Output of Java program | Set 26

11 Nov, 2021
Ques1: What is the output of this program?
class A {    public int i;    private int j;}  class B extends A {    void display()    {        super.j = super.i + 1;        System.out.println(super.i + " " + super.j);    }}  class inheritance {    public static void main(String args[])    {        B obj = new B();        obj.i = 1;        obj.j = 2;        obj.display();    }}
a) 2 2b) 3 3c) Runtime Errord) Compilation Error
 Answer: d 
Explanation: Class A contains a private member variable j, this cannot be inherited by subclass B. So in class B we can not access j. So it will give a compile time error.
Ques2: What is the output of this program?
class selection_statements {    public static void main(String args[])    {        int var1 = 5;        int var2 = 6;        if ((var2 = 1) == var1)            System.out.print(var2);        else            System.out.print(++var2);    }}
options:a) 1b) 2c) 3d) 4
Answer: b
Explanation: In “If” statement, first var2 is initialized to 1 and then condition is checked whether var2 is equal to var1. As we know var1 is 5 and var2 is 1, so condition will be false and else part will be executed.
Ques3: What is the output of this program?
class comma_operator {    public static void main(String args[])    {        int sum = 0;        for (int i = 0, j = 0; i < 5 & j < 5; ++i, j = i + 1)            sum += i;        System.out.println(sum);    }}
options:a) 5b) 6c) 14d) compilation error
Answer: b
Explanation: Using comma operator, we can include more than one statement in the initialization and iteration portion of the for loop. Therefore both ++i and j = i + 1 is executed i gets the value : 0, 1, 2, 3, and j gets the values : 0, 1, 2, 3, 4, 5.
Ques4. What will be the output of the program?
class Geeks {    public static void main(String[] args)    {        Geeks g = new Geeks();        g.start();    }      void start()    {        long[] a1 = { 3, 4, 5 };        long[] a2 = fix(a1);        System.out.print(a1[0] + a1[1] + a1[2] + " ");        System.out.println(a2[0] + a2[1] + a2[2]);    }      long[] fix(long[] a3)    {        a3[1] = 7;        return a3;    }}
options:a) 12 15b) 15 15.c) 3 4 5 3 7 5d) 3 7 5 3 7 5
Answer: b
Explanation: The reference variables a1 and a3 refer to the same long array object. When fix() method is called, array a1 is passed as reference. Hence the value of a3[1] becomes 7 which will be reflected in a1[] as well because of call by reference. So the a1[] array become {3, 7, 5}. WHen this is returned to a2[], it becomes {3, 7, 5} as well.
So Output: 3 + 7 + 5 + ” ” + 3 + 7 + 5 = 15 15
Ques5. What will be the output of the program?
class BitShift {    public static void main(String[] args)    {        int x = 0x80000000;        System.out.print(x + " and  ");        x = x >>> 31;        System.out.println(x);    }}
options:a) -2147483648 and 1b) 0x80000000 and 0x00000001c) -2147483648 and -1d) 1 and -2147483648
Answer: a
Explanation: Option A is correct. The >>> operator moves all bits to the right, zero filling the left bits. The bit transformation looks like this:
Before: 1000 0000 0000 0000 0000 0000 0000 0000
After: 0000 0000 0000 0000 0000 0000 0000 0001
Option C is incorrect because the >>> operator zero fills the left bits, which in this case changes the sign of x, as shown.
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