Java Programming - Flow Control - Discussion
Discussion Forum : Flow Control - Finding the output (Q.No. 6)
6.
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 y: System.out.print("0 "); /* Line 11 */
case x-1: System.out.print("1 "); /* Line 12 */
case x: System.out.print("2 "); /* Line 13 */
}
}
}
}
Answer: Option
Explanation:
Case expressions must be constant expressions. Since x is marked final, lines 12 and 13 are legal; however y is not a final so the compiler will fail at line 11.
Discussion:
12 comments Page 1 of 2.
Juel Khan said:
6 years ago
In C/C++ int values are work well. But why in Java int values does not work? Please help anyone.
Animesh said:
7 years ago
Why Y is not accepted in a switch case? Explain to me.
Partha said:
7 years ago
public class Main
{
final static short x = 2;
final static int y = 0;
public static void main(String [] args)
{
for (int z=0; z < 3; z++)
{
switch (z)
{
case y: System.out.print("0 "); /* Line 11 */
case x-1: System.out.print("1 "); /* Line 12 */
case x: System.out.print("2 "); /* Line 13 */
}
}
}
}
Here, if we change the static to final then the output : 012122.
{
final static short x = 2;
final static int y = 0;
public static void main(String [] args)
{
for (int z=0; z < 3; z++)
{
switch (z)
{
case y: System.out.print("0 "); /* Line 11 */
case x-1: System.out.print("1 "); /* Line 12 */
case x: System.out.print("2 "); /* Line 13 */
}
}
}
}
Here, if we change the static to final then the output : 012122.
Neha said:
7 years ago
Here x is evaluated at compile time as x is a final variable but y is not a final variable so compiler evaluates y's value at runtime. So the value of is not considered as constant.
Raman said:
1 decade ago
I declare why as final and not remove final so my program is success.
Its mean switch case acceptable public.
Its mean switch case acceptable public.
Maxwell said:
1 decade ago
Why is y, (a public variable) not acceptable in a switch case? Thanks in Advance.
Ramya said:
1 decade ago
Why no error in line 13? I learnt type of case label should be same as 'x' (switch (x) ).
Pallavi said:
1 decade ago
Why is it a compilation error in line number 11 but not in 12 ?
Dev Singh said:
1 decade ago
@Mahesh:
In your program output will be:
1st time (at z=0) condition (case y) : 012 (because No break).
2nd time (at z=1) condition (x-1) : 12 (again after 1 No break, so 2 printed).
3rd time (at z=2) condition (x=2) : 2.
So output : 012122.
In your program output will be:
1st time (at z=0) condition (case y) : 012 (because No break).
2nd time (at z=1) condition (x-1) : 12 (again after 1 No break, so 2 printed).
3rd time (at z=2) condition (x=2) : 2.
So output : 012122.
Piya said:
1 decade ago
When I define the keyword public for why it display the op as 012122 how is it?
(1)
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