Output of Java program | Set 26

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 2
b) 3 3
c) Runtime Error
d) 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) 1
b) 2
c) 3
d) 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) 5
b) 6
c) 14
d) 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 15
b) 15 15.
c) 3 4 5 3 7 5
d) 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 1
b) 0x80000000 and 0x00000001
c) -2147483648 and -1
d) 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.