methods - Is Java "pass-by-reference" or "pass-by-value"? -

i thought java pass-by-reference; i've seen couple of blog posts (for example, this blog) claim it's not. don't think understand distinction they're making.

what explanation?

java pass-by-value. unfortunately, decided call location of object "reference". when pass value of object, passing reference it. confusing beginners.

it goes this:

public static void main( string[] args ) {     dog adog = new dog("max");     // pass object foo     foo(adog);     // adog variable still pointing "max" dog when foo(...) returns     adog.getname().equals("max"); // true, java passes value     adog.getname().equals("fifi"); // false  }  public static void foo(dog d) {     d.getname().equals("max"); // true     // change d inside of foo() point new dog instance "fifi"     d = new dog("fifi");     d.getname().equals("fifi"); // true } 

in example adog.getname() still return "max". value adog within main not changed in function foo dog "fifi" object reference passed value. if passed reference, adog.getname() in main return "fifi" after call foo.

likewise:

public static void main( string[] args ) {     dog adog = new dog("max");     foo(adog);     // when foo(...) returns, name of dog has been changed "fifi"     adog.getname().equals("fifi"); // true }  public static void foo(dog d) {     d.getname().equals("max"); // true     // changes name of d "fifi"     d.setname("fifi"); } 

in above example, fifi dog's name after call foo(adog) because object's name set inside of foo(...). operations foo performs on d such that, practical purposes, performed on adog (except when d changed point different dog instance d = new dog("boxer")).