Why Java passes objects as references passed by value

I came across a question recently on stackoverflow that was very enlightening and proves how objects are actually passed in java. It was quite an interesting discovery!

I am quoting from here

Java is always pass-by-value. The difficult thing can be to understand that Java passes objects as references passed by value.

It goes like this:

public void foo(Dog d) {
  d.name.equals("Max"); // true
  d = new Dog("Fifi");
  d.name.equals("Fifi"); // true
}

Dog aDog = new Dog("Max");
foo(aDog);
aDog.name.equals("Max"); // true

In this example aDog.name will still be “Max”. “d” is not overwritten in the function as the object reference is passed by value.

Likewise:

public void foo(Dog d) {
  d.name.equals("Max"); // true
  d.setname("Fifi");
}

Dog aDog = new Dog("Max");
foo(aDog);
aDog.name.equals("Fifi"); // true

Chris Georgoulis

Summary: - Primary programming language: Java SE (SWING, JDBC), Java EE, (JSF - PrimeFaces, JPA - EclipseLink/Hibernate, EJB, CDI ), - Secondary languages: C/C++, Web Technologies(HTML, XHTML, XML, CSS, JavaScript, jQuery, AngularJS). - Frameworks: Android SDK, Global Sensors Network (GSN), Open Computer Vision (OpenCV). - IDE's: Eclipse, Visual Studio - Other Apps: MS Office, Photoshop, Dreamweaver - Database Design - SQL(MySQL), NoSQL(MongoDB) - Modeling Tools: UML

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