiencethough I did eventually succeedto a degree. -Original Message- From: Jules Bean [mailto:[EMAIL PROTECTED] Sent: 21 December 2007 15:33 To: Nicholls, Mark Cc: haskell-cafe@haskell.org Subject: Re: [Haskell-cafe] nice simple problem for someone struggling Nicholls, Mark wrote: *instance* ShapeInterface SquareType *where* area (SquareConstructor sideLength) = sideLength * sideLength *data* SquareType a = Num a = SquareConstructor a Now you have changed your type from SquareType to SquareType a, you need to change the instance to: instance ShapeInterface (SquareType a) where... Jules ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
RE: [Haskell-cafe] nice simple problem for someone struggling....
Oh You are correct... I thought from Num numberType = SquareConstructor numberType We could deduce that (in English rather than get Haskell and FOL confusion) all values of SquareConstructor athe type of a would have be be in class Num?.. is this not correct?if notwhy not? From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of David Menendez Sent: 21 December 2007 17:05 To: Nicholls, Mark Cc: Jules Bean; haskell-cafe@haskell.org Subject: Re: [Haskell-cafe] nice simple problem for someone struggling On Dec 21, 2007 11:50 AM, Nicholls, Mark [EMAIL PROTECTED] wrote: Now I have module Main where data SquareType numberType = Num numberType = SquareConstructor numberType This is a valid declaration, but I don't think it does what you want it to. The constraint on numberType applies only to the data constructor. That is, given an unknown value of type SquareType a for some a, we do not have enough information to infer Num a. For your code, you want something like: instance (Num a) = ShapeInterface (SquareType a) where area (SquareConstructor side) = side * side -- Dave Menendez [EMAIL PROTECTED] http://www.eyrie.org/~zednenem/ ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
RE: [Haskell-cafe] nice simple problem for someone struggling....
Yes sorrybut this still fails with `numberType1' is a rigid type variable bound by From: Brent Yorgey [mailto:[EMAIL PROTECTED] Sent: 21 December 2007 17:29 To: Nicholls, Mark Cc: Jules Bean; haskell-cafe@haskell.org Subject: Re: [Haskell-cafe] nice simple problem for someone struggling class ShapeInterface shape where area :: shape-Int now looks dubiousI want it to be something like class ShapeInterface shape where area :: Num numberType = shape-Int ? Rather, I think you probably want class ShapeInterface shape where area :: Num numberType = shape - numberType -Brent ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
RE: [Haskell-cafe] nice simple problem for someone struggling....
Let me resend the code...as it stands module Main where data SquareType numberType = Num numberType = SquareConstructor numberType class ShapeInterface shape where area :: Num numberType = shape-numberType data ShapeType = forall a. ShapeInterface a = ShapeType a instance (Num a) = ShapeInterface (SquareType a) where area (SquareConstructor side) = side * side and the errors are for the instance declaration... [1 of 1] Compiling Main ( Main.hs, C:\Documents and Settings\nichom\Haskell\Shapes2\out/Main.o ) Main.hs:71:36: Couldn't match expected type `numberType' against inferred type `a' `numberType' is a rigid type variable bound by the type signature for `area' at Main.hs:38:15 `a' is a rigid type variable bound by the instance declaration at Main.hs:70:14 In the expression: side * side In the definition of `area': area (SquareConstructor side) = side * side I'm becoming lost in errors I don't comprehend What bamboozles me is it seemed such a minor enhancement. From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of David Menendez Sent: 21 December 2007 17:05 To: Nicholls, Mark Cc: Jules Bean; haskell-cafe@haskell.org Subject: Re: [Haskell-cafe] nice simple problem for someone struggling On Dec 21, 2007 11:50 AM, Nicholls, Mark [EMAIL PROTECTED] wrote: Now I have module Main where data SquareType numberType = Num numberType = SquareConstructor numberType This is a valid declaration, but I don't think it does what you want it to. The constraint on numberType applies only to the data constructor. That is, given an unknown value of type SquareType a for some a, we do not have enough information to infer Num a. For your code, you want something like: instance (Num a) = ShapeInterface (SquareType a) where area (SquareConstructor side) = side * side -- Dave Menendez [EMAIL PROTECTED] http://www.eyrie.org/~zednenem/ ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] OOP'er with (hopefully) trivial questions.....
I'm trying to teach myself HaskellI've spent a few hours going
through a few tutorialsand I sort of get the basics...
My interest in Haskell is specifically around the strength of the type
system.
After many years of OOP though my brain is wired up to construct
software in that 'pattern'a problem for me at the moment is I cannot
see how to construct programs in an OO style in HaskellI know this
is probably not the way to approach it...but I feel I need to master the
syntax before the paradigm.
