About Question enthuware.ocpjp.v7.2.1366 :

[shamran99]

Hi,

Just need to clarify on the following.
The answer says..
"Numbers 1 to 5 will be printed (some may be repeated or missed) with "N Done" anywhere in the sequence where N cannot be determined."

I answered...
"Total of 5 numbers will be printed followed by "N Done" where N cannot be determined."
considering the following scenario as well.

Since the threadCounter variable is not defined volatile, the changes might not be visible to other threads. So there is a possibility to get 5 times the number 1 printed. Am I wrong? PLease explain.

Regards,
Shamran.

[admin]

If you access a variable from within a synchronized block, then volatile is not required. Synchronization guarantees that most recent update will be visible to accessor thread.

[teodorj]

Option 4 is wrong because of the line in the main method..

System.out.println(threadcounter+" DONE");

any numbers (1-5) will be printed (from run) then the main method will print the thread counter value .
So it is not equal to 5 numbers

Am I correct?

[admin]

No, that option says, 5 numbers and then N done also. So, 6 numbers is not the problem with this option. Please read the explanation carefully. It explains why it is wrong.

[teodorj]

Option 4 is also not correct because of the bold part..
Total of 5 numbers will be printed followed by "N Done" where N cannot be determined.

This is wrong because the order in which the print method in main cannot be determined.
Please verify.

[admin]

Correct.