Output of Java Programs | Set 24 (Final Modifier)

Difficulty level: Easy

Prerequisite: final keyword in Java

Predict the output of the following Java Programs.

1. What will be the output of the following program?

class Test  
{ 
    final int MAXIMUM; 
    final double PI; 
  
public Test(int max) 
    { 
        MAXIMUM = max; 
    } 
  
public Test(double pi) 
    { 
        PI = pi; 
    } 
  
public static void main(String[] args) 
    { 
        Test t1 = new Test(1500); 
        Test t2 = new Test(3.145); 
  
        System.out.println("MAXIMUM : " + t1.MAXIMUM + " PI : " + t2.PI); 
    } 
} 

Options:

a) Compilation error
b) Runtime error
c) 1500 3.145
d) 3.145 1500

Answer: a) Compilation error

Explanation : As we know that we can initialize a blank final variable inside constructor also, but if there are more than one constructors then it is mandatory to initialize all final variables in all of them. This is because we can create an object of class by calling any one of the constructors, but if that constructor is not initializing any one of declared final variable than there is problem.

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

class Test  
{ 
    final int MAXIMUM = m1(); 
  
private int m1() 
    { 
        System.out.println(MAXIMUM); 
        return 1500; 
    } 
  
public static void main(String[] args) 
    { 
        Test t = new Test(); 
  
        System.out.println(t.MAXIMUM); 
    } 
} 

Options:

a) Compilation error
b) Runtime error
c) 0
1500
d) 1500
1500

Answer: c) 0
1500

Explanation : A final variable can only be initialized once, either via an initializer or an assignment statement. Also you might think that in above program, MAXIMUM is initialized two times.This is wrong. The output is based on the fact that JVM first initialize any(final or normal) variable with it’s default-type value and then look through assignment statement (if any).

3. What will be the output of the following program?

// filename : Test1.java 
final interface Test 
{ 
    int MAXIMUM = 1500; 
  
    void m1(); 
} 
  
class Test1 implements Test { 
  
    @Override public void m1() 
    { 
        System.out.println("From Test1 m1 method"); 
    } 
  
public static void main(String[] args) 
    { 
        new Test1().m1(); 
    } 
} 

Options:

a) Compilation error
b) Runtime error
c) From Test1 m1 method

Answer: a) Compilation error

Explanation : Interfaces can never be declared final as they are meant to be implemented in derived classes. Please see interface and inheritance.

4. What will be the output of the following program?

class Test { 
public static void main(String[] args) 
    { 
        int arr[] = { 1, 2, 3 }; 
  
        // final with for-each statement 
        for (final int i : arr) 
            System.out.print(i + " "); 
    } 
} 

Options:

a) Compilation error
b) Runtime error
c) 1 2 3

Answer: c) 1 2 3

Explanation : Since the variable i goes out of scope with each iteration of the loop, it is actually getting re-declared after each iteration, allowing the same token (i.e. i) to be used to represent multiple variables.

5. What will be the output of the following program?

class Test { 
public static void main(String[] args) 
    { 
        final StringBuilder sb = new StringBuilder("Geeks"); 
  
        sb.append("ForGeeks"); 
  
        System.out.println(sb); 
    } 
} 

Options:

a) Compilation error
b) Runtime error
c) Geeks
d) GeeksForGeeks

Answer: d) GeeksForGeeks

Explanation : In case of a reference final variable(here sb), internal state of the object pointed by that reference variable can be changed. Note that this is not re-assigning. This property of final is called non-transitivity.