Java Method Overloading Interview Programs for Practice

Here, we have listed a sample of top 8 Java method overloading interview programs for practice. All these questions are very important that can be asked in technical tests or interviews for freshers and 1 to 3 years of experience.

If you try to solve all these questions based on method overloading concept, your concept will be more clear and could be able to solve questions easily in the technical test.

Java Interview Questions on Method Overloading

Java method overloading interview programs


Question 1:

package methodOverloading; 
public class A 
{ 

 } 
public class B extends A 
{ 

 } 
public class C extends B 
{ 

 }
Consider the above program source code. What will be the output of the following scenarios?
Scenario 1:
public class OverLoadingScenarios 
{ 
void m1(A a) { 
  System.out.println(" I am in m1-A"); 
} 
void m1(B b) { 
  System.out.println("I am in m1-B"); 
} 
void m1(C c) { 
  System.out.println("I am in m1-C"); 
 } 
} 
public class OverLoadingTest { 
public static void main(String[] args) 
{ 
OverLoadingScenarios obj = new OverLoadingScenarios(); 
// Scene 1: 
   A a = new A(); 
   obj.m1(a); 
// Scene 2: 
   B b = new B(); 
   obj.m1(b); 
// Scene 3: 
   C c = new C(); 
   obj.m1(c); 
// Scene 4: 
   B bc = new C(); 
   obj.m1(bc); 
// Scene 5: 
   A ab = new B(); 
   obj.m1(ab); 
  } 
}
Output: 
Scene 1: 
       I am in m1-A 
Scene 2: 
       I am in m1-B 
Scene 3: 
       I am in m1-C 
Scene 4: 
       I am in m1-B 
Scene 5: 
       I am in m1-A

Explanation:

1. Scene 1: ‘a’ is the reference variable of class A  and pointing to object of class A. In resolving method overloading, during compilation, Java compiler does not check types of object to which particular reference variable is pointing. It just checks that which method is called through the object reference variable. It also checks method definition exists or not.

 

In scene 1, ‘a’ is the reference variable of class A. Therefore, the compiler will call m1() method of A because it is exactly matched. Similarly, for scene 2, and 3.

2. Scene 4: ‘bc’ is the reference variable of class B and it is pointing to object of class C but Java compiler will call m1() method of class B because bc is the reference variable of class B. Similarly for scene 5. 

 

Key points: 
In method overloading, method resolution takes place entirely at compile-time and It is always resolved by the compiler based on reference type only. In overloading, runtime object does not play any role.

 

Scenario 2:
public class OverLoadingScenarios2 { 
void m1(A a)
{ 
  System.out.println(" I am in m1-A"); 
} 
void m1(B b) { 
  System.out.println("I am in m1-B"); 
} 
void m1(C c) { 
  System.out.println("I am in m1-C"); 
} 
public static void main(String[] args) 
{ 
 OverLoadingScenarios2 obj = new OverLoadingScenarios2(); 
 obj.m1(null); 
 } 
}
Output: 
      I am in m1-C
Key points: 
A is the parent class B, and B is the parent class of C. So, C is the child class of B. A call to m1() method with null argument executes the third version of m1() method with argument c of type C because while resolving overloaded method, the compiler always uses the presidency for child type argument. In this kind of case, you always see the parent-child relationship.
For practice, more questions, go to this link: Automatic type promotion in method overloading Java

Note: A field of an object type can have a null value in Java. The default value for string and an object type argument is null.
Scenario 3:
public class OverLoadingScenarios3 
{ 
void m1(A a) { 
  System.out.println(" I am in m1-A"); 
} 
void m1(B b) { 
  System.out.println("I am in m1-B"); 
} 
void m1(Object o) { 
  System.out.println("I am in m1-C"); 
} 
public static void main(String[] args) 
{ 
OverLoadingScenarios3 obj = new OverLoadingScenarios3(); 
 obj.m1(null); 
 } 
}
Output: 
      I am in m1-B
Key points: 
An object is the superclass of A and B where A is the superclass of B. Thus, B is the child class. Therefore, the output is “I am in m1 B”.
Scenario 4:
public class OverLoadingScenarios4 
{ 
void m1(A a) { 
  System.out.println(" I am in m1-A"); 
} 
void m1(String s) { 
  System.out.println("I am in m1-B"); 
} 
void m1(Object o) { 
  System.out.println("I am in m1-C"); 
} 
public static void main(String[] args) 
{ 
OverLoadingScenarios2 obj = new OverLoadingScenarios2(); 
 obj.m1(null); 
 } 
}
Output: 
       Unresolved compilation problem: The method m1(A) is ambiguous for the type OverLoadingScenarios2

Key points: 
In this scenario, Object class is superclass of string class and class A at the same level. There is also no parent-child relationship between string class and class A.

When we will pass null argument for calling m1() method, we have created an ambiguous situation for the compiler that cannot determine which method to call because all three methods are exact matches for our call. 

Therefore, the compiler complains of an error that the call is ambiguous and it will generate compile-time error. 

Note: An overloaded method cannot be ambiguous its own. It only becomes ambiguous if you create an ambiguous situation.

