In same class, if name of the method remains common but the number and type of parameters are different, then it is called method overloading in Java.
let's see a example :
public class Overload {
void add(int m, int n)
{
int sum = m + n;
System.out.println( "Sum of a+b is " +sum);
}
void add(int a, int b, int c) {
int sum = a + b + c;
System.out.println("Sum of a+b+c is " +sum);
}
void add(double a, double b) {
double sum = a + b;
System.out.println("Sum of a+b is "+sum);
}
void add(String s1, String s2)
{
String s = s1+s2;
System.out.println(s);
}
}
class overloadfunc{
public static void main(String args[])
{
Overload2 obj = new Overload2();
obj.add(4,19);
obj.add(4,17,11);
obj.add(1.5,21.5);
obj.add("Life at"," the speed of rail ");
}
}
Output will be:
Sum of a+b is 23
Sum of a+b+c is 32
Sum of a+b is 23.0
Life at the speed of rail
JAVA TUTORIALS HOMEPAGE
java method overloading :: futureX |
public class Overload {
void add(int m, int n)
{
int sum = m + n;
System.out.println( "Sum of a+b is " +sum);
}
void add(int a, int b, int c) {
int sum = a + b + c;
System.out.println("Sum of a+b+c is " +sum);
}
void add(double a, double b) {
double sum = a + b;
System.out.println("Sum of a+b is "+sum);
}
void add(String s1, String s2)
{
String s = s1+s2;
System.out.println(s);
}
}
class overloadfunc{
public static void main(String args[])
{
Overload2 obj = new Overload2();
obj.add(4,19);
obj.add(4,17,11);
obj.add(1.5,21.5);
obj.add("Life at"," the speed of rail ");
}
}
Output will be:
Sum of a+b is 23
Sum of a+b+c is 32
Sum of a+b is 23.0
Life at the speed of rail
JAVA TUTORIALS HOMEPAGE
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