Comments by Member 13508014

Member 13508014 8-Apr-19 5:09am View

What I am surprised was, though the signature of foo() method is different in derived class, why should it hide foo() method of base class.

Member 13508014 1-Apr-19 13:50pm View

Yes. I too realized and agree that the point that t3 should be exactly the sum of t1 and t2. Thanks CPallini for your time in answering my question.

Member 13508014 31-Mar-19 3:41am View

Yes. Copy of b will be altered. I agree that.

But, I think, my question is not understood here.

My question is, in "a = b + c", lets assume that the object "a" is created a bit long and its state is modified now and when i call "a = b + c", i expect the addition operation to happen on only few members and those will get altered. Other member of object "a" for which "+" operation does not make sense should be same as before calling "a = b + c".

Member 13508014 28-Mar-19 7:40am View

I read "Binary arithmetic operators" section of operator overloading - cppreference.com[^].
When we write a = b + c, according to the article, here 'b' will also get altered isn't it.

Member 13508014 16-Oct-18 12:09pm View

Excellent. It clarified my doubt.

Member 13508014 16-Oct-18 5:13am View

I understood the examples given there. All examples given there makes it to understand as python maintains the list as single linked list. That is why L[4:1] gives empty list, because a singly linked list can not be traversed in reverse way. But if that is the case how does L[start:end:-1] can traverse in reverse direction is what I did not understand. This is not explained in the article.

Member 13508014 16-Oct-18 4:38am View

I read the article. I still have the same doubt.

Member 13508014 14-Dec-17 9:23am View

That would have been a perfect solution if it worked. But there is no success when thread::join() is called before return in main(). That clears my question that I had in my mind. If thread::join() is NOT executed, then how does it can effect. That is what it happens when join() is called before return right?

And when thread::join() is called before the while loop in main, main thread does not get a chance at all to execute the loop which is as expected.

Member 13508014 14-Dec-17 5:18am View

thread::join is it a blocking call? If it is so I may not want to use it. Because the intention is not to block main thread and allow it to do its work.

Comments by Member 13508014 (Top 9 by date)