(This email is a literate haskell program that fails to compile without -fglasgow-exts.) I'm sure I'm missing something lame here, but can someone tell me why we apparently can't declare a list to be an instance of a class in Haskell 98? Or is there perhaps some other syntax by which I'd declare this instance? If so, is this slated for fixing in Haskell'? $ ghc Test.lhs Test.lhs:6:1: Illegal instance declaration for `Vec [Double]' (The instance type must be of form (T a b c) where T is not a synonym, and a,b,c are distinct type variables) In the instance declaration for `Vec [Double]'
module Vec where
class Vec v where (.+.) :: v -> v -> v
instance Vec [Double] where xs .+. ys = zipWith (+) xs ys
instance Vec Double where x .+. y = x + y
feeling very stupid, David P.S. This is with ghc 6.4.1. And oddly enough, if you make the instance instance Num a => Vec [a] where xs .+. ys = zipWith (+) xs ys it works fine, but this strikes me as quite an ugly hack. I really want only Doubles to be instances of this class (which I've abbreviated for this email).
On Mon, 10 Jul 2006, David Roundy wrote:
class Vec v where (.+.) :: v -> v -> v
instance Vec [Double] where xs .+. ys = zipWith (+) xs ys
instance Vec Double where x .+. y = x + y
P.S. This is with ghc 6.4.1. And oddly enough, if you make the instance
instance Num a => Vec [a] where xs .+. ys = zipWith (+) xs ys
What about instance Vec a => Vec [a] where (.+.) = zipWith (.+.) ?
The problem isn't with lists specifically, but with any instance that applies types (rather than type variables) to a type constructor
From section 4.3.2 of The Haskell 98 Report: "The type (T u1 ... uk) must take the form of a type constructor T applied to simple type variables u1, ... uk". I've run into this restriction several times myself, and I'm also curious whether this will change in Haskell'.
Spencer Janssen
On 7/10/06, David Roundy
(This email is a literate haskell program that fails to compile without -fglasgow-exts.)
I'm sure I'm missing something lame here, but can someone tell me why we apparently can't declare a list to be an instance of a class in Haskell 98? Or is there perhaps some other syntax by which I'd declare this instance? If so, is this slated for fixing in Haskell'?
$ ghc Test.lhs
Test.lhs:6:1: Illegal instance declaration for `Vec [Double]' (The instance type must be of form (T a b c) where T is not a synonym, and a,b,c are distinct type variables) In the instance declaration for `Vec [Double]'
module Vec where
class Vec v where (.+.) :: v -> v -> v
instance Vec [Double] where xs .+. ys = zipWith (+) xs ys
instance Vec Double where x .+. y = x + y
feeling very stupid, David
P.S. This is with ghc 6.4.1. And oddly enough, if you make the instance
instance Num a => Vec [a] where xs .+. ys = zipWith (+) xs ys
it works fine, but this strikes me as quite an ugly hack. I really want only Doubles to be instances of this class (which I've abbreviated for this email). _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
From: haskell-cafe-bounces@haskell.org [mailto:haskell-cafe-bounces@haskell.org] On Behalf Of Spencer Janssen
The problem isn't with lists specifically, but with any instance that applies types (rather than type variables) to a type constructor
From section 4.3.2 of The Haskell 98 Report: "The type (T u1 ... uk) must take the form of a type constructor T applied to simple type variables u1, ... uk". I've run into this restriction several times myself, and I'm also curious whether this will change in Haskell'.
Spencer Janssen
Sorry, I'm struggling with this. Why is [] not of the form (T a b c ...) ? I assume that [] is syntactic sugar for something like: data List a = Cons a | Nil ===>? data [a] = (:) a | [] so [Double] is just sugar for List Double, which appears to me to be of the form (T a b c ...). What subtlety am I missing? Thanks, Alistair ***************************************************************** Confidentiality Note: The information contained in this message, and any attachments, may contain confidential and/or privileged material. It is intended solely for the person(s) or entity to which it is addressed. Any review, retransmission, dissemination, or taking of any action in reliance upon this information by persons or entities other than the intended recipient(s) is prohibited. If you received this in error, please contact the sender and delete the material from any computer. *****************************************************************
On 7/10/06, Bayley, Alistair
From: haskell-cafe-bounces@haskell.org [mailto:haskell-cafe-bounces@haskell.org] On Behalf Of Spencer Janssen
The problem isn't with lists specifically, but with any instance that applies types (rather than type variables) to a type constructor
From section 4.3.2 of The Haskell 98 Report: "The type (T u1 ... uk) must take the form of a type constructor T applied to simple type variables u1, ... uk". I've run into this restriction several times myself, and I'm also curious whether this will change in Haskell'.
Spencer Janssen
Sorry, I'm struggling with this. Why is [] not of the form (T a b c ...) ?
I assume that [] is syntactic sugar for something like:
data List a = Cons a | Nil ===>? data [a] = (:) a | []
so [Double] is just sugar for List Double, which appears to me to be of the form (T a b c ...).
What subtlety am I missing?
Double is not a type variable. /S -- Sebastian Sylvan +46(0)736-818655 UIN: 44640862
David Roundy wrote:
I'm sure I'm missing something lame here, but can someone tell me why we apparently can't declare a list to be an instance of a class in Haskell 98?
I think it is a "feature" of H98 intended to disallow any possibility of overlapping instances. If you have... instance Vec [Double] ...there's nothing from stopping you from also declaring... instance Num a => Vec [a] ...but since Double is a member of Num, which instance should the compiler use?
Or is there perhaps some other syntax by which I'd declare this instance?
Not by the looks of section 4.3.2 of the Haskell Report (at least by my reading). Greg Buchholz
participants (7)
-
Bayley, Alistair -
David House -
David Roundy -
Greg Buchholz -
Henning Thielemann -
Sebastian Sylvan -
Spencer Janssen