2009/3/12 R J
Part of my problem is that "e" is defined as a function that takes one argument, I don't see how that fits in with the usual scheme for foldr, which, as I understand it, is:
foldr f e [x1, x2, ...] = f x1 (f x2 (f x3 ...(f xn e)))...
It's pretty easy, actually. Lets rewrite the type signatures a bit:
foldr :: (a -> b -> b) -> b -> ([a] -> b) zip :: [x] -> [y] -> [(x,y)] zip = foldr f e where e _ = [] f _ _ [] = [] f x g (y:ys) = (x,y) : g ys
So, from the signature for foldr, we can derive:
f :: (a -> b -> b) e :: b zip :: [a] -> b
And from the two type signatures for zip, we derive:
b ~ [y] -> [(x,y)] a ~ x
(~ is type equality) This gives
e :: [y] -> [(x,y)] f :: x -> ([y] -> [(x,y)]) -> ([y] -> [(x,y)])
or, removing the extra parenthesis
f :: x -> ([y] -> [(x,y)]) -> [y] -> [(x,y)]
that is, f takes *three* arguments, the second of which is a function of type [y] ->[(x,y)] What happens is that the *partially applied* f is getting chained through the fold; so you get zip [1,2,3] ['a','b','c'] = foldr f e [1,2,3] ['a','b','c'] = f 1 (f 2 (f 3 e)) ['a', 'b', 'c'] Then, in the first application of f, "g" is (f 2 (f 3 e)): = (1, 'a') : (f 2 (f 3 e)) ['b','c'] Now, there are two termination conditions; if the first list ends, we reach "e", which eats the remainder of the second list, returning []. In fact, e is the only total function of its type (forall x y. [y] -> [(x,y)]). If the second list ends, then f sees that and doesn't call g; that is, the rest of the foldr is unevaluated and unused! foldr f e [1,2,3] [] => f 1 (foldr f e [2,3]) [] => [] -- ryan