Java Method Overloading: Definition and Examples

Rumman Ansari   Software Engineer   2024-07-04 03:36:52   10779  Share
Subject Syllabus DetailsSubject Details 5 Questions 1 Program
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Table of Content:

What is Method Overloading

In Java it is possible to define two or more methods within the same class that share the same name, as long as their parameter declarations are different. When this is the case, the methods are said to be overloaded, and the process is referred to as method overloading.

If a class has multiple methods having same name but different in parameters, it is known as Method Overloading.


Method overloading is one of the ways that Java supports polymorphism.

Problem without method overloading

Suppose you have to perform addition of the given numbers but there can be any number of arguments, if you write the method such as addition(int,int) for two parameters, and additionMethod(int,int,int) for three parameters then it may be difficult for you as well as other programmers to understand the behavior of the method because its name differs.
So, we perform method overloading to figure out the program quickly.

Advantage of method overloading

  1. The main advantage of this is cleanliness of code.
  2. Method overloading increases the readability of the program.
  3. Flexibility
  4. Overloaded methods give programmers the flexibility to call a similar method for different types of data. If you are working on a mathematics program, for example, you could use overloading to create several multiply classes, each of which multiplies a different number of type of argument: the simplest multiply(int a, int b) multiplies two integers; the more complicated method multiply(double a, int b, int c) multiplies one double by two integers -- you could then call "multiply" on any combination of variables that you created an overloaded method for and receive the proper result.

Different ways to overload the method

There are two ways to overload the method in java

  1. By changing number of arguments
  2. By changing the data type

Method Overloading: changing no. of arguments

In this example, we have created two methods, first addition() method performs addition of two numbers and second addition() method performs addition of three numbers.
In this example, we are creating static methods so that we don't need to create instance for calling methods. 

class OverloadingExample{
	public static void main(String[] args){
		System.out.println(AdderClass.addition(12,11));
		System.out.println(AdderClass.addition(11,15,13));
	}
}

class AdderClass{
	static int addition(int a,int b){return a+b;}
	static int addition(int a,int b,int c){return a+b+c;}
}

Output

23
39
Press any key to continue . . .

Method Overloading: changing data type of arguments

In this example, we have created two methods that differs in data type. The first add method addition(int a, int a) first receives two integer arguments and second add method addition(double a, double b)first receives two double arguments, then it calculate.
In this example, we are creating static methods so that we don't need to create instance for calling methods.

class OverloadingExample{
	public static void main(String[] args){
		System.out.println(AdderClass.addition(11,12));
		System.out.println(AdderClass.addition(11.3,15.6));
	}
}

class AdderClass{
 static int addition(int a, int b){
	return a+b;
 }

 static double addition(double a, double b){
	return a+b;
 }
}

Output

23
26.9
Press any key to continue . . .

Method Overloading: changing data type of arguments

As you can see, method( ) is overloaded four times. The first version takes no parameters, the second takes one integer parameter, the third takes two integer parameters, and the fourth takes one double parameter. The fact that the fourth version of method( ) also returns a value is of no consequence relative to overloading, since return types do not play a role in overload resolution. 

// Demonstrate method overloading.

class OverloadDemo {
	void method() {
	System.out.println("No parameters");
}

// Overload test for one integer parameter.
void method(int a) {
	System.out.println("a: " + a);
	}
// Overload test for two integer parameters.
void method(int a, int b) {
	System.out.println("a and b: " + a + " " + b);
	}
// overload test for a double parameter
double method(double a) {
	System.out.println("double a: " + a);
	return a*a;
	}
}

class OverloadExample {
	public static void main(String args[]) {
		OverloadDemo ob = new OverloadDemo();
		double result;
		// call all versions of test()
		ob.method();
		ob.method(10);
		ob.method(10, 20);
		result = ob.method(123.25);
		System.out.println("Result of ob.test(123.25): " + result);
	}
}

Output

No parameters
a: 10
a and b: 10 20
double a: 123.25
Result of ob.test(123.25): 15190.5625
Press any key to continue . . .