I'm not fussed about mutability, and I'm not fussed about implementation
inheritance (as apposed to subtyping) my concerns are encapsulation
of 'state' and abstraction, and extension/expansiona simple example
would be.
interface IShape
{
Int GetArea();
}
class Square : IShape
{
Readonly int length;
Public Square(int length) { this.length = length; }
Int GetArea() { return length * length;}
}
Class Rectangle : IShape
{
Readonly int lengthA;
Readonly int lengthB;
Public Rectangle (int lengthA,int lengthB) { this.lengthA = lengthA;
this.lengthB = lengthB; }
Int GetArea() { return lengthA * lengthB;}
}
Class Circle : IShape
{
Readonly int radius;
Public Circle(int radius) { this.radius = radius; }
Int GetArea() { return pi * radius * radius;}
}
Client code.
Void Foo(IShape shape)
{
// look!I know nothing except its of type IShape.
Int area = shape.GetArea();
}
I can obviously at a later date add a new class Triangle, and not have
to touch any of the above code
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RE: [Haskell-cafe] OOP'er with (hopefully) trivial questions.....
OK I'll have to digest this and mess about a bitbut can I make an observation at this point If I define Shape like data Shape = Circle Int | Rectangle Int Int | Square Int Isn't this now closed...i.e. the statement is effectively defining that shape is this and only ever thisi.e. can I in another module add new types of Shape? (sorry about all the quotation marks, but it's a minefield of potential confusions over types, classes etc). My other observation is...are the things on the right hand side of the the ='s sign not types? The lower version makes more sense to me...I'll have to give it a go A P.S. would be...I tend to write code rather than mess about in the GHCi shell.is there a way in code to output the type of a value..i.e. the :t operation? -Original Message- From: Thomas Davie [mailto:[EMAIL PROTECTED] Sent: 17 December 2007 11:04 To: Nicholls, Mark Cc: haskell-cafe@haskell.org Subject: Re: [Haskell-cafe] OOP'er with (hopefully) trivial questions. On 17 Dec 2007, at 10:46, Nicholls, Mark wrote: I can obviously at a later date add a new class Triangle, and not have to touch any of the above code Yes, and you can indeed do a similar thing in Haskell. The natural thing to do here would be to define a type Shape... data Shape = Circle Int | Rectangle Int Int | Square Int area :: Shape - Int -- Note, this is an interesting type if you want the area of circles area (Circle r) = pi * r^2 area (Rectangle h w) = h * w area (Square l) = area (Rectangle l l) If however, you *really* want to keep your shapes as being seperate types, then you'll want to invoke the class system (note, not the same as OO classes). class Shape a where area :: a - Int newtype Circle = C Int instance Shape Circle where area (C r) = pi * r^2 newtype Rectangle = R Int Int instance Shape Rectangle where area (R h w) = h * w newtype Square = Sq Int instance Shape Square where area (Sq l) = l * l -- Now we can do something with our shapes doubleArea :: Shape a = a - Int doubleArea s = (area s) * 2 Hope that helps Bob ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
RE: [Haskell-cafe] OOP'er with (hopefully) trivial questions.....
Ooo The constructor of a newtype must have exactly one field but `R' has two In the newtype declaration for `Rectangle' It doesn't like newtype Rectangle = R Int Int -Original Message- From: Thomas Davie [mailto:[EMAIL PROTECTED] Sent: 17 December 2007 11:04 To: Nicholls, Mark Cc: haskell-cafe@haskell.org Subject: Re: [Haskell-cafe] OOP'er with (hopefully) trivial questions. On 17 Dec 2007, at 10:46, Nicholls, Mark wrote: I can obviously at a later date add a new class Triangle, and not have to touch any of the above code Yes, and you can indeed do a similar thing in Haskell. The natural thing to do here would be to define a type Shape... data Shape = Circle Int | Rectangle Int Int | Square Int area :: Shape - Int -- Note, this is an interesting type if you want the area of circles area (Circle r) = pi * r^2 area (Rectangle h w) = h * w area (Square l) = area (Rectangle l l) If however, you *really* want to keep your shapes as being seperate types, then you'll want to invoke the class system (note, not the same as OO classes). class Shape a where area :: a - Int newtype Circle = C Int instance Shape Circle where area (C r) = pi * r^2 newtype Rectangle = R Int Int instance Shape Rectangle where area (R h w) = h * w newtype Square = Sq Int instance Shape Square where area (Sq l) = l * l -- Now we can do something with our shapes doubleArea :: Shape a = a - Int doubleArea s = (area s) * 2 Hope that helps Bob ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
RE: [Haskell-cafe] OOP'er with (hopefully) trivial questions.....
