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 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: [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
>
>
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