I see now that what I thought was happening wasn't happening at all.
top (push 1 s1) gives me the top of the new Stack returned by PUSH. s1 remains unchanged.
Thanks,
Michael
--- On Fri, 2/5/10, minh thu
Not using Stack for anything, just trying to understand how things can be done in Haskell.
To that end...
What's going on here? I'm not even calling function POP.
Michael
======================
module Data.Stack (Stack, emptyStack, isEmptyStack, push, pop, top) where
newtype Stack a = Stack [a]
emptyStack = Stack [] isEmptyStack (Stack xs) = null xs push x (Stack xs) = Stack (x:xs) pop (Stack (_:xs)) = Stack xs top (Stack (x:_)) = x
======================
[michael@localhost ~]$ ghci Stack.hs GHCi, version 6.10.4: http://www.haskell.org/ghc/ :? for help Loading package ghc-prim ... linking ... done. Loading package integer ... linking ... done. Loading package base ... linking ... done. [1 of 1] Compiling Data.Stack ( Stack.hs, interpreted ) Ok, modules loaded: Data.Stack. *Data.Stack> let s1 = emptyStack *Data.Stack> top (push 1 s1) 1 *Data.Stack> top (push 2 s1) 2 *Data.Stack> top (push 3 s1) 3 *Data.Stack> let s2 = pop s1 *Data.Stack> top s2 *** Exception: Stack.hs:8:0-28: Non-exhaustive patterns in function pop
When you write push 1 s1 you get a new stack value. you can view it by typing 'it' in ghci (provided you have an instance of Show for it). s1 is still s1, the empty stack. When you write let s2 = pop s1 Nothing happens yet, but if you want to evaluate s2, e.g. by typing it in ghci, pop will be applied to the empty stack, which is not taken care of in its definition. And you do want evaluate s2 when you eventually write top s2. HTH, Thu