Object References
Remarks#
This should help you understand a “Null Pointer Exception” — one gets one of those because an object reference is null, but the program code expects the program to use something in that object reference. However, that deserves its own topic…
Object References as method parameters
This topic explains the concept of an object reference; it is targeted at people who are new to programming in Java. You should already be familiar with some terms and meanings: class definition, main method, object instance, and the calling of methods “on” an object, and passing parameters to methods.
public class Person {
private String name;
public void setName(String name) { this.name = name; }
public String getName() { return name; }
public static void main(String [] arguments) {
Person person = new Person();
person.setName("Bob");
int i = 5;
setPersonName(person, i);
System.out.println(person.getName() + " " + i);
}
private static void setPersonName(Person person, int num) {
person.setName("Linda");
num = 99;
}
}
To be fully competent in Java programming, you should be able to explain this example to someone else off the top of your head. Its concepts are fundamental to understanding how Java works.
As you can see, we have a main
that instantiates an object to the variable person
, and calls a method to set the name
field in that object to "Bob"
. Then it calls another method, and passes person
as one of two parameters; the other parameter is an integer variable, set to 5.
The method called sets the name
value on the passed object to “Linda’, and sets the integer variable passed to 99, then returns.
So what would get printed?
Linda 5
So why does the change made to person
take effect in main
, but the change made to the integer does not?
When the call is made, the main method passes an object reference for person
to the setPersonName
method; any change that setAnotherName
makes to that object is part of that object, and so those changes are still part of that object when the method returns.
Another way of saying the same thing: person
points to an object (stored on the heap, if you’re interested). Any change the method makes to that object are made “on that object”, and are not affected by whether the method making the change is still active or has returned. When the method returns, any changes made to the object are still stored on that object.
Contrast this with the integer that is passed. Since this is a primitive int (and not an Integer object instance), it is passed “by value”, meaning its value is provided to the method, not a pointer to the original integer passed in. The method can change it for the method’s own purposes, but that does not affect the variable used when the method call is made.
In Java, all primitives are passed by value. Objects are passed by reference, which means that a pointer to the object is passed as the parameter to any methods that take them.
One less-obvious thing this means: it is not possible for a called method to create a new object and return it as one of the parameters. The only way for a method to return an object that is created, directly or indirectly, by the method call, is as a return value from the method. Let’s first see how that would not work, and then how it would work.
Let’s add another method to our little example here:
private static void getAnotherObjectNot(Person person) {
person = new Person();
person.setName("George");
}
And, back in the main
, below the call to setAnotherName
, let’s put a call to this method and another println call:
getAnotherObjectNot(person);
System.out.println(person.getName());
Now the program would print out:
Linda 5
Linda
What happened to the object that had George? Well, the parameter that was passed in was a pointer to Linda; when the getAnotherObjectNot
method created a new object, it replaced the reference to the Linda object with a reference to the George object. The Linda object still exists (on the heap), the main
method can still access it, but the getAnotherObjectNot
method wouldn’t be able to do anything with it after that, because it has no reference to it. It would appear that the writer of the code intended for the method to create a new object and pass it back, but if so, it didn’t work.
If that is what the writer wanted to do, he would need to return the newly created object from the method, something like this:
private static Person getAnotherObject() {
Person person = new Person();
person.setName("Mary");
return person;
}
Then call it like this:
Person mary;
mary = getAnotherObject();
System.out.println(mary.getName());
And the entire program output would now be:
Linda 5
Linda
Mary
Here is the entire program, with both additions:
public class Person {
private String name;
public void setName(String name) { this.name = name; }
public String getName() { return name; }
public static void main(String [] arguments) {
Person person = new Person();
person.setName("Bob");
int i = 5;
setPersonName(person, i);
System.out.println(person.getName() + " " + i);
getAnotherObjectNot(person);
System.out.println(person.getName());
Person person;
person = getAnotherObject();
System.out.println(person.getName());
}
private static void setPersonName(Person person, int num) {
person.setName("Linda");
num = 99;
}
private static void getAnotherObjectNot(Person person) {
person = new Person();
person.setMyName("George");
}
private static person getAnotherObject() {
Person person = new Person();
person.setMyName("Mary");
return person;
}
}