Haskell如何生成这个无限列表？

Question

I saw this code to generate Fibonacci numbers.

fibs = 1:1:(zipWith (+) fibs (tail fibs))

Can a similar styled code be written to generate the infinite list [1..]?

I saw this link on cyclic structures on Haskell site.

There an example is given

cyclic = let x = 0 : y
         y = 1 : x
     in  x

I tried to define a list for my problem in a cyclic manner, but could not succeed. What I want is a list defined in terms of itself and which evaluates to [1..] in Hasekll.

Note: The Haskell [1..] evaluates to [1,2,3,4,5...] and not to [1,1,1...] .

Answer 1

The following should give you the desired result:

nats = 1 : map (+1) nats

Or, more idiomatically:

nats = iterate (+1) 1

It's easy to see why the first snippet evaluates to [1,2,3...] by using equational reasoning:

nats = 1 : map (+1) nats 
     = 1 : map (+1) (1 : map (+1) nats) 
     = 1 : map (+1) (1 : map (+1) (1 : map (+1) nats))
     = 1 : 1 + 1 : 1 + 1 + 1 : .... 
     = [1,2,3...]

Answer 2

Yes.

Think about how you could write out each element of your list:

1
1 + 1
1 + 1 + 1
1 + 1 + 1 + 1
1 + 1 + 1 + 1 + 1

After each entry, every subsequent entry has an extra + 1 . So we want to start at 1 and then add 1 to each subsequent element. Then we want to take the second element and add 1 to everything after that.

Here's how we can do this:

let xs = 1 : map (+ 1) xs

This expands like this:

1 : map (+ 1) xs
1 : (1 + 1) : map (+ 1) xs
1 : (1 + 1) : ((1 + 1) + 1) : map (+ 1) xs

and so on.