scala - 连接两个不可变映射 - 首选哪些元素？

Question

当连接两个不可变映射时，似乎右侧操作数的元素将“覆盖”左侧的元素：

scala> List((1, 2), (5, 6)).toMap ++ List((5, 9)).toMap
res13: scala.collection.immutable.Map[Int,Int] = Map(1 -> 2, 5 -> 9)

scala> List((5, 9)).toMap ++ List((1, 2), (5, 6)).toMap
res14: scala.collection.immutable.Map[Int,Int] = Map(5 -> 6, 1 -> 2)

我想知道，这是否是 Scala 的规则？

从 Scala API 我无法弄清楚这个问题。

Answer 1

是的，这种行为是不变的

Answer 2

Map.++ 定义为：

override def ++[B1 >: B](xs: GenTraversableOnce[(A, B1)]): immutable.Map[A, B1] =
    ((repr: immutable.Map[A, B1]) /: xs.seq) (_ + _)

repr您当前的地图在哪里，并xs.seq为您提供存储在您传递到的地图中的一系列对/映射++。

Map./: 被描述为：

def /:[B](z: B)(op: (B, (A, B)) ⇒ B): B

Applies a binary operator to a start value and all elements of this
immutable map, going left to right.

Note: /: is alternate syntax for foldLeft;
z /: xs is the same as xs foldLeft z.

请注意，未指定“从左到右”对于无序映射的含义。

下面通过使用调试语句重新实现++和扩充它来 说明幕后发生的事情：println

val m1 = Map(1 -> "A", 2 -> "B", 3 -> "C")
val m2 = Map(2 -> "X", 3 -> "Y", 4 -> "Z")

println(m1.repr)
  /* Map(1 -> A, 2 -> B, 3 -> C) */
println(m1.repr.getClass.getName)
  /* scala.collection.immutable.Map$Map3 */

def ++[K, V](ts: Map[K, V], xs: Map[K, V]): Map[K, V] =
  (ts /: xs)  {case (acc, entry) =>
                println("acc = " + acc)
                println("entry = " + entry)
                acc + entry
              }

val m3 = ++(m1, m2)
  /*
    acc = Map(1 -> A, 2 -> B, 3 -> C)
    entry = (2,X)
    acc = Map(1 -> A, 2 -> X, 3 -> C)
    entry = (3,Y)
    acc = Map(1 -> A, 2 -> X, 3 -> Y)
    entry = (4,Z)
  */

println(m3)
  /* Map(1 -> A, 2 -> X, 3 -> Y, 4 -> Z) */