scala - 并行集合中scala折叠的行为

Question

让我们多次运行以下代码行:

Set(1,2,3,4,5,6,7).par.fold(0)(_ - _)

结果非常有趣:

scala> Set(1,2,3,4,5,6,7).par.fold(0)(_ - _)
res10: Int = 8
scala> Set(1,2,3,4,5,6,7).par.fold(0)(_ - _)
res11: Int = 20

但是显然，它应该类似于其顺序版本:

scala> Set(1,2,3,4,5,6,7).fold(0)(_ - _)
res15: Int = -28

我知道 -操作在整数上是非关联的，这就是这种行为的原因，但是我的问题很简单:这并不意味着在集合的 fold实现中不应该将 .par并行化吗？

Answer 1

当您查看standard library documentation时，您会发现fold在这里不确定:

Folds the elements of this sequence using the specified associative binary operator. The order in which operations are performed on elements is unspecified and may be nondeterministic.

另外，还有 foldLeft:

Applies a binary operator to a start value and all elements of this sequence, going left to right. Applies a binary operator to a start value and all elements of this collection or iterator, going left to right.

Note: might return different results for different runs, unless the underlying collection type is ordered or the operator is associative and commutative.

由于 Set不是有序集合，因此没有可以折叠元素的规范顺序，因此即使是 foldLeft，标准库也允许自己不确定。如果您在此处使用有序序列，则 foldLeft在这种情况下将是确定性的。