Tuesday, January 25, 2011

Chapter 12: Constructors and Instantiation

Objects are constructed. You can’t make a new object without invoking a constructor. In fact, you can’t make a new object without invoking not just the constructor of the object’s actual class type, but also the constructor of each of its superclasses including the Object class itself! Constructors are the code that runs whenever you use the keyword new. To be a bit more accurate, there can also be initialization blocks that run when you say new (init blocks), but we’re going to cover them, and their static initialization counterparts, in the next chapter.
We already have a lot to discuss here. So let’s get started.

Constructor Basics

Every class, including abstract classes, MUST have a constructor. Hard Code that into your brain. But just because a class must have one, doesn’t mean the programmer has to type it. A constructor looks like this:

class Car {
Car() { } // The constructor for the Car class

You notice anything missing in the declaration above? There’s no return type! Two key points to remember about constructors are that they have no return type and their names must exactly match the class name. Typically, constructors are used to initialize instance variable state, as follows:

class Car {
int size;
String name;
Car(String name, int size) {
this.name = name;
this.size = size;

In the preceding code example, the Car class does not have a no-arg constructor. That means the following will fail to compile:
Car f = new Car(); // Won't compile, no matching constructor

but the following will compile:

Car f = new Car("Ford", 43); // No problem. Arguments match
// the Car constructor.

So it’s very common for a class to have a no-arg constructor, regardless of how many other overloaded constructors are in the class (constructors can be overloaded just like methods). You can’t always make that work for your classes; occasionally you have a class where it makes no sense to create an instance without supplying information to the constructor. A java.awt.Color object, for example, can’t be created by calling a no-arg constructor, because that would be like saying to the JVM, “Make me a new Color object, and I really don’t care what color it is....” Do you seriously want the JVM making your color choices?

Constructor Chaining

We know that constructors are invoked at runtime when you say new on some class type as follows:
Lamborghini h = new Lamborghini();

But what really happens when you say new Lamborghini() ? (Assume Lamborghini extends Car and Car extends Object.)

1. Lamborghini constructor is invoked. Every constructor invokes the constructor of its superclass with an (implicit) call to super(),
2. Car constructor is invoked (Car is the superclass of Lamborghini).
3. Object constructor is invoked (Object is the ultimate superclass of all classes, so class Car extends Object even though you don’t actually type “extends Object” into the Car class declaration. It’s implicit.) At this point we’re on the top of the hierarchy.
4. Object instance variables are given their explicit values. By explicit values, we mean values that are assigned at the time the variables are declared, like “int x = 27”, where “27” is the explicit value (as opposed to the default value) of the instance variable.
5. Object constructor completes.
6. Car instance variables are given their explicit values (if any).
7. Car constructor completes.
8. Lamborghini instance variables are given their explicit values (if any).
9. Lamborghini constructor completes.

Rules for Constructors

The following list summarizes the rules you’ll need to know about Constructors. There are very important and you MUST remember these to answer the questions related to Constructors in the SCJP Exam.

