Language Fundamentals - Finding the output Questions and Answers

Q: What is the output of the following Java code ?JavaCopy codeclass Animal { void makeSound() { System.out.println(“Animal makes a sound”); }}class Dog extends Animal { void makeSound() { System.out.println(“Dog barks”); }}public class Test { public static void main(String[ ] args) { Animal obj = new Dog(); obj.makeSound(); }}

The output is Dog barks.The reference obj has the type Animal but points to a Dog object created with new Dog(). Dog overrides makeSound(), and Java chooses an overridden instance method by the object's actual class at run time (dynamic dispatch), so obj.makeSound() runs the Dog version and prints Dog barks.'Animal makes a sound' would be right only if Java used the reference type, which is the rule for static or private methods and for fields, not for overridden instance methods. There is no compilation error: Dog is a subclass, so assigning it to an Animal reference is legal, and the call is valid because Animal declares makeSound(). There is no runtime error either, since the object is not null and the method exists.

Q: What will be the output of the program? and the command-line invocation is public class CommandArgsThree { public static void main(String [] args) { String [][] argCopy = new String[2][2]; int x; argCopy[0] = args; x = argCopy[0].length; for (int y = 0; y < x; y++) { System.out.print(" " + argCopy[0][y]); } } } > java CommandArgsThree 1 2 3

Correct answer: 1 2 3Assigning args to argCopy[0] replaces the two-element array that was there with a reference to the command line array itself - array references are assigned, not copied element by element, and the declared inner length of 2 is not a constraint on what may later be stored there. So x becomes 3 and the loop prints all three arguments. The new String[2][2] is a red herring, and the array it created for row 0 is simply discarded.

Q: What will be the output of the program? public class CommandArgs { public static void main(String [] args) { String s1 = args[1]; String s2 = args[2]; String s3 = args[3]; String s4 = args[4]; System.out.print(" args[2] = " + s2); } } and the command-line invocation is > java CommandArgs 1 2 3 4

Correct answer: An exception is thrown at runtime.Four arguments give args indices 0 to 3, so args[4] is out of bounds and throws ArrayIndexOutOfBoundsException. The throw happens on the fourth assignment, before any output is produced, so nothing is printed at all - not even the value of args[2], which had been read successfully a moment earlier.

Q: What will be the output of the program ? public class Test { public static void main(String [] args) { signed int x = 10; for (int y=0; y<5; y++, x--) { System.out.print(x + ", "); } } }

Correct answer: Compilation fails.Java has no signed keyword. Its integral types are signed by definition - there is no unsigned counterpart to declare against - so the word is not part of the language and signed int x = 10; is rejected. It is a C and C++ construct, and the same applies to unsigned. Nothing runs, so neither of the printed sequences arises.

Q: What will be the output of the program? public class X { public static void main(String [] args) { String names [] = new String[5]; for (int x=0; x < args.length; x++) names[x] = args[x]; System.out.println(names[2]); } } and the command line invocation is > java X a b

The names array is initialized with five null elements. Then elements 0 and 1 are assigned the String values "a" and "b" respectively (the command-line arguments passed to main). Elements of names array 2, 3, and 4 remain unassigned, so they have a value of null.

Q: What will be the output of the program? public class TestDogs { public static void main(String [] args) { Dog [][] theDogs = new Dog[3][]; System.out.println(theDogs[2][0].toString()); } } class Dog { }

The second dimension of the array referenced by theDogs has not been initialized. Attempting to access an uninitialized object element (System.out.println(theDogs[2][0].toString());) raises a NullPointerException.

Q: What will be the output of the program? public class CommandArgsTwo { public static void main(String [] argh) { int x; x = argh.length; for (int y = 1; y <= x; y++) { System.out.print(" " + argh[y]); } } } and the command-line invocation is > java CommandArgsTwo 1 2 3

An exception is thrown because at some point in (System.out.print(" " + argh[y]);), the value of x will be equal to y, resulting in an attempt to access an index out of bounds for the array. Remember that you can access only as far as length - 1, so loop logical tests should use x < someArray.length as opposed to x <= someArray.length.

