scala构造函数参数的可访问性

Question

Clarity needed in determining the scope of Scala's constructor parameter

As per this link https://alvinalexander.com/scala/how-to-control-visibility-constructor-fields-scala-val-var-private#comment-13237 , whenever a constructor parameter is labelled as private, then no getter and setter methods will be created. But the code I have provided here works fine even though the parameter is labelled as private. I went through this StackOverflow link Do scala constructor parameters default to private val? . This one & the above contradicts. Can someone please explain. The code segment is, in fact, available in the StackOverflow link.

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

The below line runs fine:

val a = new Foo(1)
a.otherBar(new Foo(3))

It prints 3.

As per the first link, the code should result in compile error because the parameter is private.

Answer 1

If you have a look at the scala language specification the private modifier allows access

from within the directly enclosing template and its companion module or companion class.

To allow access only from your inside the instance you can use the modifier private[this] . The code

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

results in

[error] .../src/main/scala/sandbox/Main.scala:9:17: value bar is not a member of sandbox.Foo
[error]       println(f.bar) // access bar of another foo
[error]                 ^
[error] one error found

Answer 2

If you want a constructor parameter that is only visible inside the class, don't make it a member, just make it a normal function parameter:

class Foo(bar: Int) {
  def otherBar(f: Foo) {
    println(f.bar) // Fails: value bar is not a member of Foo
  }
}

All the code inside the constructor can still access bar , but it is not a member of Foo and therefore is not visible outside the constructor.