Java – How to copy an object in Java

The usual way to check if the value of a property is the special value undefined, is:

if(o.myProperty === undefined) {
  alert("myProperty value is the special value `undefined`");
}

To check if an object does not actually have such a property, and will therefore return undefined by default when you try and access it:

if(!o.hasOwnProperty('myProperty')) {
  alert("myProperty does not exist");
}

To check if the value associated with an identifier is the special value undefined, or if that identifier has not been declared. Note: this method is the only way of referring to an undeclared (note: different from having a value of undefined) identifier without an early error:

if(typeof myVariable === 'undefined') {
  alert('myVariable is either the special value `undefined`, or it has not been declared');
}

In versions of JavaScript prior to ECMAScript 5, the property named "undefined" on the global object was writeable, and therefore a simple check foo === undefined might behave unexpectedly if it had accidentally been redefined. In modern JavaScript, the property is read-only.

However, in modern JavaScript, "undefined" is not a keyword, and so variables inside functions can be named "undefined" and shadow the global property.

If you are worried about this (unlikely) edge case, you can use the void operator to get at the special undefined value itself:

if(myVariable === void 0) {
  alert("myVariable is the special value `undefined`");
}

Java is always pass-by-value. Unfortunately, when we deal with objects we are really dealing with object-handles called references which are passed-by-value as well. This terminology and semantics easily confuse many beginners.

It goes like this:

public static void main(String[] args) {
    Dog aDog = new Dog("Max");
    Dog oldDog = aDog;

    // we pass the object to foo
    foo(aDog);
    // aDog variable is still pointing to the "Max" dog when foo(...) returns
    aDog.getName().equals("Max"); // true
    aDog.getName().equals("Fifi"); // false
    aDog == oldDog; // true
}

public static void foo(Dog d) {
    d.getName().equals("Max"); // true
    // change d inside of foo() to point to a new Dog instance "Fifi"
    d = new Dog("Fifi");
    d.getName().equals("Fifi"); // true
}

In the example above aDog.getName() will still return "Max". The value aDog within main is not changed in the function foo with the Dog "Fifi" as the object reference is passed by value. If it were passed by reference, then the aDog.getName() in main would return "Fifi" after the call to foo.

Likewise:

public static void main(String[] args) {
    Dog aDog = new Dog("Max");
    Dog oldDog = aDog;

    foo(aDog);
    // when foo(...) returns, the name of the dog has been changed to "Fifi"
    aDog.getName().equals("Fifi"); // true
    // but it is still the same dog:
    aDog == oldDog; // true
}

public static void foo(Dog d) {
    d.getName().equals("Max"); // true
    // this changes the name of d to be "Fifi"
    d.setName("Fifi");
}

In the above example, Fifi is the dog's name after call to foo(aDog) because the object's name was set inside of foo(...). Any operations that foo performs on d are such that, for all practical purposes, they are performed on aDog, but it is not possible to change the value of the variable aDog itself.

For more information on pass by reference and pass by value, consult the following SO answer: https://stackoverflow.com/a/430958/6005228. This explains more thoroughly the semantics and history behind the two and also explains why Java and many other modern languages appear to do both in certain cases.