如何在haskell中定义一个lazier sum函数？

Question

我怎么能在"不太严格"的表达式中使左手总和,以便我不评估整个列表xs.在该示例中,只有前3个元素足以知道第二个表达式(True)的结果.

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

ghci的:

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

Answer 1

使用懒惰的天然.

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 _,_,_,_,_,_,_,_]

为了更加懒惰,使用foldr而不是foldl这样sum做:

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 _ : _