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# Operators and Assignments - Finding the output

1.

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);
}
}

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.

#### Enter details here

2.

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);
}
}

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.

#### Enter details here

3.

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;
}
}

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.

#### Enter details here

4.

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;
}
}

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.

#### Enter details here

5.

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";
}
}