• Constructors can use any access modifier, including private. (We will deal with private constructors in one of the later chapters)
• The constructor name must match the name of the class. (100% Case Sensitive)
• Constructors must not have a return type.
• It’s legal (but probably stupid) to have a method with the same name as the class, but that doesn’t make it a constructor. If you see a return type, it’s a method and not a constructor. In fact, you could have both a method and a constructor with the same name—the name of the class—in the same class, and that’s not a problem for Java. Be careful not to mistake a method for a constructor—be sure to look for a return type.
• If you don’t type a constructor into your class code, a default constructor will be automatically generated by the compiler.
• The default constructor is ALWAYS a no-arg constructor. (Obviously the compiler has no clue what all arguments you might want for your class. So it takes the safe way out with a no argument constructor)
• If you want a no-arg constructor and you’ve typed any other constructor(s) into your class code, the compiler won’t provide the no-arg constructor (or any other constructor) for you. In other words, if you’ve typed in a constructor with arguments, you won’t have a no-arg constructor unless you type it in yourself!
• Every constructor has, as its first statement, either a call to an overloaded constructor “this()” or a call to the superclass constructor “super()”. (You cant do both in the same constructor)
• If you do type in a constructor, and you do not type in the call to super() or a call to this(), the compiler will insert a no-arg call to super() for you, as the very first statement in the constructor.
• A call to super() can be either a no-arg call or can include arguments passed to the super constructor. (The compiler will not insert the super with arguments call by itself. You have to do it manually)
• A no-arg constructor is not necessarily the default (i.e., compiler-supplied) constructor, although the default constructor is always a no-arg constructor. The default constructor is the one the compiler provides! While the default constructor is always a no-arg constructor, you’re free to put in your own no-arg constructor.
• You cannot make a call to an instance method, or access an instance variable, until after the super constructor runs.
• Only static variables and methods can be accessed as part of the call to super() or this(). (Example: super(Car.NAME) is OK, because NAME is declared as a static variable.)
• Abstract classes have constructors, and those constructors are always called when a concrete subclass is instantiated.
• Interfaces do not have constructors. Interfaces are not part of an object’s inheritance tree.
• The only way a constructor can be invoked is from within another constructor. In other words, you can’t write code that actually calls a constructor explicitly as follows:

class Lamborghini {
Lamborghini() { } // constructor
void doStuff() {
Lamborghini(); // calling the constructor - illegal!

Determine Whether a Default Constructor Will Be Created

The following example shows a Lamborghini class with two constructors:
class Lamborghini {
Lamborghini() { }
Lamborghini(String name) { }

Will the compiler put in a default constructor for the class above? “No!” What about for the following variation of the class?
class Lamborghini {
Lamborghini(String name) { }

Now will the compiler insert a default constructor? “No Again!” What about this class?
class Lamborghini { }

Now we’re talking. The compiler will generate a default constructor for the preceding class, because the class doesn’t have any constructors defined. OK, what about this one below?
class Lamborghini {
void Lamborghini() { }

You might be tempted to say that, the compiler won’t create one, since there already is a constructor in the Lamborghini class? Take another look at the Lamborghini class.
What’s wrong with the Lamborghini() constructor? It isn’t a constructor at all! It’s simply a method that happens to have the same name as the class. Remember, the return type is a dead straight way to tell us that we’re looking at a method, and not a constructor. So, here again the compiler will put the default no-arg constructor in the class.

Some Questions:

Q: How do you know for sure whether a default constructor will be created?
A: Because you didn’t write any constructors in your class.

Q: How do you know what the default constructor will look like?
A: Because...
• The default constructor has the same access modifier as the class.
• The default constructor has no arguments.
• The default constructor includes a no-arg call to the super constructor (super()).

What happens if the super constructor has arguments?

Constructors can have arguments just as methods can, and if you try to invoke a method that takes, say, an int, but you don’t pass anything to the method, the compiler will complain.
So if your super constructor (that is, the constructor of your immediate superclass/parent) has arguments, you must type in the call to super(), supplying the appropriate arguments. Crucial point: if your superclass does not have a no-arg constructor, you must type a constructor in your class (the subclass) because you need a place to put in the call to super with the appropriate arguments.
The following is an example of the problem:
class Car {
Car(String name) { }

class Lamborghini extends Car {
Lamborghini() {
super(); // Problem!

And once again the compiler treats us with the stunningly lucid:
Lamborghini.java:7: cannot resolve symbol
symbol : constructor Car ()
location: class Car
super(); // Problem!

If you’re lucky (and it’s a full moon), your compiler might be a little more explicit. But again, the problem is that there just isn’t a match for what we’re trying to invoke with super()—an Car constructor with no arguments.
Another way to put this is that if your superclass does not have a no-arg constructor, then in your subclass you will not be able to use the default constructor supplied by the compiler. It’s that simple. Because the compiler can only put in a call to a no-arg super(), you won’t even be able to compile something like this:

class Dress {
Dress(String s) { }
class Skirt extends Dress { }

Trying to compile this code gives us exactly the same error we got when we put a constructor in the subclass with a call to the no-arg version of super():

Dress.java:4: cannot resolve symbol
symbol : constructor Dress ()
location: class Dress
class Skirt extends Dress { }

In fact, the preceding Dress and Skirt code is implicitly the same as the following code, where we’ve supplied a constructor for Skirt that’s identical to the default constructor supplied by the compiler:

class Dress {
Dress(String s) { }
class Skirt extends Dress {
// Constructor identical to compiler-supplied
// default constructor
Skirt() {
super(); // Won't work!
} // Invokes a no-arg Dress() constructor,
} // but there isn't one!

