INTERVIEW QUESTIONS MICROSOFT VC++ DETAILS

Question: How to Killi an Object Prematurely..

Answer: August 30th, 2009 by sd | Posted in Uncategorized

Sometimes, you need to force an object to destroy itself because its destructor performs an operation needed immediately. For example, when you want to release a mutex or close a file:

void func(Modem& modem)
{
Mutex mtx(modem); // lock modem
Dial(”1234567″);
/* at this point, you want to release the
modem lock by invoking Mutex’s dtor */
do_other_stuff(); //modem is still locked
} //Mutex dtor called here
After the function Dial() has finished, the modem no longer needs to be locked. However, the Mutex object will release it only when func() exits, and in the meantime, other users will not be able to use the modem. Before you suggest to invoke the destructor explicitly, remember that the destructor will be called once again when func() exits, with undefined behavior:

void func(Modem& modem)
{
Mutex mtx(modem); // lock modem
Dial(”1234567″);
mtx->~Mutex(); // very bad idea!
do_other_stuff();
} //Mutex dtor called here for the second time
There is a simple and safe solution to this. You wrap the critical code in braces:

void func(Modem& modem)
{
{
Mutex mtx(modem); // lock modem
Dial(”1234567″);
} //mtx destroyed here
//modem is not locked anymore
do_some_other_stuff();
}
VC++ FAQ’s And Interview Questions and Answers

Popularity: 6% [?]

August 30th, 2009 by sd | Posted in Uncategorized

Sometimes, you need to force an object to destroy itself because its destructor performs an operation needed immediately. For example, when you want to release a mutex or close a file:

void func(Modem& modem)
{
Mutex mtx(modem); // lock modem
Dial(”1234567″);
/* at this point, you want to release the
modem lock by invoking Mutex’s dtor */
do_other_stuff(); //modem is still locked
} //Mutex dtor called here
After the function Dial() has finished, the modem no longer needs to be locked. However, the Mutex object will release it only when func() exits, and in the meantime, other users will not be able to use the modem. Before you suggest to invoke the destructor explicitly, remember that the destructor will be called once again when func() exits, with undefined behavior:

void func(Modem& modem)
{
Mutex mtx(modem); // lock modem
Dial(”1234567″);
mtx->~Mutex(); // very bad idea!
do_other_stuff();
} //Mutex dtor called here for the second time
There is a simple and safe solution to this. You wrap the critical code in braces:

void func(Modem& modem)
{
{
Mutex mtx(modem); // lock modem
Dial(”1234567″);
} //mtx destroyed here
//modem is not locked anymore
do_some_other_stuff();
}
VC++ FAQ’s And Interview Questions and Answers

Popularity: 6% [?]
Source: CoolInterview.com

If you have the better answer, then send it to us. We will display your answer after the approval.

Rules to Post Answers in CoolInterview.com:-

• There should not be any Spelling Mistakes.
• There should not be any Gramatical Errors.
• Answers must not contain any bad words.
• Answers should not be the repeat of same answer, already approved.
• Answer should be complete in itself.

Name :*
Email Id :*
Answer :*
Verification Code	Not readable? Load New Code
Enter the above shown code: *
	Inform me about updated answers to this question