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Operators and Assignments - Finding the output

1.

What will be the output of the program?

class Equals 
{
    public static void main(String [] args) 
    {
        int x = 100;
        double y = 100.1;
        boolean b = (x = y); /* Line 7 */
        System.out.println(b);
    }
}

Answer: C

The code will not compile because in line 7, the line will work only if we use (x==y) in the line. The == operator compares values to produce a boolean, whereas the = operator assigns a value to variables.

Option A, B, and D are incorrect because the code does not get as far as compiling. If we corrected this code, the output would be false.

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2.

What will be the output of the program?

class Test 
{
    public static void main(String [] args) 
    {
        int x= 0;
        int y= 0;
        for (int z = 0; z < 5> 2 ) && (++y > 2)) 
            {
                x++;
            }
        }
        System.out.println(x + " " + y);
    }
}

Answer: C

In the first two iterations x is incremented once and y is not because of the short circuit && operator. In the third and forth iterations x and y are each incremented, and in the fifth iteration x is doubly incremented and y is incremented.

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3.

What will be the output of the program?

class Test 
{
    public static void main(String [] args) 
    {
        Test p = new Test();
        p.start();
    }

    void start() 
    {
        boolean b1 = false;
        boolean b2 = fix(b1);
        System.out.println(b1 + " " + b2);
    }

    boolean fix(boolean b1) 
    {
        b1 = true;
        return b1;
    }
}

Answer: B

The boolean b1 in the fix() method is a different boolean than the b1 in the start() method. The b1 in the start() method is not updated by the fix() method.

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4.

What will be the output of the program?

class Two 
{
    byte x;
}

class PassO 
{
    public static void main(String [] args) 
    {
        PassO p = new PassO();
        p.start();
    }

    void start() 
    {
        Two t = new Two();
        System.out.print(t.x + " ");
        Two t2 = fix(t);
        System.out.println(t.x + " " + t2.x);
    }

    Two fix(Two tt) 
    {
        tt.x = 42;
        return tt;
    }
}

Answer: C

In the fix() method, the reference variable tt refers to the same object (class Two) as the t reference variable. Updating tt.x in the fix() method updates t.x (they are one in the same object). Remember also that the instance variable x in the Two class is initialized to 0.

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5.

What will be the output of the program?

class BoolArray 
{
    boolean [] b = new boolean[3];
    int count = 0;

    void set(boolean [] x, int i) 
    {
        x[i] = true;
        ++count;
    }

    public static void main(String [] args) 
    {
        BoolArray ba = new BoolArray();
        ba.set(ba.b, 0);
        ba.set(ba.b, 2);
        ba.test();
    }

    void test() 
    {
        if ( b[0] && b[1] | b[2] )
            count++;
        if ( b[1] && b[(++count - 2)] )
            count += 7;
        System.out.println("count = " + count);
    }
}

Answer: C

The reference variables b and x both refer to the same boolean array. count is incremented for each call to the set() method, and once again when the first if test is true. Because of the && short circuit operator, count is not incremented during the second if test.

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6.

What will be the output of the program?\

class Bitwise 
{
    public static void main(String [] args) 
    {
        int x = 11 & 9;
        int y = x ^ 3;
        System.out.println( y | 12 );
    }
}

Answer: D

The & operator produces a 1 bit when both bits are 1. The result of the & operation is 9. The ^ operator produces a 1 bit when exactly one bit is 1; the result of this operation is 10. The | operator produces a 1 bit when at least one bit is 1; the result of this operation is 14.

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7.

What will be the output of the program?

class SSBool 
{
    public static void main(String [] args) 
    {
        boolean b1 = true;
        boolean b2 = false;
        boolean b3 = true;
        if ( b1 & b2 | b2 & b3 | b2 ) /* Line 8 */
            System.out.print("ok ");
        if ( b1 & b2 | b2 & b3 | b2 | b1 ) /*Line 10*/
            System.out.println("dokey");
    }
}

 

Answer: B

The & operator has a higher precedence than the | operator so that on line 8 b1 and b2 are evaluated together as are b2 & b3. The final b1 in line 10 is what causes that if test to be true. Hence it prints "dokey".

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8.

What will be the output of the program?

class Test 
{
    static int s;
    public static void main(String [] args) 
    {
        Test p = new Test();
        p.start();
        System.out.println(s);
    }

    void start() 
    {
        int x = 7;
        twice(x);
        System.out.print(x + " ");
    }

    void twice(int x) 
    {
        x = x*2;
        s = x;
    }
}

Answer: B

The int x in the twice() method is not the same int x as in the start() method. Start()'s x is not affected by the twice() method. The instance variable s is updated by twice()'s x, which is 14.

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9.

What will be the output of the program?

class PassS 
{
    public static void main(String [] args) 
    {
        PassS p = new PassS();
        p.start();
    }

    void start() 
    {
        String s1 = "slip";
        String s2 = fix(s1);
        System.out.println(s1 + " " + s2);
    }

    String fix(String s1) 
    {
        s1 = s1 + "stream";
        System.out.print(s1 + " ");
        return "stream";
    }
}

Answer: D

When the fix() method is first entered, start()'s s1 and fix()'s s1 reference variables both refer to the same String object (with a value of "slip"). Fix()'s s1 is reassigned to a new object that is created when the concatenation occurs (this second String object has a value of "slipstream"). When the program returns to start(), another String object is created, referred to by s2 and with a value of "stream".

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10.

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);
    }
}

Answer: A

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.

Option B is incorrect because the output method print() always displays integers in base 10.

Option D is incorrect because this is the reverse order of the two output numbers.

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