One last point on the whole default constructor thing, constructors are never inherited. They aren’t methods. They can’t be overridden (because they aren’t methods and only instance methods can be overridden). So the type of constructor(s) your superclass has in no way determines the type of default constructor you’ll get. Some folks mistakenly believe that the default constructor somehow matches the super constructor, either by the arguments the default constructor will have (remember, the default constructor is always a no-arg), or by the arguments used in the compiler-supplied call to super().

So, although constructors can’t be overridden, you’ve already seen that they can be overloaded, and typically are.

Overloaded Constructors

Overloading a constructor means typing in multiple versions of the constructor, each having a different argument list, like the following examples:

class Car {
Car() { }
Car(String s) { }

The preceding Car class has two overloaded constructors, one that takes a string, and one with no arguments. Because there’s no code in the no-arg version, it’s actually identical to the default constructor the compiler supplies, but remember—since there’s already a constructor in this class (the one that takes a string), the compiler won’t supply a default constructor. If you want a no-arg constructor to overload the with-args version you already have, you’re going to have to type it yourself, just as in the Car example.

Overloading a constructor is typically used to provide alternate ways for clients to instantiate objects of your class. For example, if a client knows the Car name, they can pass that to a Car constructor that takes a string. But if they don’t know the name, the client can call the no-arg constructor and that constructor can supply a default name. Here’s what it looks like:

1. public class Car {
2. String name;
3. Car(String name) {
4. this.name = name;
5. }
7. Car() {
8. this(makeRandomName());
9. }
11. static String makeRandomName() {
12. int x = (int) (Math.random() * 5);
13. String name = new String[] {"Ferrari", "Lamborghini",
"Rover", "Spyker",
14. return name;
15. }
17. public static void main (String [] args) {
18. Car a = new Car();
19. System.out.println(a.name);
20. Car b = new Car("Proton");
21. System.out.println(b.name);
22. }
23. }

Running the code four times produces this output:

% java Car

% java Car

% java Car

% java Car

There’s a lot going on in the preceding code. Figure 2-7 shows the call stack for constructor invocations when a constructor is overloaded. Take a look at the call stack, and then let’s walk through the code straight from the top.

• Line 2 Declare a String instance variable name.
• Lines 3–5 Constructor that takes a String, and assigns it to instance variable name.
• Line 7. Assume every Car needs a name, but the client (calling code) might not always know what the name should be, so you’ll assign a random name. The no-arg constructor generates a name by invoking the makeRandomName() method.
• Line 8 The no-arg constructor invokes its own overloaded constructor that takes a String, in effect calling it the same way it would be called if client code were doing a new to instantiate an object, passing it a String for the name. The overloaded invocation uses the keyword this, but uses it as though it were a method name, this(). So line 8 is simply calling the constructor on line 3, passing it a randomly selected String rather than a client-code chosen name.
• Line 11 Notice that the makeRandomName() method is marked static! That’s because you cannot invoke an instance (in other words, nonstatic) method (or access an instance variable) until after the super constructor has run. And since the super constructor will be invoked from the constructor on line 3, rather than from the one on line 7, line 8 can use only a static method to generate the name. If we wanted all Cars not specifically named by the caller to have the same default name, say, “Ford,” then line 8 could have read this("Ford"); rather than calling a method that returns a string with the randomly chosen name.
• Line 12 This doesn’t have anything to do with constructors, but since we’re all here to learn...it generates a random integer between 0 and 4.
• Line 13 We’re creating a new String object (just a single String instance), but we want the string to be selected randomly from a list. Except we don’t have the list, so we need to make it. So in that one line of code we

