### Finding The Output

Question 1 :

What will be the output of the program?

``````int i = l, j = -1;
switch (i)
{
case 0, 1: j = 1; /* Line 4 */
case 2: j = 2;
default: j = 0;
}
System.out.println("j = " + j);``````

A). j = -1
B). j = 0
C). j = 1
D). Compilation fails.
Answer : Option D

Explanation :

The case statement takes only a single argument. The case statement on line 4 is given two arguments so the compiler complains.

Question 2 :

What will be the output of the program?

``````int i = 1, j = 10;
do
{
if(i > j)
{
break;
}
j--;
} while (++i < 5);
System.out.println("i = " + i + " and j = " + j);``````

A). i = 6 and j = 5
B). i = 5 and j = 5
C). i = 6 and j = 4
D). i = 5 and j = 6
Answer : Option D

Explanation :

This loop is a do-while loop, which always executes the code block within the block at least once, due to the testing condition being at the end of the loop, rather than at the beginning. This particular loop is exited prematurely if i becomes greater than j.

The order is, test i against j, if bigger, it breaks from the loop, decrements j by one, and then tests the loop condition, where a pre-incremented by one i is tested for being lower than 5. The test is at the end of the loop, so i can reach the value of 5 before it fails. So it goes, start:

1, 10
2, 9
3, 8
4, 7
5, 6 loop condition fails.

Question 3 :

What will be the output of the program?

``````public class Switch2
{
final static short x = 2;
public static int y = 0;
public static void main(String [] args)
{
for (int z=0; z < 3; z++)
{
switch (z)
{
case x: System.out.print("0 ");
case x-1: System.out.print("1 ");
case x-2: System.out.print("2 ");
}
}
}
}``````

A). 0 1 2
B). 0 1 2 1 2 2
C). 2 1 0 1 0 0
D). 2 1 2 0 1 2
Answer : Option D

Explanation :

The case expressions are all legal because x is marked final, which means the expressions can be evaluated at compile time. In the first iteration of the for loop case x-2 matches, so 2 is printed. In the second iteration, x-1 is matched so 1 and 2 are printed (remember, once a match is found all remaining statements are executed until a break statement is encountered). In the third iteration, x is matched. So 0 1 and 2 are printed.

Question 4 :

What will be the output of the program?

``````public class SwitchTest
{
public static void main(String[] args)
{
System.out.println("value =" + switchIt(4));
}
public static int switchIt(int x)
{
int j = 1;
switch (x)
{
case l: j++;
case 2: j++;
case 3: j++;
case 4: j++;
case 5: j++;
default: j++;
}
return j + x;
}
}``````

A). value = 2
B). value = 4
C). value = 6
D). value = 8
Answer : Option D

Explanation :

Because there are no break statements, once the desired result is found, the program continues though each of the remaining options.

Question 5 :

What will be the output of the program?

``````public class If2
{
static boolean b1, b2;
public static void main(String [] args)
{
int x = 0;
if ( !b1 ) /* Line 7 */
{
if ( !b2 ) /* Line 9 */
{
b1 = true;
x++;
if ( 5 > 6 )
{
x++;
}
if ( !b1 )
x = x + 10;
else if ( b2 = true ) /* Line 19 */
x = x + 100;
else if ( b1 | b2 ) /* Line 21 */
x = x + 1000;
}
}
System.out.println(x);
}
}``````

A). 0
B). 1
C). 101
D). 111
Answer : Option C

Explanation :

As instance variables, b1 and b2 are initialized to false. The if tests on lines 7 and 9 are successful so b1 is set to true and x is incremented. The next if test to succeed is on line 19 (note that the code is not testing to see if b2 is true, it is setting b2 to be true). Since line 19 was successful, subsequent else-if's (line 21) will be skipped.