Do scala constructor parameters default to private val?

问题

I have been trying:

class Foo(bar: Int)

vs:

class Foo(private val bar: Int)

and they seem to behave the same although I couldn\'t find anywhere saying that (bar: Int) expands to (private val bar: Int) so my question is, are these identical/similar?

On a side note, I have been trying to use -Xprint:typer on these code pieces and they produce the same code except for an extra line in the second one. How do I read that extra line?

..
class Foo extends scala.AnyRef {
  <paramaccessor> private[this] val bar: Int = _;
  def <init>(bar: Int): this.Foo = {
    Foo.super.<init>();
    ()
  }
}
..


..
class Foo extends scala.AnyRef {
  <paramaccessor> private[this] val bar: Int = _;
  <stable> <accessor> <paramaccessor> private def bar: Int = Foo.this.bar;
  def <init>(bar: Int): this.Foo = {
    Foo.super.<init>();
    ()
  }
}
..

回答1:

bar: Int

This is barely a constructor parameter. If this variable is not used anywhere except the constructor, it remains there. No field is generated. Otherwise private val bar field is created and value of bar parameter is assigned to it. No getter is created.

private val bar: Int

Such declaration of parameter will create private val bar field with private getter. This behavior is the same as above no matter if the parameter was used beside the constructor (e.g. in toString() or not).

val bar: Int

Same as above but Scala-like getter is public

bar: Int in case classes

When case classes are involved, by default each parameter has val modifier.

回答2:

In the first case, bar is only a constructor parameter. Since the main constructor is the content of the class itself, it is accessible in it, but only from this very instance. So it is almost equivalent to:

class Foo(private[this] val bar:Int)

On the other hand, in the second case bar is a normal private field, so it is accessible to this instance and other instances of Foo. For example, this compiles fine:

class Foo(private val bar: Int) {
  def otherBar(f: Foo) {
    println(f.bar) // access bar of another foo
  }
}

And runs:

scala> val a = new Foo(1)
a: Foo = Foo@7a99d0af

scala> a.otherBar(new Foo(3))
3

But this doesn't:

class Foo(bar: Int) {
  def otherBar(f: Foo) {
    println(f.bar) // error! cannot access bar of another foo
  }
}

来源：https://stackoverflow.com/questions/14694712/do-scala-constructor-parameters-default-to-private-val

标签

scala

scala-primary-constructor