swapping of objects in java [duplicate]

Answer 1

1. First and foremost thing, Java is pass by Value.( ie Pass by Copy).

2. You are passing the reference Not the Object to the method.

Eg:

public class Test{

        private String x;

            public void go(String s){

                this.x = s;

           }

         public static void main(String[] args){

                  Test t = new Test();

                  String a = new String("Bye");

                  t.go(a);


             }
      }

The above code is not to show how swapping is done, but to make an important point visible.

• When go(String s) method is called by passing an Argument "a" which is an Object reference variable of type String to the Parameter "s" which is also an Object reference variable of type String, its just the reference that is passed by Value / Copy.

• Any changes done on the reference will effect the String object on the Heap, NOT the reference.

Answer 2

Java is Strictly Pass by value.

Let's modify your code as below from the explanation point of view

static void swap(String str1, String str2){
String temp = str1;
str1=str2;
str2=temp;
}

public static void main(String[] args) {
String s1 = "Hello", s2 = "world";
swap(s1, s2);
System.out.println(s1 + s2);

}

S1 and S2 are the references which hold the address of the string object. For ex: S1=123 , S2= 234

When we call swap method, copy of S1 and S2 is created and will be assigned to str1 and str2.

Which will be str1=123 and str2=234.

Swap method will only swap the values present in str1 and str2. (str1=234 str2=123) whereas S1 and S2 still points to the same memory location.

This is the reason why the swap is not reflected in the main method.