获取位于初始列表偶数位置的元素列表

Question

The task is to use just one passthrough using foldr or foldl . I use this decision:

evenOnly = fst . foldr (\x (y1, y2) -> (y2, x:y1)) ([],[])

Pretty good for finite lists. If I try evenOnly [1,2..] I'll got a fail. I know it's because of suspended caclulations, but how can I otherwise split the list or how to pass additional information about list positions or something to the calculations which begin at the end of the list?

Answer 1

You can use a lazy pattern ~(x1, y2) :

> let evenOnly = fst . foldr (\x ~(y1, y2) -> (y2, x:y1)) ([],[])
> take 20 $ evenOnly [1..]
[2,4,6,8,10,12,14,16,18,20,22,24,26,28,30,32,34,36,38,40]

This is essentially the same as

> let evenOnly = fst . foldr (\x y -> (snd y, x:fst y)) ([],[])

which has the advantage of not forcing the pair constructor too early. That is, the lambda above will produce the (,) constructor in its output before it demands y (the "recursive" result).