visual basic barcode printing 5: Object Orientation, Overloading and Overriding, Constructors, and Return Types in Java

Paint PDF417 in Java 5: Object Orientation, Overloading and Overriding, Constructors, and Return Types

5: Object Orientation, Overloading and Overriding, Constructors, and Return Types
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Compiler-Generated Constructor Code (continued)
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Class Code (What You Type) class Foo { Foo() { } }
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Compiler-Generated Constructor Code (In Bold Type)
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class Foo { Foo() { super(); } } class Foo { public Foo() { super(); } } class Foo { Foo(String s) { super(); } }
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public class Foo {}
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class Foo { Foo(String s) { super(); } } class Foo { void Foo() {} }
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class Foo { void Foo() {} Foo() { super(); } } (void Foo() is a method, not a constructor)
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 as follows:
class Bar { void takeInt(int x) { } } class UseBar {
Constructors and Instantiation (Exam Objectives 1.3, 6.3, 6.2)
public static void main (String [] args) { Bar b = new Bar(); b.takeInt(); // Try to invoke a no-arg takeInt() method } }
The compiler will complain that you can t invoke takeInt() without passing an int. Of course, the compiler enjoys the occasional riddle, so the message it spits out on some versions of the JVM (your mileage may vary) is less than obvious:
UseBar.java:7: takeInt(int) in Bar cannot be applied to () b.takeInt(); ^
But you get the idea. The bottom line is that there must be a match for the method. And by match, we mean that the argument types must be able to accept the values or variables you re passing, and in the order you re passing them. Which brings us back to constructors (and here you were thinking we d never get there), which work exactly the same way. 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 Animal { Animal(String name) { } } class Horse extends Animal { Horse() { super(); // Problem! } }
And once again the compiler treats us with the stunningly lucid:
Horse.java:7: cannot resolve symbol symbol : constructor Animal () location: class Animal super(); // Problem! ^
5: Object Orientation, Overloading and Overriding, Constructors, and Return Types
If you re lucky (and eat all your vegetables, including broccoli), 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 Animal constructor with no arguments. Another way to put this and you can bet your favorite Grateful Dead t-shirt it ll be on the exam is 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 the following:
class Clothing { Clothing(String s) { } } class TShirt extends Clothing { }
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():
Clothing.java:4: cannot resolve symbol symbol : constructor Clothing () location: class Clothing class TShirt extends Clothing { } ^
In fact, the preceding Clothing and TShirt code is implicitly the same as the following code, where we ve supplied a constructor for TShirt that s identical to the default constructor supplied by the compiler:
class Clothing { Clothing(String s) { } } class TShirt extends Clothing { // Constructor identical to compiler-supplied default constructor TShirt() { super(); } }
One last point on the whole default constructor thing (and it s probably very obvious, but we have to say it or we ll feel guilty for years), constructors are never inherited. They aren t methods. They can t be overridden (because they aren t methods and only 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
Constructors and Instantiation (Exam Objectives 1.3, 6.3, 6.2)
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 compilersupplied call to super(). But although constructors can t be overridden, you ve already seen that they can be overloaded, and typically are.
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