Enum Question

[Lao Kinsuyi]

Hi all,
This is a question from master exam accompanying Bert holy bible book.

Given,
Class Stepper {
Enum Roman {I,V,X,L,C,M}
public static void main(String...bang){
int x =7;
int z =2;
Roman r=Roman.x;
do{
Switch(r){
case C:r=Roman.L; break;
case X:r=Roman.C;
case L:if (r.ordinal)>2z+=5;
case M:x++;
}z++;
}
While (x>10);
System.out.println(Z);
}
}

What is the result?
a 5
b v
c 18
d 21
e 22
f compilation fails



But,i think the answer should be compilation fails. Because, not all the alphabelts in Enum Roman block were used in Switch block(case)
Only case C,X,L,M were used.

Please, i want someone to explain with reasons why it is not compilation fails.

[marc weber]

Originally posted by Lao Kinsuyi:
...i think the answer should be compilation fails...

Have you tested this code?

There are 7 errors (typos) that will prevent this from compiling. Once you fix these, try experimenting with values that are not represented as cases (for example, Roman.V).
[ July 21, 2007: Message edited by: marc weber ]

[Lao Kinsuyi]

Please, someone should explain this thread(enum question)
Over to you Chandra Bhatt for your usual effort.

[Padma Asrani]

hello Lao

There are lot of typos in the program and also the correct output 8 is not list, so it leads to mark for compilation fails but was the program really testing us to spot typos. If it is then I guess it is really a poor example.
Anyways i have corrected the program

You can manually find out the difference.

    
public class Stepper {
enum Roman {I,V,X,L,C,M};
public static void main(String... bang){
int x =7;
int z =2;
Roman r=Roman.X;
do {
switch(r) {
case C:r=Roman.L; break;
case X:r=Roman.C;
case L:if (r.ordinal()>2) z+=5;
case M:x++;
}
z++;
}
while (x>10);
System.out.println(z);
}
}
 

Regards
Padma

[ram shah]

I'll try to explain...

  1. public class Stepper {
  2. enum Roman {I,V,X,L,C,M};
  3. public static void main(String... bang){
  4. int x =7;
  5. int z =2;
  6. Roman r=Roman.X;
  7. do {
  8.   switch(r) {
  9.     case C:r=Roman.L; break;
  10.    case X:r=Roman.C;
  11.    case L:if (r.ordinal()>2) z+=5;    
  12.    case M:x++;
  13.    }
  14.    z++;
  15.    }
  16. while (x>10);
  17. System.out.println(z);
  18.  }
  19.} 
 

In line 6, "r" is set to value X. In line 10,Case X is executed, which sets "r" to C.Since, there is no break statement, all other cases are also executed.In line 11, as "r.ordinal()" returns 5, z is incremented to 7.
In line 14, z is again incremented by 1. value of z is now 8. In line 16, the while condition is evaluated and found to be false.So, the execution comes out of the loop and value of z is printed as 8.

Somebody please correct me if I'm wrong.

[Lao Kinsuyi]

I've corrected the missing codes. Thank you.Now, it's well understood.