Haskell列表长度替代

Question

Hi I've got a list on Haskell with close to 10^15 Int's in it and I'm trying print the length of the list.

let list1 = [1..1000000000000000]   --  this is just a dummy list I dont
print list1 length                  --  know the actual number of elements

printing this takes a very long time to do, is there another way to get the number of elements in the list and print that number?

Answer 1

I've occasionally gotten some value out of lists that carry their length. The poor man's version goes like this:

import Data.Monoid

type ListLength a = (Sum Integer, [a])

singletonLL :: a -> ListLength a
singletonLL x = (1, [x])

lengthLL :: ListLength a -> Integer
lengthLL (Sum len, _) = len

The Monoid instance that comes for free gives you empty lists, concatenation, and a fromList -alike. Other standard Prelude functions that operate on lists like map , take , drop aren't too hard to mimic, though you'll need to skip the ones like cycle and repeat that produce infinite lists, and filter and the like are a bit expensive. For your question, you would also want analogs of the Enum methods; eg perhaps something like:

enumFromToLL :: Integral a => a -> a -> ListLength a
enumFromToLL lo hi = (fromIntegral hi-fromIntegral lo+1, [lo..hi])

Then, in ghci, your example is instant:

> lengthLL (enumFromToLL 1 1000000000000000)
1000000000000000