What is the output of the following Java code ?

Java
Copy code
class Animal {
    void makeSound() {
        System.out.println(“Animal makes a sound”);
    }
}
class Dog extends Animal {
    void makeSound() {
        System.out.println(“Dog barks”);
    }
}
public class Test {
    public static void main(String[ ] args) {
        Animal obj = new Dog();
        obj.makeSound();
    }
}

A.Animal makes a sound
B.Dog barks
C.Compilation error
D.Runtime error

Exam Year(s)

Tripura Public Service Commission - 2025,

Answer: B

The output is Dog barks.

The reference obj has the type Animal but points to a Dog object created with new Dog(). Dog overrides makeSound(), and Java chooses an overridden instance method by the object's actual class at run time (dynamic dispatch), so obj.makeSound() runs the Dog version and prints Dog barks.

'Animal makes a sound' would be right only if Java used the reference type, which is the rule for static or private methods and for fields, not for overridden instance methods. There is no compilation error: Dog is a subclass, so assigning it to an Animal reference is legal, and the call is valid because Animal declares makeSound(). There is no runtime error either, since the object is not null and the method exists.

Discuss in Forum

What will be the output of the program?

and the command-line invocation is

public class CommandArgsThree 
{
    public static void main(String [] args) 
    {
        String [][] argCopy = new String[2][2];
        int x;
        argCopy[0] = args;
        x = argCopy[0].length;
        for (int y = 0; y < x; y++) 
        {
            System.out.print(" " + argCopy[0][y]);
        }
    }
}

> java CommandArgsThree 1 2 3

A.0 0
B.1 2
C.0 0 0
D.1 2 3

Answer: D

Correct answer: 1 2 3

Assigning args to argCopy[0] replaces the two-element array that was there with a reference to the command line array itself - array references are assigned, not copied element by element, and the declared inner length of 2 is not a constraint on what may later be stored there. So x becomes 3 and the loop prints all three arguments. The new String[2][2] is a red herring, and the array it created for row 0 is simply discarded.

Discuss in Forum

What will be the output of the program?

public class CommandArgs 
{
    public static void main(String [] args) 
    {
        String s1 = args[1];
        String s2 = args[2];
        String s3 = args[3];
        String s4 = args[4];
        System.out.print(" args[2] = " + s2);
    }
}

and the command-line invocation is

> java CommandArgs 1 2 3 4

A.args[2] = 2
B.args[2] = 3
C.args[2] = null
D.An exception is thrown at runtime.

Answer: D

Correct answer: An exception is thrown at runtime.

Four arguments give args indices 0 to 3, so args[4] is out of bounds and throws ArrayIndexOutOfBoundsException. The throw happens on the fourth assignment, before any output is produced, so nothing is printed at all - not even the value of args[2], which had been read successfully a moment earlier.

Discuss in Forum

What will be the output of the program ?

public class Test 
{
    public static void main(String [] args) 
    {
        signed int x = 10;
        for (int y=0; y<5; y++, x--) 
        {
            System.out.print(x + ", ");
        }
    }
}

A.10, 9, 8, 7, 6,
B.9, 8, 7, 6, 5,
C.Compilation fails.
D.An exception is thrown at runtime.

Answer: C

Correct answer: Compilation fails.

Java has no signed keyword. Its integral types are signed by definition - there is no unsigned counterpart to declare against - so the word is not part of the language and signed int x = 10; is rejected. It is a C and C++ construct, and the same applies to unsigned. Nothing runs, so neither of the printed sequences arises.

