Scala Java与Val的互操作性

Question

In scala, if we have a class say,

case class B(b: Int)
object B {
    def main(args: Array[String] {
        val b = B(1)
        b.b = 2 // ** compile time error here **
    }
}

would give an error saying, when compile with scalac B.scala

error: reassignment to val
b.b = 2 
    ^

Everything is fine till here.

However, if I have a java class A.java :

public class A {
     public int a;
     public A(int a) {this.a = a;}
     public void setA(int a_) {this.a = a_;}
     public String toString() {return "A: " + a;}
}

first compile it with javac A.java , then use it in scala with val :

object B {
    def main(args: Array[String] {
        val a = new A(1)
        println("Before assignment: " + a)
        a.a = 2 // ** No error here **
        println("After assignment: " + a)
    }
}

compile B.scala with scalac B.scala and run with scala B . No error occurs.

Does it break intention of val in scala?

If I want to, when using java class in scala, maintain the fixed value of instance a , what can I do?

Or there actually is some documentation mentioning this behavior?

Environment: Scala version 2.11.5 (OpenJDK 64-Bit Server VM, Java 1.7.0_75).
Centos 6.6

Edit

Thank you everyone for your reply.

While I am well aware that by making the field final: public final int a; , immutability can be achieved.

I'm not hoping to solve this problem by modifying java code. Imagine that I'm using third party java libraries that I can no way modify their code, yet I want making sure my code won't accidentally modify third-party's internal state, hence come the val .

In addition, by disassembling the .class file using javap -c , it can be seen clearly that scala is not making Bb immutable by inserting final to class B .

Or maybe someone explaining the "magic" behind val would be helpful

Answer 1

Clearly in your Java class A , the member a is both public and mutable. To treat a Java variable as Scala val , you can make it final in your Java class.

Also remember that Scala too has var which has essentially same behaviour as Java variables ( except for final ones ).

EDIT 1

Define B like so:

case class B(val b: Int, var c: Int)

And then check the compiled class definition

$ javap B
Compiled from "B.scala"
public class B implements scala.Product,scala.Serializable {
  public static scala.Option<scala.Tuple2<java.lang.Object, java.lang.Object>> unapply(B);
  public static B apply(int, int);
  public static void main(java.lang.String[]);
  public int b();
  public int c();
  public int copy$default$1();
  public int copy$default$2();
  public void c_$eq(int);
  public B copy(int, int);
  public java.lang.String productPrefix();
  public int productArity();
  public java.lang.Object productElement(int);
  public scala.collection.Iterator<java.lang.Object> productIterator();
  public boolean canEqual(java.lang.Object);
  public int hashCode();
  public java.lang.String toString();
  public boolean equals(java.lang.Object);
  public B(int, int);
}

Thing to note here is that for member 'c' you will have public void c_$eq(int); but for member b there is no such method. In fact in Scala, the assignment turns out to be a method call.

I hope that makes it clearer.

Answer 2

Your confusion comes from line val a = new A(1) in object B . It means that you cannot reassign value to your variable a but it's completely ok to reassign any field of a . This can be limited by marking a filed be final

Try making public int a final in your java class as following

public class A {
   public final int a;
   public A(int a) {this.a = a;}
   //public void setA(int a_) {this.a = a_;} //Compilation error here. Cannot assign to a final field
   public String toString() {return "A: " + a;}
}

It will enforce that field a of class A cannot be reassigned.