Operators and Assignments - Finding the output

What will be the output of the program?

public class Test
{
public static void leftshift(int i, int j)
{
i <<= j;
}
public static void main(String args[])
{
int i = 4, j = 2;
leftshift(i, j);
System.out.println(i);
}
}

A.	2
B.	4
C.	8
D.	16

Answer: Option B

Explanation:

Java only ever passes arguments to a method by value (i.e. a copy of the variable) and never by reference. Therefore the value of the variable i remains unchanged in the main method.

If you are clever you will spot that 16 is 4 multiplied by 2 twice, (4 * 2 * 2) = 16. If you had 16 left shifted by three bits then 16 * 2 * 2 * 2 = 128. If you had 128 right shifted by 2 bits then 128 / 2 / 2 = 32. Keeping these points in mind, you don't have to go converting to binary to do the left and right bit shifts.

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);
}
}

A.	count = 0
B.	count = 2
C.	count = 3
D.	count = 4

Answer: Option C

Explanation:

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.

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;
}
}

A.	null null 42
B.	0 0 42
C.	0 42 42
D.	0 0 0

Answer: Option C

Explanation:

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.

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;
}
}

A.	7 7
B.	7 14
C.	14 0
D.	14 14

Answer: Option B

Explanation:

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.

What will be the output of the program?

class SC2
{
public static void main(String [] args)
{
SC2 s = new SC2();
s.start();
}

void start()
{
int a = 3;
int b = 4;
System.out.print(" " + 7 + 2 + " ");
System.out.print(a + b);
System.out.print(" " + a + b + " ");
System.out.print(foo() + a + b + " ");
System.out.println(a + b + foo());
}

String foo()
{
return "foo";
}
}

A.	9 7 7 foo 7 7foo
B.	72 34 34 foo34 34foo
C.	9 7 7 foo34 34foo
D.	72 7 34 foo34 7foo

Answer: Option D

Explanation:

Because all of these expressions use the + operator, there is no precedence to worry about and all of the expressions will be evaluated from left to right. If either operand being evaluated is a String, the + operator will concatenate the two operands; if both operands are numeric, the + operator will add the two operands.

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");
}
}

A.	ok
B.	dokey
C.	ok dokey
D.	No output is produced
E.	Compilation error

Answer: Option B

Explanation:

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".

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 );
}
}

A.	0
B.	7
C.	8
D.	14

Answer: Option D

Explanation:

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.

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; z++)
{
if (( ++x > 2 ) || (++y > 2))
{
x++;
}
}
System.out.println(x + " " + y);
}
}

A.	5 3
B.	8 2
C.	8 3
D.	8 5

Answer: Option B

Explanation:

The first two iterations of the for loop both x and y are incremented. On the third iteration x is incremented, and for the first time becomes greater than 2. The short circuit or operator || keeps y from ever being incremented again and x is incremented twice on each of the last three iterations.

Operators and Assignments - Finding the output

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