Question

I wonder why I can’t use the same name for function parameter and the name used in a class. Please refer to the following example.

scala> class Person() {var name = "bob" }
defined class Person

scala> val p = new Person
p: Person = Person@486f8860

scala> p.name
res0: java.lang.String = bob

scala> p.name = "alice"

scala> p.name
res1: java.lang.String = alice

scala> def chngName(name:String) = new Person() {this.name= name}
chngName: (name: String)Person

scala> val p = chngName("aa")
p: Person = $anon$1@3fa1732d

scala> p.name
res2: java.lang.String = bob

scala> def chngName(n:String) = new Person() {name= n}
chngName: (n: String)Person

scala> val p = chngName("aa")
p: Person = $anon$1@19d2052b

scala> p.name
res3: java.lang.String = aa

Of course, I can use a different name but I want to why I can’t or there is something I am miss here. Thanks

Answer 1

(Why do want to use var when you return a new object in chngName?)

As has been said, the class’s field of the same name shadows the method argument.

You could of course rename it before entering the class’s scope:

def chngName(name:String) = {
  val _name = name
  new Person() { name = _name }
}

For this use case, there are a few other options, however. It depends, though, whether you want to copy the object, ie. return a new Person, or if a simple change of the var suffices.

If you want to return a completely new object, you may consider using a case class which adds a copy method with the same semantics and identically named method arguments. (So you can use named arguments):

case class Person(name: String)
val p = Person("bob")
p.name // bob
val q= p.copy(name = "aa")
q.name // aa
q == p // false