IndexedSeq [Int]与Array [Int]

来自Java背景,我正在学习Scala,以下内容使我感到非常困惑。为什么在这两个(非常相似但不同)的构造中返回的类型不同,它们仅在构建源集合的方式上有所不同-

  val seq1: IndexedSeq[Int] = for (i <- 1 to 10) yield i

  val seq2: Array[Int] = for (i <- Array(1, 2, 3)) yield i

请为我指出正确的文献,以便在这里我可以理解核心基础。

评论
  • 樱花
    樱花 回复

    发生这种情况是因为构造:

    for (x <-someSeq) yield x
    

    是相同的:

    someSeq.map(x => x)
    

    for () yield is just syntactic sugar for flatMap/map function.

    As we know map function doesn't change type of object-container, it changes element inside container. So, in your example 1 to 10 has a type Range.Inclusive which extends Range, and Range extends IndexedSeq. Mapped IndexedSeq has the same type IndexedSeq.

    for (i <- 1 to 10) yield i the same as (1 to 10).map(x => x)

    In second case: for (i <- Array(1, 2, 3)) yield i you have Array, and mapped Array type also Array.

    for (i <- Array(1, 2, 3)) yield i the same as Array(1, 2, 3).map(x => x)

  • zrerum
    zrerum 回复

    通常,有两种不同样式的收集操作库:

    • 保留类型:这就是您在问题中感到困惑的地方
    • 通用的(不是“参数多态性”的意思,而是单词的标准英语意义)或“同质的”

    Type-preserving collection operations try to preserve the type exactly for operations like filter, take, drop, etc. that only take existing elements unmodified. For operations like map, it tries to find the closest super type that can still hold the result. E.g. mapping over an IntSet with a function from Int to String can obviously not result in an IntSet, but only in a Set. Mapping an IntSet to Boolean could be represented in a BitSet, but I know of no collections framework that is clever enough to actually do that.

    Generic / homogeneous collection operations always return the same type. Usually, this type is chosen to be very general, to accommodate the widest range of use cases. For example, In .NET, collection operations return IEnumerable, in Java, they return Streams, in C++, they return iterators, in Ruby, they return arrays.

    Until recently, it was only possible to implement type-preserving collection operations by duplicating all operations for all types. For example, the Smalltalk collections framework is type-preserving, and it does this by having every single collections class re-implement every single collections operation. This results in a lot of duplicated code and is a maintenance nightmare. (It is no coincidence that many new object-oriented abstractions that get invented have their first paper written about how it can be applied to the Smalltalk collections framework. See Traits: Composable Units of Behaviour for an example.)

    To my knowledge, the Scala 2.8 re-design of the collections framework (see also this answer on SO) was the first time someone managed to create type-preserving collections operations while minimizing (though not eliminating) duplication. However, the Scala 2.8 collections framework was widely criticized as being overly complex, and it has required constant work over the last decade. In fact, it actually lead to a complete re-design of the Scala documentation system as well, just to be able to hide the very complex type signatures that the type-preserving operations require. But, this still wasn't enough, so the collections framework was completely thrown out and re-designed yet again in Scala 2.13. (And this re-design took several years.)

    因此,对您的问题的简单回答是:Scala尝试尽可能保留集合的类型。

    In your second case, the type of the collection is Array, and when you map over an Array, you get back an Array.

    In your first case, the type of the collection is Range. Now, a Range doesn't actually have elements, though. It only has a beginning and an end and a step, and it produces the elements on demand while you are iterating over it. So, it is not that easy to produce a new Range with the new elements. The map function would basically need to be able to "reverse engineer" your mapping function to figure out what the new beginning and end and step should be. (Which is equivalent to solving the Halting Problem, or in other words impossible.) And what if you do something like this:

    val seq1: IndexedSeq[Int] = for (i <- 1 to 10) yield scala.util.Random.nextInt(i)
    

    Here, there isn't even a well-defined step, so it is actually impossible to build a Range that does this.

    So, clearly, mapping over a Range cannot return a Range. So, it does the next best thing: It returns the most precise super type of Range that can contain the mapped values. In this case, that happens to be IndexedSeq.

    There is a wrinkle, in that type-preserving collections operations challenge what we consider to be part of the contract of certain operations. For example, most people would argue that the cardinality of a collection should be invariant under map, in other words, map should map each element to exactly one new element and thus map should never change the size of the collection. But, what about this code:

    Set(1, 2, 3).map { _ % 2 == 0 }
    //=> Set(true, false)
    

    Here, you get back a collection with fewer elements from a map, which is only supposed to transform elements, not remove them. But, since we decided we want type-preserving collections, and a Set cannot have duplicate values, the two false values are actually the same value, so there is only one of them in the set.

    [It could be argued that this actually only demonstrates that Sets aren't collections and shouldn't be treated as collections. Sets are predicates ("Is this element a member?") rather than collections ("Give me all your elements!")]