Method Overloading: changing data type of arguments. Example find maximum value from two number

Let’s consider the example discussed earlier chapter for finding maximum numbers of integer type. If, let’s say we want to find the maximum number of double type. Then the concept of overloading will be introduced to create two or more methods with the same name but different parameters.
The following example explains the same ? 

public class ExampleOverloading {

   public static void main(String[] args) {
      int a = 12;
      int b = 6;
      double c = 10.3;
      double d = 5.4;
      int result1 = maxFunction(a, b);

      // same function name with different parameters
      double result2 = maxFunction(c, d);
      System.out.println("Maximum Value = " + result1);
      System.out.println("Maximum Value = " + result2);
   }

   // for integer
   public static int maxFunction(int n1, int n2) {
      int max;
      if (n1 > n2)
         max = n1;
      else
         max = n2;

      return max;
   }

   // for double
   public static double maxFunction(double n1, double n2) {
     double max;
      if (n1 > n2)
         max = n1;
      else
         max = n2;

      return max;
   }
}

Output

Maximum Value = 12
Maximum Value = 10.3
Press any key to continue . . .

Java’s automatic type conversions can play a role in overloading Method

When an overloaded method is called, Java looks for a match between the arguments used to call the method and the method’s parameters. However, this match need not always be exact. 

// Automatic type conversions apply to overloading.
class OverloadExample {

	void testMethod() {
		System.out.println("No parameters");
	}

// Overload test for two integer parameters.
	void testMethod(int a, int b) {
		System.out.println("a and b: " + a + " " + b);
	}
// overload test for a double parameter
	void testMethod(double a) {
		System.out.println("Inside test(double) a: " + a);
	}
}

class Overload {
	public static void main(String args[]) {
		OverloadExample obj = new OverloadExample();
		int p = 57;
		obj.testMethod();
		obj.testMethod(5, 10);
		obj.testMethod(p); // this will invoke test(double)
		obj.testMethod(158.6); // this will invoke test(double)
	}
}

Output

No parameters
a and b: 5 10
Inside test(double) a: 57.0
Inside test(double) a: 158.6
Press any key to continue . . .

As you can see, this version of OverloadExample does not define testMethod(int). Therefore, when testMethod( ) is called with an integer argument inside Overload, no matching method is found. However, Java can automatically convert an integer into a double, and this conversion can be used to resolve the call. Therefore, after testMethod(int) is not found, Java elevates p to double and then calls testMethod(double). Of course, if testMethod(int) had been defined, it would have been called instead. Java will employ its automatic type conversions only if no exact match is found.

Method Overloading with Type Promotion if matching found

If there are matching type arguments in the method, type promotion is not performed. 

class OverloadingAutoCast{

  	  void division(int a,int b){

	  	System.out.println("int arg method invoked: Ans: "+(a/b));
	  }

  	 void division(long a,long b){
	 	 System.out.println("long arg method invoked: Ans: "+(a/b));
	  }

  public static void main(String args[]){
  OverloadingAutoCast obj=new OverloadingAutoCast();
  obj.division(20,5);//now int arg sum() method gets invoked
  }
}

Output: 

int arg method invoked: Ans: 4
Press any key to continue . . .

Method Overloading with Type Promotion in case of ambiguity

If there are no matching type arguments in the method, and each method promotes similar number of arguments, there will be ambiguity. 

class OverloadingAutoCast{

  	  void division(long a,int b){

	  	System.out.println("int arg method invoked: Ans: "+(a/b));
	  }

  	 void division(int a,long b){
	 	 System.out.println("long arg method invoked: Ans: "+(a/b));
	  }

  public static void main(String args[]){
  OverloadingAutoCast obj=new OverloadingAutoCast();
  obj.division(20,5);//now int arg sum() method gets invoked
  }
}

Output: 

OverloadingAutoCast.java:14: error: reference to division is ambiguous
  obj.division(20,5);//now int arg sum() method gets invoked
     ^
  both method division(long,int) in OverloadingAutoCast and method division(int,long) in OverloadingAutoCast match
1 error

Why Method Overloading is not possible by changing the return type of method only?

Method overloading is not possible by changing the return type of the method only because of ambiguity. 

class AdderClass{

	static int addme(int a,int b){
	return a+b;
	}

	static double addme(int a,int b){
	return a+b;
	}
}

class OverloadingSampleTest{
	public static void main(String[] args){
	System.out.println(AdderClass.addme(15,21));//ambiguity
	}
}

Output: 

OverloadingSampleTest.java:7: error: method addme(int,int) is already defined in class AdderClass
	static double addme(int a,int b){
	              ^
1 error

Can we overload java main() method?

Yes, by method overloading. You can have any number of main methods in a class by method overloading. But JVM calls main() method which receives string array as arguments only. Let's see the simple example: 

class MainMethodOverloading{

	public static void main(){
			System.out.println("main without args");
	}
	public static void main(String[] args){
		System.out.println("main with String[]");
	}

	public static void main(String args){
		System.out.println("main with String");
	}


}

Output: 

main with String[]
Press any key to continue . . .