Explain Method Overriding In Java With Program And Examples With Explanation, What Is Method Overriding Explain
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Method Overriding
In a class hierarchy, when a method in a subclass has the same name and type signature as
a method in its superclass, then the method in the subclass is said to override the method in
the superclass. When an overridden method is called from within a subclass, it will always
refer to the version of that method defined by the subclass. The version of the method defined
by the superclass will be hidden. Consider the following:
// Method overriding.
class A {
int i, j;
A(int a, int b) {
i = a;
j = b;
}
// display i and j
void show() {
System.out.println("i and j: " + i + " " + j);
}
}
class B extends A {
int k;
B(int a, int b, int c) {
super(a, b);
k = c;
}
// display k – this overrides show() in A
void show() {
System.out.println("k: " + k);
}
}
class Override {
public static void main(String args[]) {
B subOb = new B(1, 2, 3);
subOb.show(); // this calls show() in B
}
}
The output produced by this program is shown here:
k: 3
When
show( )
is invoked on an object of type
B
, the version of
show( )
defined within
B
is used. That is, the version of
show( )
inside
B
overrides the version declared in
A
.
If you wish to access the superclass version of an overridden method, you can do so by
using
super
. For example, in this version of
B
, the superclass version of
show( )
is invoked
within the subclass’ version. This allows all instance variables to be displayed.
class B extends A {
int k;
B(int a, int b, int c) {
super(a, b);
k = c;
}
void show() {
super.show(); // this calls A's show()
System.out.println("k: " + k);
}
}
If you substitute this version of
A
into the previous program, you will see the following
output:
i and j: 1 2
k: 3
Here,
super.show( )
calls the superclass version of
show( )
.
172
Part I:
The Java Language
Method overriding occurs only when the names and the type signatures of the two
methods are identical. If they are not, then the two methods are simply overloaded. For
example, consider this modified version of the preceding example:
// Methods with differing type signatures are overloaded – not
// overridden.
class A {
int i, j;
A(int a, int b) {
i = a;
j = b;
}
// display i and j
void show() {
System.out.println("i and j: " + i + " " + j);
}
}
// Create a subclass by extending class A.
class B extends A {
int k;
B(int a, int b, int c) {
super(a, b);
k = c;
}
// overload show()
void show(String msg) {
System.out.println(msg + k);
}
}
class Override {
public static void main(String args[]) {
B subOb = new B(1, 2, 3);
subOb.show("This is k: "); // this calls show() in B
subOb.show(); // this calls show() in A
}
}
The output produced by this program is shown here:
This is k: 3
i and j: 1 2
The version of
show( )
in
B
takes a string parameter. This makes its type signature
different from the one in
A
, which takes no parameters. Therefore, no overriding (or name
hiding) takes place. Instead, the version of
show( )
in
B
simply overloads the version of
show( )
in
A
.
Chapter 8:
Inheritance
173