Hi,
See inline comments
On Wed, Dec 1, 2010 at 9:34 PM, John Moore
<snip>
data Expression = Val Double | Add Expression Expression | Subtract Expression Expression | Multiply Expression Expression | Divide Expression Expression deriving Show
evalStep :: Expression -> Expression evalStep (Val x)= (Val x)
We can pattern match to a constructor. Val is a constructor for Expression, our evalStep receives an Expression as input and produces another one, presumably the evaluated input. In the above case, we know that if we try to evaluate a value (that's what Val stands for) we obtain the exact same value back.
evalStep (Add x y) = case x of (Val a) -> case y of (Val b) -> Val (a+b) left -> Add x (evalStep y) right -> Add (evalStep x)y <snip>
This case is a little more complicated. We want to do a single step of evaluation for expressions like 2+3 and 2+3+4 (expressed in our format, they would be Add (Double 2) (Double 3) and Add (Double 2) (Add (Double 3) (Double 4)) or Add (Add (Double 2) (Double 3)) (Double 4)). because the later case in ambiguous, we have two imbricated cases instead of only one. Each of the cases must look to see if we can narrow down x or y to (Val something) because this will make the evaluation progress. Hope it's useful, -- MM