Why is this TreeSet passing by reference, in Java?

Question

Why does this code change the value of set1 in the last print statement, I was under the impression that in Java arguments were passed by value?

Am I missing something?

import java.util.Set;
import java.util.TreeSet;


   public class Testing {
      public static void main(String args[]) {

       Set<String> set1 = new TreeSet<String>() ;
       set1.add("A") ;
       set1.add("B") ;
       set1.add("C") ;
       set1.add("D") ;
       set1.add("F") ;
       set1.add("G") ;
       System.out.println("set1, the tree set: " + set1) ;

       Set<String> set2 = new TreeSet<String>() ;
       set2.add("B") ;
       set2.add("D") ;
       set2.add("E") ;
       set2.add("F") ;
       set2.add("G") ;
       System.out.println("set2, the tree set: " + set2) ;

       Set set3 = difference(set1, set2);

       System.out.println("Difference: " + set1 + " - " + set2 
                    + " = " + set3) ;



    }
    public static Set<String> difference(Set<String> x, Set<String> y)
    {

        x.removeAll(y);

        return x;
  }}

score 1 · Answer 1 · answered Mar 25 '17 at 03:48

1

Java is pass-by-value, but the value it is passing in this case is a reference to a Set, so it can modify the original object.

answered Mar 25 '17 at 03:48

nbarriga

36
4

Is this the case specifically for sets? Because if you did the same sort of thing with Strings, this does not occur. – Jurgen Mar 25 '17 at 03:50
1

@gencode Strings are immutable. – Powerlord Mar 25 '17 at 03:51

score 0 · Answer 2 · answered Mar 25 '17 at 03:47

0

You confused with the term pass by value. pass by value in the sense reference passed as value. Reference value passed, and the set gets effected.

answered Mar 25 '17 at 03:47

Suresh Atta

120,458
37
198
307

Why is this TreeSet passing by reference, in Java?

2 Answers2