doubt in kathy sierra question(Flow Control)?

[Hennry Smith]

3. public class Ebb {
4. static int x = 7;
5. public static void main(String[] args) {
6. String s = "";
7. for(int y = 0; y < 3; y++) {
8. x++;
9. switch(x) {
10. case 8: s += "8 ";
11. case 9: s += "9 ";
12. case 10: { s+= "10 "; break; }
13. default: s += "d ";
14. case 13: s+= "13 ";
15. }
16. }
17. System.out.println(s);
18. }
19. static { x++; }
20. }

What is the result?
A. 9 10 d
B. 8 9 10 d
C. 9 10 10 d
D. 9 10 10 d 13
E. 8 9 10 10 d 13
F. 8 9 10 9 10 10 d 13
G. Compilation fails

Answer:
D is correct. Did you catch the static initializer block? Remember that switches work on
"fall-thru" logic, and that fall-thru logic also applies to the default case, which is used when
no other case matches.
A, B, C, E, F, and G are incorrect based on the above.


Please tell how D is correct...???

According to me in line 3 we have declared a static variable x=7, then we have used a for loop,so according to the for loop it will run thrice,so the value of x will be increased 3 times, as we run the loop first time the value is increased from 7 to 8 and matched with switch (case 8) so first 8 should be printed....but the answer is 9 10 10 d 13....please explain where i am wrong....also tell why 10 is printed twice....???

[Meena Ajay]

Hi Hennry ,

As given in the answer from K&B,Did you catch the static initializer block given at the end of the code - line 19.?

After the line 4. executes, the value of x is set to 7.Since there is a static initializer block ( given in line number 19.), this would be the next thing to run. So, the value of x would be incremented to 8 after the static initializer block executes.

Now, the code enters the for loop with value of x having 8.After the x++ at line 8. is executed, the value of x becomes 9 and this enters the switch and case 9: would be executed. As per the looping logic, D is the answer"9 10 10 d 13 "

Hope this helps.

[Hennry Smith]

Meena Ajay wrote:Hi Hennry ,

As given in the answer from K&B,Did you catch the static initializer block given at the end of the code - line 19.?

After the line 4. executes, the value of x is set to 7.Since there is a static initializer block ( given in line number 19.), this would be the next thing to run. So, the value of x would be incremented to 8 after the static initializer block executes.

Now, the code enters the for loop with value of x having 8.After the x++ at line 8. is executed, the value of x becomes 9 and this enters the switch and case 9: would be executed. As per the looping logic, D is the answer"9 10 10 d 13 "

Hope this helps.

Thank you so much....i got your answer...

[Henry Wong]

Riaz ahmed pinjari,

Please don't hijack topics. Your post was moved to a new topic.

Henry