haskell中两个列表上的两个操作

Question

I'm a bit of a Haskell newbie. Let's say that i have two infinite lists of numbers A and B and I want to create another infinite list C which contains a sequence of x+y 's and xy 's where x <- A and y <- B , ie, C grows by 2 in every loop; what's the most clever way to do this?

Answer 1

制作由两个元素组成的列表的无限列表并将其连接起来。

concat [[x+y, x-y] | (x, y) <- zip listA listB]

Answer 2

You probably don't want the most clever way, since by definition it would be too clever for you to debug :-)

An obvious way would be to zip the infinite streams together, like so:

zipWith k as bs
    where
        k a b = (a + b, a - b)

Answer 3

For infinite lists only it's just

mkList (x:xs) (y:ys) = x+y : x-y : mkList xs ys

and to support finite lists, too, you've to add the base case

mkList _ _ = []

Answer 4

f xs = concat . zipWith go xs where 
       go x y = map (($y).($x)) [(+),(-)]