带有函数和初始值的 Scala 折叠

Question

给定一个泛型函数，你如何实现折叠函数？

def dofold[A, B](list: List[A], func: (B, A) => B, initial: B): B 

Answer 1

你可以看看它是如何在 Scala 标准库或 Haskell 中实现的

https://github.com/scala/scala/blob/2.12.x/​​src/library/scala/collection/immutable/List.scala#L407 https://github.com/scala/scala/blob/2.12.x /src/library/scala/collection/LinearSeqOptimized.scala#L122 https://github.com/scala/scala/blob/2.12.x/​​src/library/scala/collection/LinearSeqOptimized.scala#L133

https://wiki.haskell.org/Foldr_Foldl_Foldl%27 （也许这个链接是最好的）

http://hackage.haskell.org/package/base-4.12.0.0/docs/src/Data.Foldable.html#foldr http://hackage.haskell.org/package/base-4.12.0.0/docs/src /Data.Foldable.html#foldl