Ok... Thanks I need to revisit data and newtype to work out what the difference is I think. -Original Message- From: Jed Brown [mailto:[EMAIL PROTECTED] On Behalf Of Jed Brown Sent: 17 December 2007 12:04 To: Nicholls, Mark Cc: haskell-cafe@haskell.org Subject: Re: [Haskell-cafe] OOP'er with (hopefully) trivial questions. On 17 Dec 2007, [EMAIL PROTECTED] wrote: Ooo The constructor of a newtype must have exactly one field but `R' has two In the newtype declaration for `Rectangle' It doesn't like newtype Rectangle = R Int Int You want data Rectangle = R Int Int A newtype declaration will be completely erased at compile time. That is, when you have a declaration like newtype Circle = C Int the compiled code will not be able to distinguish between a Circle and an Int. You do, however, get all the benefits of a separate entity in the type system. When your type only has one constructor, newtype is preferred over data, but they are semantically equivalent. There are extensions which provide impressive newtype-deriving-foo (getting the compiler to write fairly non-trivial instance declarations for you). Jed ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
RE: [Haskell-cafe] OOP'er with (hopefully) trivial questions.....
No that's fineits all as clear as mud!..but that's not your fault. To recap... type introduces a synonym for another type, no new type is createdit's for readabilities sake. Newtype introduces an isomorphic copy of an existing type...but doesn't copy it's type class membership...the types are disjoint/distinct but isomorphic (thus only 1 constructor param). data introduces a new type, and defines a composition of existing types to create a new one based on - and (. class introduces a constraint that any types declaring themselves to be a member of this class...that functions must exist to satisfy the constraint. I'm sure that's wrong, but it's a good as I've got at the moment. And to a degree it's all upside downwhat Haskell thinks are types...I think are singnatures and what Haskell thinks is a type class I think of as a type.it's not going to be easy. -Original Message- From: Thomas Davie [mailto:[EMAIL PROTECTED] Sent: 17 December 2007 12:35 To: Nicholls, Mark Cc: Haskell Cafe Subject: Re: [Haskell-cafe] OOP'er with (hopefully) trivial questions. On 17 Dec 2007, at 12:22, Nicholls, Mark wrote: Ok... Thanks I need to revisit data and newtype to work out what the difference is I think. Beware in doing so -- type, and newtype are not the same either. type creates a type synonim. That is, if I were to declare type Jam = Int then Jam and Int from that point on become completely interchangable, the only thing this does is make things readable. For example, a parser might be described as a function that takes a list of tokens, and outputs a parse tree, and a list of unparsed tokens: type Parser = [Token] - (ParseTree, [Token]) if I write some parser combinators, I can now give them clear types like | :: Parser - Parser - Parser I could however still write this, and it would have *exactly* the same meaning. | :: ([Token] - (ParseTree, [Token])) - ([Token] - (ParseTree, [Token])) - [Token] - (ParseTree, [Token]) newtype on the other hand introduces a new type to the type system. Because of this, the type system has to be able to tell when you're using your new type, so a tag gets attached. newtype Ham = Ham Int This creates a type that contains only an integer, but is different from Int (and Jam) in the type system's eyes. Thus, I cannot for example write (Ham 5) + (Ham 6) Because Ham is not Int and thus (+) does not work (or actually, more specifically, Ham is not a member of the class Num, the numeric types, and therefore (+) doesn't work). This can of course be fixed thus: newtype Ham = Ham Int deriving Num Hope that helps Tom Davie p.s. Sorry for the slip with the newtype Rectangle. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
RE: [Haskell-cafe] Re: OOP'er with (hopefully) trivial questions.....
