Scala如何处理方法返回类型中的Any to AnyVal协方差？

Question

An example is worth a thousand words:

class A { def foo: Any = new Object }

class B extends A {
  override def foo: AnyVal = 42
}

In Java, the signature @Override public int foo() wouldn't even be allowed, and the overridden method foo in B could only return the wrapper integer type ( @Override java.lang.Integer foo() ).

Is Scala able to avoid the boxing/unboxing of AnyVal values in the overridden def foo: AnyVal method above?

Answer 1

No, it does not. Scala has to adhere to emitting the correct bytecode:

λ scalac -Xprint:jvm Bar.scala
[[syntax trees at end of                       jvm]] // Bar.scala
package yuval.tests {
  class A extends Object {
    def foo(): Object = new Object();
    def <init>(): yuval.tests.A = {
      A.super.<init>();
      ()
    }
  };
  class B extends yuval.tests.A {
    override def foo(): Object = scala.Int.box(42);
    def <init>(): yuval.tests.B = {
      B.super.<init>();
      ()
    }
  }
}

You can see that although AnyVal was permitted in Scala, the actual method signature for the emitted foo is Object and not AnyVal , and Int is boxed.

Answer 2

Yuval's answer can be generalized: erasure of AnyVal is Object (you can see this eg by entering classOf[AnyVal] in the REPL), so whenever you have AnyVal in Scala, you can expect Object in bytecode.

Eg if you change A to

class A { def foo: AnyVal = 0 }

it's still Object .

Maybe there is some situation in which using AnyVal itself will avoid boxing, but I would be surprised. It was created pretty much for compiler's convenience, and picked up another use ( value classes ) later, but it's rarely useful in user code (except for defining value classes).