Question

我怎样才能在下面的表达式中对左手总和“不严格”，这样我就不会评估整个列表xs。在该示例中，只有前3个元素足以知道第二个表达式的结果（True）。

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

ghci中：

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

要变得更加懒惰，请使用foldr而不是foldl sum的方式：

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 _ : _

如何在haskell中定义一个lazier sum函数？

1 个答案: