How to define a more lazy sum function in haskell?

Question

How can I make the left hand amount in the expression below “less restrictive” so that I do not evaluate the entire list xs. In this example, only the first three elements are enough to find out the result of the second expression ( True).

xs=[1..10]
sum xs > 3

GHCi:

λ> let xs = [1..10]
λ> :sp xs
xs = _
λ> sum xs > 3
True
λ> :sp xs
xs = [1,2,3,4,5,6,7,8,9,10] 

Answer 1

Use lazy natural .

Prelude Data.Number.Natural> let xs = [1..10] :: [Natural]
Prelude Data.Number.Natural> :sp xs
xs = _
Prelude Data.Number.Natural> sum xs > 3
True
Prelude Data.Number.Natural> :sp xs
xs = [Data.Number.Natural.S Data.Number.Natural.Z,
      Data.Number.Natural.S
        (Data.Number.Natural.S Data.Number.Natural.Z),
      Data.Number.Natural.S _,_,_,_,_,_,_,_]

To make it even more lazy, use the method foldrinstead :foldlsum

Prelude Data.Number.Natural> let xs = [1..10] :: [Natural]
Prelude Data.Number.Natural> let lazySum = foldr (+) 0
Prelude Data.Number.Natural> lazySum xs > 3
True
Prelude Data.Number.Natural> :sp xs
xs = Data.Number.Natural.S Data.Number.Natural.Z :
     Data.Number.Natural.S
       (Data.Number.Natural.S Data.Number.Natural.Z) :
     Data.Number.Natural.S _ : _