Hello, I wonder if someone could answer
the following…
The short question is what does <at> mean in
mulNat a b
| a <= b = mulNat' a b b
| otherwise = mulNat' b a a
where
mulNat' x <at> (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 <at> (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 <at> (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 <at> (S a)"
mean? in
mulNat' x <at> (S a) y orig
From:
haskell-cafe-bounces <at> haskell.org [mailto:haskell-cafe-bounces <at> haskell.org] On Behalf Of Nicholls, Mark
Sent: 21 December 2007 17:47
To: David Menendez
Cc: Jules Bean;
haskell-cafe <at> haskell.org
Subject: RE: [Haskell-cafe] nice
simple problem for someone struggling....
Let me
resend the code…as it stands….
module Mainwhere
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 <at> gmail.com [mailto:d4ve.menendez <at> gmail.com] On Behalf Of David Menendez
Sent: 21 December 2007 17:05
To: Nicholls, Mark
Cc: Jules Bean;
haskell-cafe <at> haskell.org
Subject: Re: [Haskell-cafe] nice
simple problem for someone struggling....
On Dec 21, 2007 11:50 AM,
Nicholls, Mark <Nicholls.Mark <at> mtvne.com>
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 <dave <at> zednenem.com>
<http://www.eyrie.org/~zednenem/ >
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