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 fair....I almost understand the errors....which is not bad for me.....surely "class ShapeInterface shape where area :: shape->Int" now looks dubious....I 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 in....and reminds me of trying to learn C++ once....not a pleasant experience....though I did eventually succeed....to a degree. -----Original Message----- From: Jules Bean [mailto:jules@jellybean.co.uk] 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

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 a"....the type of a would have be be in class Num?.. is this not correct?....if not....why not? ________________________________ From: d4ve.menendez@gmail.com [mailto:d4ve.menendez@gmail.com] 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 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 <http://www.eyrie.org/~zednenem/ >

David Menendez wrote:

...

That's a reasonable thing to assume. It just happens that Haskell doesn't work that way. There's an asymmetry between constructing and pattern-matching, and it's one that many people have complained about.

With GADTs turned on (-XGADTS in 6.8, -fglasgow-exts in 6.6) pattern matchings will give rise to class contexts as you would naively expect. Contexts on constructors aren't actualy haskell98, it is a bug that GHC 6.6 accepts them without any extensions being activated. Or that's my understanding, see http://hackage.haskell.org/trac/ghc/ticket/1901 Jules

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: d4ve.menendez@gmail.com [mailto:d4ve.menendez@gmail.com] 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 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 <http://www.eyrie.org/~zednenem/ >

On Dec 21, 2007 12:47 PM, Nicholls, Mark wrote:

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

Part of the problem is that GHC's error messages have to account for a lot of complex typing extensions you aren't using, so they aren't clear. I'll try to explain what's going wrong. If you look at the function, area (SquareConstructor side) = side * side in isolation (that is, not as part of the class instance), it has the type "forall a. Num a => SquareConstructor a -> a". The function in the class declaration has type "forall a b. (ShapeInterface a, Num b) => a -> b". The problem is that a and b are independent variables, but the instance declaration for SquareType a requires them to be related. I'm not sure which way you were trying to parameterize things, but here are some possibilities: (1) If every shape type is parameteric, then you can make ShapeInterface a class of type constructors. class ShapeInterface shape where area :: Num a => shape a -> a instance ShapeInterface SquareType where area (SquareConstructor side) = side * side (2) If only some shape types are parametric, you can use a multi-parameter type class to express a relationship between the shape type and the area type: class ShapeInterface shape area where area :: shape -> area instance (Num a) => ShapeInterface (SquareType a) a where area (SquareConstructor side) = side * side (3) If you only need to be parameteric over some subset of numeric types, you can use conversion functions: class ShapeIterface shape where area :: shape -> Rational class (Real a) => ShapeInterface (SquareType a) where area (SquareConstructor side) = toRational (side * side) (Equivalently, you can replace Rational and Real with Integer and Integral.) It may be that none of these are what you want. There are other, more complex possibilities, but I expect they're overkill. -- Dave Menendez http://www.eyrie.org/~zednenem/

@ 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 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 verbose....but 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 = 0....fine...

0 * x = 0....fine..

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: haskell-cafe-bounces@haskell.org [mailto:haskell-cafe-bounces@haskell.org] 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: d4ve.menendez@gmail.com [mailto:d4ve.menendez@gmail.com] 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 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 <http://www.eyrie.org/~zednenem/ > _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe

2007/12/28, Nicholls, Mark :

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

Lets say I've got Interface IFoo 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: haskell-cafe-bounces@haskell.org [mailto:haskell-cafe-bounces@haskell.org] 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?..... Lovely....thank you very much....another small step forward. -----Original Message----- From: Chaddaï Fouché [mailto:chaddai.fouche@gmail.com] 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 :

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

2007/12/28, Alfonso Acosta :

@ 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.

And in plenty of case it will greatly boost your speed because you won't be reconstructing the objects against which you matched every time you want to return them unchanged. -- Jedaï

ChrisK wrote:

zeroNothing Nothing = Nothing zeroNothing (Just n) = if n == 0 then Nothing else (Just n)

versus

zeroNothing Nothing = Nothing zeroNothing x@(Just n) = if n == 0 then Nothing else x

versus zeroNothing Nothing = Nothing zeroNothing x = let (Just n) = x in if n == 0 then Nothing else x so, @ is kind of like a let, just with its arguments flipped.

Ben Franksen wrote:

Achim Schneider wrote:

...

ChrisK wrote:

...

zeroNothing Nothing = Nothing zeroNothing (Just n) = if n == 0 then Nothing else (Just n)

versus

zeroNothing Nothing = Nothing zeroNothing x@(Just n) = if n == 0 then Nothing else x

versus

zeroNothing Nothing = Nothing zeroNothing x = let (Just n) = x in if n == 0 then Nothing else x

so, @ is kind of like a let, just with its arguments flipped.

However, if x@(Just n) fails to match, the next clause is chosen, whereas the variable pattern x matches always. Thus, the last version works only because the other possible case (Nothing) has already been handled. IOW, in the second version of zeroNothing you may swap the order of patterns, but not in the third one.

Actually, I considered working it out to nothingIf :: (a -> Bool) -> Maybe a -> Maybe a nothingIf f m = m >>= (\j -> if f j then Nothing else Just j) nothingIf' :: (a -> Bool) -> Maybe a -> Maybe a nothingIf' f m = m >>= (\j -> if f j then Nothing else m) zeroNothing = nothingIf (== 0) zeroNothing' = nothingIf' (== 0) , but was too lazy. Hopefully only because it completely messes up my point. OTOH, zeroIf :: MonadPlus m => (a -> Bool) -> m a -> m a zeroIf f m = m >>= (\nz -> if f nz then mzero else m) zeroZero :: (MonadPlus m, Num a) => m a -> m a zeroZero = zeroIf (==0) makes it interesting again as you can't construct a Just value with it.

"class ShapeInterface shape where area :: shape->Int"

now looks dubious....I 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