I'm trying to understand lazy evaluation as well. I created an example for myself and I'm wondering if I've got it right.
let adder n = \x -> n + x in (map (adder 12) [1,2,3]) !! 1
So the first thing I did is draw a graph to represent my expression. (map (adder 12) [1,2,3]) !! 1 = | (!!) ----/ \ / 1 map / \ / \ adder (:)--(:)--(:) | | | | \ 12 1 2 3 [] In order to proceed, I need definitions for adder, map and (!!). adder n = \x -> n + x map f [] = [] map f (x:xs) = (f x) : xs (x:xs) !! 0 = x (x:xs) !! n = xs !! (n-1) Here is how I think the evaluation would proceed: 0 => (map (adder 12) [1,2,3]) !! 1 The top node is (!!). In order to evaluate (!!), I need to expand the left hand side to get at least the first node of a list. I'll expand "map". 1 => ( (adder 12 1) : (map (adder 12) [2,3]) ) !! 1 I evaluated "map" once to get a list with an expression for the head and an expression for the tail. head: adder 12 1 tail: map (adder 12) [2,3] I proceed to evaluate (!!) for one step; this time n /= 0 so I extract the tail of the list and recursively call (!!). 2 => ( (map (adder 12) [2,3]) ) !! 0 The top node is (!!) again. I need to expand "map" again. 3 => ( (adder 12 2) : (map (adder 12) [3]) ) !! 0 I evaluate (!!) and this time n == 0 so I match the base case for (!!) and take the head of the list. 4 => adder 12 2 => (adder 12) 2 In order to proceed, I need to expand (adder 12). The adder function take one argument, "12", and produces a closure. I'll express it as a "let" expressions. Note: It is at this point (steps 4 to 7) that I'm confused about what's supposed to happen with closures and let expressions. 5 => (let n = 12 in \x -> n + x) 2 I'll substitute "2" into the let statement. 6 => (let n = 12 in n + 2) I'll substitute "12" for "n". 7 => 12 + 2 8 => 14 Can anyone tell me if I've got this right? -- Ron