method overloading in JAVA
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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