Haskell Help

[scarface]

Hi people, i have some haskell problems with expressions. What i am trying to do is add two expressions together and simplify the output. I have this code

module X where

type Variable = String

data X = Const Integer
	| Var Variable
	| Plus X X
	deriving (Eq, Show)

What i am trying to achieve is when i write for example Plus (Var "x")(Const 0) in Hugs, to get "Var x". The 0 is disregarded as it adds nothing to x (0 + x = x). If the Const is 1 again the output should be "Var x" as 1 + x = x.

Any help is appreciated

Thnx

[Twey]

What? Since when does 1 + x = x? I think the universe would explode or something if that were true.

-- code removed by request

[scarface]

thnx for the code i will try it later when i come back home..actually i have done a mistake..not 1+x but rather 1*x = x..i apologize for the confusion ..

[Twey]

But you have no multiplication operator. You'd have to add one:

-- code removed by request

[scarface]

yes i was just giving example with the multiplication to make it clearer..also is there a way to add the 2 Expr together in case they are both Const? example : if we have Plus(Const 2)(Const 5) = 7. What i did was i wrote a separate function

[/CODE]add :: Expr -> Integer
add (Plus (Const x) (Const y)) = x + y[/CODE]

and it works but i was wondering if i can have this in the simplify function? It gives me an error to do with the Type. I have defined simplify to be Expr -> Expr but with the adding i want to get an integer back and thats where the error comes.

[Twey]

What we were doing before was merely normalisation; now, you're talking about evaluation. It will work something like this (rather dumb version, I'm afraid):

-- code removed by request

The error with your original was that you were trying to return an Integer, but the function was meant to return an Expr.

[scarface]

Thanks for the code but i think i will just stick to my function. I am pretty bad at haskell and i dont completely understand the code you gave so i rather not use it (in case i get asked questions on it). I will try and do it the way i can and understand it and just keep it as simple as possible even if my program is not going to behave as it is expected.

I have another question concerning converting an expression to string. I had a look on the web and "Show" function is mentioned.I tried it with simple input but i couldnt convert it to string. Anyone could point me to something more useful?

[Twey]

You're probably confused by the on function. on is (roughly) defined thusly:

on :: (b -> b -> b) -> (a -> b)
on g f = \x y -> f x `g` f y

So,

liftConst = Const . on (fromConst . evaluate)

is the same as (adding in all the points)

liftConst op a b = Const $ op (fromConst $ evaluate a) (fromConst $ evaluate b)

It's just a little more elegant.

[scarface]

do u know why it doesnt let me load the data.function module?
I typed it exactly as posted and it gave me error saying i cant find the module.

[Twey]

What compiler and version are you using?