Not really with this... The open case (as in OO) seems to be more like the Haskell class construct, i.e. if new types declare themselves to be members of a class then they must satisfy certain constaintsI can then specify equals with the class and leave the onus on the implementor to implement itthe data construct seems more analogous to a OO class definition...which is closed in the same way. The approach is deliberate...but I accept may be harder than it needs to be...I'm interested in Haskell because of the alleged power/formality of it's type system against the relatively weakness of OO ones...the irony at the moment is that they do not really seem to correspond directlyand OO type system seems to (loosely) correlate to Haskell type class system, and an OO class system (loosely) to Haskels type system, though in OOP's they are unpleasantly tangled. -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of apfelmus Sent: 17 December 2007 12:34 To: haskell-cafe@haskell.org Subject: [Haskell-cafe] Re: OOP'er with (hopefully) trivial questions. Nicholls, Mark wrote: data Shape = Circle Int | Rectangle Int Int | Square Int Isn't this now closed...i.e. the statement is effectively defining that shape is this and only ever thisi.e. can I in another module add new types of Shape? Yes, but in most cases, this is actually a good thing. For instance, you can now define equality of two shapes: equal :: Shape - Shape - Bool equal (Circle x)(Circle y)= x == y equal (Rectangle x1 x2) (Rectangle y1 y2) = x1 == x2 y1 == y2 equal (Square x)(Square y)= x == y In general, the open approach is limited to functions of the form Shape - ... - Shape / Int / something else with no Shape occurring in the other ... arguments. I'm trying to teach myself HaskellI've spent a few hours going through a few tutorialsand I sort of get the basics... [...] After many years of OOP though my brain is wired up to construct software in that 'pattern'a problem for me at the moment is I cannot see how to construct programs in an OO style in HaskellI know this is probably not the way to approach it...but I feel I need to master the syntax before the paradigm. This approach is probably harder than it could be, you'll have a much easier time learning it from a paper-textbook like http://www.cs.nott.ac.uk/~gmh/book.html http://web.comlab.ox.ac.uk/oucl/publications/books/functional/ http://haskell.org/soe/ After all, it's like learning programming anew. Regards, apfelmus ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
RE: [Haskell-cafe] OOP'er with (hopefully) trivial questions.....
Ahhh I'll give it a read. thanks -Original Message- From: Henning Thielemann [mailto:[EMAIL PROTECTED] Sent: 17 December 2007 13:05 To: Nicholls, Mark Cc: haskell-cafe@haskell.org Subject: Re: [Haskell-cafe] OOP'er with (hopefully) trivial questions. On Mon, 17 Dec 2007, Nicholls, Mark wrote: After many years of OOP though my brain is wired up to construct software in that 'pattern'a problem for me at the moment is I cannot see how to construct programs in an OO style in HaskellI know this is probably not the way to approach it...but I feel I need to master the syntax before the paradigm. This issue is rather a FAQ. Let's look what our Wiki provides: http://www.haskell.org/haskellwiki/OOP_vs_type_classes ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
RE: [Haskell-cafe] OOP'er with (hopefully) trivial questions.....
No neither do II think we can drop that bitI think I got
confused about it for a second.not unsurprisingly.
From: Brent Yorgey [mailto:[EMAIL PROTECTED]
Sent: 17 December 2007 15:38
To: Nicholls, Mark
Cc: Thomas Davie; Haskell Cafe
Subject: Re: [Haskell-cafe] OOP'er with (hopefully) trivial
questions.
On Dec 17, 2007 8:04 AM, Nicholls, Mark [EMAIL PROTECTED] wrote:
No that's fineits all as clear as mud!..but that's not your
fault.
To recap...
type introduces a synonym for another type, no new type is
createdit's for readabilities sake.
Newtype introduces an isomorphic copy of an existing type...but
doesn't copy it's type class membership...the types are
disjoint/distinct but isomorphic (thus only 1 constructor param).
data introduces a new type, and defines a composition of existing
types to create a new one based on - and (.
class introduces a constraint that any types declaring themselves to
be a member of this class...that functions must exist to satisfy the
constraint.
I'm sure that's wrong, but it's a good as I've got at the moment.
And to a degree it's all upside downwhat Haskell thinks are
types...I think are singnatures and what Haskell thinks is a type
class I think of as a type.it's not going to be easy.
I think you've got it pretty well! The one quibble I would have with
your recap is that I'm not sure what you mean by saying that data
creates a new type 'based on - and ('. Other than that it seems
pretty spot-on. =)
-Brent
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RE: Re[2]: [Haskell-cafe] OOP'er with (hopefully) trivial questions.....
My Haskell is not up to understanding themI'm still writing hello world programswhat I read, gave me a good initial hint as to whats going on...I just need to get my Haskell going before I can jump in the deep end. -Original Message- From: Bulat Ziganshin [mailto:[EMAIL PROTECTED] Sent: 17 December 2007 16:37 To: Nicholls, Mark Cc: Henning Thielemann; haskell-cafe@haskell.org Subject: Re[2]: [Haskell-cafe] OOP'er with (hopefully) trivial questions. Hello Mark, Monday, December 17, 2007, 4:47:50 PM, you wrote: I'll give it a read. http://www.haskell.org/haskellwiki/OOP_vs_type_classes i recommend you to read two papers mentioned in References section there. at least i'm one of this page authors and i don't think that i had very good understanding of type classes on the moment when this page was written. OTOH it also contains section written by John Meacham - it should be better because John is author of one Haskell compiler -- Best regards, Bulatmailto:[EMAIL PROTECTED] ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