Discuss in Forum

What will be the output of the program?

public class X 
{
    public static void main(String [] args) 
    {
        String names [] = new String[5];
        for (int x=0; x < args.length; x++)
            names[x] = args[x];
        System.out.println(names[2]);
    }
}

and the command line invocation is

> java X a b

A.names
B.null
C.Compilation fails
D.An exception is thrown at runtime

Answer: B

The names array is initialized with five null elements. Then elements 0 and 1 are assigned the String values "a" and "b" respectively (the command-line arguments passed to main). Elements of names array 2, 3, and 4 remain unassigned, so they have a value of null.

Discuss in Forum

In the given program, how many lines of output will be produced?

public class Test 
{
    public static void main(String [] args) 
    {
    int [] [] [] x = new int [3] [] [];
    int i, j;
    x[0] = new int[4][];
    x[1] = new int[2][];
    x[2] = new int[5][];
    for (i = 0; i < x.length; i++) 
    {
        for (j = 0; j < x[i].length; j++) 
        {
            x[i][j] = new int [i + j + 1];
            System.out.println("size = " + x[i][j].length);
        }
    }
    }
}

A.7
B.9
C.11
D.13
E.Compilation fails

Answer: C

The loops use the array sizes (length).

It produces 11 lines of output as given below.

D:\Java>javac Test.java

D:\Java>java Test
size = 1
size = 2
size = 3
size = 4
size = 2
size = 3
size = 3
size = 4
size = 5
size = 6
size = 7

Therefore, 11 is the answer.

Discuss in Forum

public class F0091 
{    
    public void main( String[] args ) 
    {  
        System.out.println( "Hello" + args[0] ); 
    } 
}

What will be the output of the program, if this code is executed with the command line:

> java F0091 world

A.Hello
B.Hello Foo91
C.Hello world
D.The code does not run.

Answer: D

Option D is correct. A runtime error will occur owning to the main method of the code fragment not being declared static:

Exception in thread "main" java.lang.NoSuchMethodError: main

The Java Language Specification clearly states: "The main method must be declared public, static, and void. It must accept a single argument that is an array of strings."

Discuss in Forum

What will be the output of the program?


public class TestDogs 
{
    public static void main(String [] args) 
    {
        Dog [][] theDogs = new Dog[3][];
        System.out.println(theDogs[2][0].toString());
    }
}
class Dog { }

A.null
B.theDogs
C.Compilation fails
D.An exception is thrown at runtime

Answer: D

The second dimension of the array referenced by theDogs has not been initialized. Attempting to access an uninitialized object element (System.out.println(theDogs[2][0].toString());) raises a NullPointerException.

Discuss in Forum

What will be the output of the program?

public class CommandArgsTwo 
{
    public static void main(String [] argh) 
    {
        int x;
        x = argh.length;
        for (int y = 1; y <= x; y++) 
        {
            System.out.print(" " + argh[y]);
        }
    }
}

and the command-line invocation is

> java CommandArgsTwo 1 2 3

A.0 1 2
B.1 2 3
C.0 0 0
D.An exception is thrown at runtime

Answer: D

An exception is thrown because at some point in (System.out.print(" " + argh[y]);), the value of x will be equal to y, resulting in an attempt to access an index out of bounds for the array. Remember that you can access only as far as length - 1, so loop logical tests should use x < someArray.length as opposed to x <= someArray.length.

Discuss in Forum

Loading…

Topics

Other Subjects

Tags: Language Fundamentals - Finding the output Questions and Answers || Language Fundamentals - Finding the output MCQ Questions and Answers || Language Fundamentals - Finding the output GK Questions and Answers || Language Fundamentals - Finding the output GK MCQ Questions || Language Fundamentals - Finding the output Multiple Choice Questions and Answers || Language Fundamentals - Finding the output GK || GK on Language Fundamentals - Finding the output || Java Programming Questions and Answers || Java Programming MCQ Questions and Answers || Java Programming GK Questions and Answers || GK on Java Programming

Language Fundamentals - Finding the output Questions and Answers

Topics

Other Subjects

About Us

Stay informed about all important Dates & Notifications