Question 2. What will be the output of the following program?

package methodOverloading; 
public class XYZ 
{ 
void msg(Object obj) { 
  System.out.println("Good"); 
} 
void msg(String str) { 
  System.out.println("Better"); 
} 
void msg(Integer itr) { 
  System.out.println("Best"); 
} 
public static void main(String[] args) 
{ 
XYZ obj = new XYZ(); 
 obj.msg(new Object()); // Exact matched to call m1() method with Object type argument. 
 obj.msg("Scientech Easy"); // Exact matched to call m1() method with String type argument. 
 obj.msg(new XYZ()); // Exact matched to call m1() method with Object type argument. 
 obj.msg(new String()); 
 obj.msg(10); // Exact matched to call m1() with Integer type argument. 
 obj.msg(new Integer(0)); 
 } 
}
Output: 
       Good 
       Better 
       Good 
       Better 
       Best 
       Best

Question 3: What will be the output of the below program?

package methodOverloading; 
public class XYZ 
{ 
void msg(Object obj) { 
 System.out.println("Good"); 
} 
void msg(String str) { 
 System.out.println("Better"); 
} 
void msg(Integer itr) { 
 System.out.println("Best"); 
} 
public static void main(String[] args) 
{ 
 XYZ obj = new XYZ(); 
 obj.msg(null); // Created an ambiguous situation for the compiler to call 
 } 
}
Output: 
      Compile time error: The method msg(Object) is ambiguous for the type XYZ

Question 4: What is the output of the below program?

package methodOverloading; 
public class Overloaded 
{ 
public static void m1(int a) { 
 System.out.println("int"); 
} 
public static void m1(short a) { 
 System.out.println("short"); 
} 
public static void m1(Object a) { 
 System.out.println("object"); 
} 
public static void m1(String a) 
{
 System.out.println("String"); 
}
public static void main(String[] args) 
{ 
 byte b = 5; 
  m1(b); // First call 
  m1(5); // Second call Integer 
 i = 10; 
  m1(i); // Third call 
 m1("10"); // Fourth call 
 m1(null); // Fifth call 
  } 
}
Output: 
       short 
       int 
       object 
       String 
       String

Key points: 
1. When we will call m1() method with byte argument, there is no m1() definition that will take byte as an argument. Since the smaller data type in size is short than int data type which is larger size as a comparison to short. Therefore, the compiler will promote byte to short. The result will be short.

2. In the second method call, an exact match is found to call m1() with int argument.

3. The third method call is m1(i) where i is a type of Integer. There is no such m1() method that takes an integer type argument.

Therefore, the compiler will implicitly upcast from Integer to Object type argument and it will call to m1() method with Object type argument. 

Remember that the compiler allows implicit upcast, not downcast. Therefore, m1(int a) will not be considered.

4. In the fourth method call, an exact match is found to call m1() with String argument. 

5. The last method call is m1(null). Since null is valid for both object and string. So, which method will be called? m1(String a) will be called because string is child class of object class.

Question 5: What is the output of the below program source code?

package methodOverloading; 
public class Overloaded 
{ 
public static void msg(long a, int b) { 
  System.out.println("Hello"); 
} 
public static void msg(int a, long b) { 
  System.out.println("Hi"); 
} 
public static void main(String[] args) { 
  msg(5l, 10); 
  msg(10,11); 
 } 
}
Output: 
       Hello 
       Unresolved compilation problem: The method msg(long, int) is ambiguous for the type Overloaded

Question 6: What will be the outcome of the below program?

package methodOverloading; 
public class Overloaded 
{ 
public static void test(int[] intArr) { 
 System.out.println("int array"); 
} 
public static void test(char[] charArr) { 
 System.out.println("char array"); 
} 
public static void main(String[] args) 
{ 
 test(null); 
 } 
}
Output: 
      Unresolved compilation problem: The method test(int[]) is ambiguous for the type Overloaded

Key points: 
int[ ] and char[ ] both are not primitive type in Java. Both are classes that extend object class at the same level. Therefore, the compiler is unable to resolve overloaded method to call.

Question 7: What will be the outcome of the below program?

package methodOverloading; 
public class Overloaded 
{ 
public void test(int i) { 
 System.out.println("int"); 
} 
public void test(Number n) { 
 System.out.println("Number"); 
} 
public void test(Integer i) { 
 System.out.println("Integer"); 
} 
public static void main(String[] args) 
{ 
 Overloaded o = new Overloaded(); 
 o.test(null); 
 o.test(10); // Exact matched. 
 } 
}
Output: 
       Integer 
       int

Key points: 
Object class is the superclass of Number whereas Number is parent class of Integer. Thus, Integer is a child class. Therefore, the result is an Integer.

Question 8: What will be the output of the following program?

package methodOverloading; 
public class Overloaded 
{ 
public void test(int i) { 
 System.out.println("Int"); 
} 
public void test(int... i) { 
 System.out.println("Int"); 
} 
public void test(char... c) { 
 System.out.println("Char varargs"); 
} 
public static void main(String[] args) 
{ 
 Overloaded obj = new Overloaded(); 
 obj.test('a'); 
 obj.test(10); // Exact matched. 
 } 
}
Output: 
       Int 
       Int

Key points: 
Methods with varargs (…) have the lowest priority. That’s why output is Int.

Recommended Post: 

1. Top 15 Java Method Overriding Interview Programs for Practice

Final words 
Hope that this tutorial has covered almost all variety of Java method overloading interview programs that you will have practiced and enjoyed them.
Thanks for reading!!!



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