haskell - 如何在haskell中定义一个惰性求和函数？

Question

我怎样才能在“不太严格”下面的表达式中进行左手求和，这样我就不会评估整个列表xs。在示例中，仅前 3 个元素就足以知道第二个表达式 ( True) 的结果。

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

使用懒惰自然。

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

为了更懒惰，使用foldr而不是foldl方式sum：

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 _ : _