Hi John, You've several errors. What is isDigit? Let's start reading your code like the compiler does:

evalStep (Add x y) = do Fine Add x y thus both x and y are of type Expression because the constructor Add has been defined this way: | Add Expression Expression

Now let's have a look at this line:

if x = isDigit

if x = isDigit What is isDigit ? Hoogle shows there is a isDigit function you didn't import. http://haskell.org/ghc/docs/latest/html/libraries/base-4.2.0.0/Data-Char.htm... It's type is isDigit :: Char -> Bool So your if statement reads as if x is a function ? You can't compare a function with a value "x". Anyway you want == not = to compare for equality. At this point I stop parsing your code because it doesn't make sense at all. You usually use pattern matching for this kind of stuff: evalStep (Add x y) = case x of Val xval -> case y of Val yval -> (evalStep xval) + (evalStep yval) _ -> error "don't know what to do" _ -> error "don't know what to do" This is too verbose. It doesn't make sense to write it this way. Let's do it right ( I feel like doing ones homework here...) but I don't implement every case so I don't feel guilty :) Step by step: evalStep (Add x y) = let xEvaled = evalStep x yEvaled = evalStep y in Val (xEvaled + yEvaled) evalStep( Multiply a b) = error "TODO implement multiply" evalStep _ = error "TODO (everything else I didn't match explicitely)" shorter (that's what I would do:) evalStep (Add x y) = Val $ (evalStep x) + (evalStep y) or even shorter: evalStep (Add x y) = Val $ evalStep `on` (+) x y `on` applies a function to both operands before applying the operand. Summary: Whenever you want to use if .. else or a switch .. case .. thing remember that Haskell does have pattern matching and you should use it. Your isDigit function could be implemented using pattern matching (in fact it's the only way I know about). It looks like this: isDigit (Val _) = true -- Val is a digit isDigit _ = false -- everything else is not Then you could use if isDigit val then .. else .. but it still doesn't make much sense because you still have to get the val out of the constructor. Example: if isDigit val then let Val x = val -- now you can use x (which is Int) to use primitive +/-* operations in x * 2 else .. It doesn't hurt visiting the haskell.org page and reading some tutorials. HTH Marc Weber

Am Samstag 23 Januar 2010 18:40:10 schrieb John Moore:

Hi I'm trying to get a if statement to work but not sure if this is the right approach,

I think pattern matching and case .. of would be better suited.

what I want to do in simple english is to evaluate one expression at a time for example. (3+2)+(4+5) I want it to return 5 + (4+5) then 5+9.

data Expression = Val Integer

| Add Expression Expression | Subtract Expression Expression | Multiply Expression Expression | Divide Expression Expression

deriving Show evalStep :: Expression -> Expression evalStep (Val x)= (Val x)

evalStep (Add x y) = do

if x = isDigit then if y isDigit else evalStep x,y

-- evaluate one step only evalStep (Add x y) = case x of (Val v) -> case y of (Val w) -> Val (v+w) other -> Add x (evalStep y) other -> Add (evalStep x) y evalStep (Subtract x y) ... -- analogously, also Multiply and Divide evalStep e = e -- only Val left. Voilà, the leftmost innermost compound expression is reduced if the expression isn't fully evaluated. To completely evaluate: eval :: Expression -> Expression eval e@(Val _) = e eval e = eval (evalStep e)

I thinking about using recursion down a tree left then right but that seems very complicated. Any thoughts welcome! John