scala - 为什么 Scala 在解包 Tuple 时要构造一个新的 Tuple？

Question

为什么这段 Scala 代码是这样的:

class Test
{
  def foo: (Int, String) =
  {
    (123, "123")
  }

  def bar: Unit =
  {
    val (i, s) = foo
  }
}

为bar()生成以下字节码，构造一个新的Tuple2，并从foo()传递Tuple2 code> 到它然后从中获取值？

public void bar();
Code:
 0:   aload_0
 1:   invokevirtual   #28; //Method foo:()Lscala/Tuple2;
 4:   astore_2
 5:   aload_2
 6:   ifnull  40
 9:   new     #7; //class scala/Tuple2
 12:  dup
 13:  aload_2
 14:  invokevirtual   #32; //Method scala/Tuple2._1:()Ljava/lang/Object;
 17:  aload_2
 18:  invokevirtual   #35; //Method scala/Tuple2._2:()Ljava/lang/Object;
 21:  invokespecial   #20; //Method scala/Tuple2."<init>":(Ljava/lang/Object;Ljava/lang/Object;)V
 24:  astore_1
 25:  aload_1
 26:  invokevirtual   #39; //Method scala/Tuple2._1$mcI$sp:()I
 29:  istore_3
 30:  aload_1
 31:  invokevirtual   #35; //Method scala/Tuple2._2:()Ljava/lang/Object;
 34:  checkcast       #41; //class java/lang/String
 37:  astore  4

这是因为编译器没有检查 foo() 的返回值不是元组吗？

JVM 会优化构造吗？

Answer 1

这似乎是根据spec (在 4.1 值声明和定义中 - 为 stackoverflow 显示稍微重新格式化):

Value definitions can alternatively have a pattern (§8.1) as left-hand side. If p is some pattern other than a simple name or a name followed by a colon and a type, then the value definition val p = e is expanded as follows:

1. If the pattern p has bound variables x1, . . . , xn, where n >= 1: Here, $x is a fresh name.

  val $x = e match {case p => (x1, . . . , xn)}
  val x1 = $x._1
  . . .
  val xn = $x._n

因此元组创建发生在解析器阶段。因此 val (i, s) = (1, "s") 在解析器阶段结束时消耗为:

private[this] val x$1 = scala.Tuple2(1, "s"): @scala.unchecked match {    
  case scala.Tuple2((i @ _), (s @ _)) => scala.Tuple2(i, s)
};
val i = x$1._1;
val s = x$1._2

通过一百万次迭代的简单测试来测量这一点:

def foo: (Int, String) = (123, "123")
def bar: Unit = { val (i, s) = foo }
def bam: Unit = { val f = foo; val i = f._1; val s = f._2 }

产量

foo: Elapsed: 0.030
bar: Elapsed: 0.051
._1 ._2 access: Elapsed: 0.040

并带有 -optimize 标志:

foo: Elapsed: 0.027
bar: Elapsed: 0.049
._1 ._2 access: Elapsed: 0.029