Java is pass by value or pass by reference ?

Java is always pass by value only. The Java Spec says that everything in Java is pass-by-value. There is no such thing as "pass-by-reference" in Java.

Java manipulates objects data by reference. But Java doesn't pass method arguments by reference; it passes them by value. 

class Box{
 int width, height;
 
 static void swap(Box b1, Box b2){
  Box temp = b1;
  b1 = b2;
  b2 = b1;
 }
 
 Box(int width, int height){
  this.width = width;
  this.height = height;
 }
 
 public static void main(String args[]){
  Box b1 = new Box(5, 5);
  Box b2 = new Box(10, 10);
  
  System.out.println("Before swapping");
  System.out.print("for b1: width=" +b1.width);
  System.out.println(" height=" + b1.height);
  System.out.print("for b2: width=" +b2.width);
  System.out.println(" height=" + b2.height);
  
  swap(b1, b2);
  
  System.out.println("After swapping");
  System.out.print("for b1: width=" +b1.width);
  System.out.println(" height=" + b1.height);
  System.out.print("for b2: width=" +b2.width);
  System.out.println(" height=" + b2.height);
 } 
}

Output

Before swapping
for b1: width=5 height=5
for b2: width=10 height=10
After swapping
for b1: width=5 height=5
for b2: width=10 height=10

As you observe the output, the swap method doesn't swap the data since b1 and b2 are just references, Java passes the references by value just like any other parameter.  

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