1. Declare a String variable, name.
2. Create a String array (anonymously—we don’t assign the array itself to anything).
3. Retrieve the string at index [x] (x being the random number generated on line 12) of the newly created String array.
4. Assign the string retrieved from the array to the declared instance variable name. We could have made it much easier to read if we’d just written
5. String[] nameList = {"Ferrari", "Lamborghini", "Rover", "Spyker","Lotus"};
6. String name = nameList[x];
• Line 18 We’re invoking the no-arg version of the constructor (causing a random name from the list to be passed to the other constructor).
• Line 20 We’re invoking the overloaded constructor that takes a string representing the name.

The key point to get from this code example is in line 8. Rather than calling super(), we’re calling this(), and this() always means a call to another constructor in the same class. OK, fine, but what happens after the call to this()? Sooner or later the super() constructor gets called, right? Yes indeed. A call to this() just means you’re delaying the inevitable. Some constructor, somewhere, must make the call to super().

Key Rule: The first line in a constructor must be a call to super() or a call to this().
No exceptions. If you have neither of those calls in your constructor, the compiler will insert the no-arg call to super(). In other words, if constructor A() has a call to this(), the compiler knows that constructor A() will not be the one to invoke super().

The preceding rule means a constructor can never have both a call to super() and a call to this(). Because each of those calls must be the first statement in a constructor, you can’t legally use both in the same constructor. That also means the compiler will not put a call to super() in any constructor that has a call to this().

Thought question: What do you think will happen if you try to compile the following code?
class A {
A() {
A(String s) {

Your compiler may not actually catch the problem (it varies depending on your compiler, but most won’t catch the problem). It assumes you know what you’re doing. Can you spot the flaw? Given that a super constructor must always be called, where would the call to super() go? Remember, the compiler won’t put in a default constructor if you’ve already got one or more constructors in your class. And when the compiler doesn’t put in a default constructor, it still inserts a call to super() in any constructor that doesn’t explicitly have a call to the super constructor—unless, that is, the constructor already has a call to this(). So in the preceding code, where can super() go? The only two constructors in the class both have calls to this(), and in fact you’ll get exactly what you’d get if you typed the following method code:

public void go() {

public void doStuff() {

Now can you see the problem? Of course you can. The stack explodes! Two overloaded constructors both calling this() are two constructors calling each other. Over and over and over, resulting in

% java A
Exception in thread "main" java.lang.StackOverflowError

The benefit of having overloaded constructors is that you offer flexible ways to instantiate objects from your class. The benefit of having one constructor invoke another overloaded constructor is to avoid code duplication. In the Car example, there wasn’t any code other than setting the name, but imagine if after line 4 there was still more work to be done in the constructor. By putting all the other constructor work in just one constructor, and then having the other constructors invoke it, you don’t have to write and maintain multiple versions of that other important constructor code. Basically, each of the other not-the-real-one overloaded constructors will call another overloaded constructor, passing it whatever data it needs (data the client code didn’t supply).

Constructors and instantiation become even more exciting (just when you thought this chapter is going to end), when you get to inner classes, but I know you can stand to have only so much fun in one chapter, so I am holding the rest of the discussion on instantiating inner classes until a later point in time (A Future Chapter)

Previous Chapter: Chapter 11 - Legal Return Types

Next Chapter: Chapter 13 - Statics


  1. class Super1
    {int ab;

    Super1(int ab)
    class Sub extends Super1
    int ab;
    int d;
    int e;
    System.out.println("Sub() cons");
    Sub(int d,int e)
    Super1 h=new Super1();
    System.out.println("Super object"+h.ab);
    class Sub1
    public static void main(String[] arg)
    Sub obj=new Sub(5,6);
    }how it is worked//send me discription in my gmail sinhaankit409@gmail.com

    1. Anonymous,
      This looks like some class assignment to me. I would suggest you download Eclipse and use the debug option to step through the code one line at a time and understand the control flow.

      I can see that you are calling a constructor and instantiating another object inside a constructor which technically is not a good coding practice. Nonetheless, the code syntax looks fine and hence, I see no issues here.



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