Remove From My Forums

problem with vector

Question
0

Sign in to vote
Hi all i am using following code with vector

class myc { static int i; public: myc() { cout<<++i<<" ctor \n"; } ~myc() { cout<<--i<<" dtor \n"; } }; int myc::i = 0; int main() { { vector<myc> vec(5); } return 0; }

and got output

1 ctor
0 dtor
1 ctor
0 dtor
1 ctor
0 dtor
1 ctor
0 dtor
1 ctor
0 dtor
-1 dtor
-2 dtor
-3 dtor
-4 dtor
-5 dtor
Press any key to continue . . .

why i got this output as per c++ 11 no copying should be done

thanks in advance here is oreginal post
Monday, February 20, 2012 11:57 AM

Answers

2

Sign in to vote
This is down to the implementation of vector itself and not the compiler. If you have any issues with this then submit a suggestion on the connect website. But one thing to note, this isn't a bug.

This particular constructor is not required to not make copies, all internal members that this implementation calls are also allowed to make copies. If you are having problems because of this then it is really just exposing a weakness in your class design anyway.

If this behaviour is causing problems for your class then it means that it will be causing problems in a situation where the trivial constructor is not sufficient for what you want to do. For example, if you are dealing with pointers and memory allocated using new, then you are going to have to problems. This problem has been known for a long time, long enough for someone to come up with a snappy name for it. So you should look at the Rule of three.

The same will be true with the move constructor, lets face it, how does the compiler know which members should be moved and those which shouldn't. How does it know the difference between pointer A which contains all of the data for the class and that means that it can be safely moved to the new class and set to null, and pointer B which is a pointer to a parent class and if it does get moved then it will end up with the destructor failing.

So basically, unless your class is made up entirely of value types and trivial members you will need to write any copy/move constructors yourself. The defaults only do a shallow copy after all.
This is a signature

Any samples given are not meant to have error checking or show best practices. They are meant to just illustrate a point. I may also give inefficient code or introduce some problems to discourage copy/paste coding. This is because the major point of my posts is to aid in the learning process.
- Proposed as answer by Helen Zhao Wednesday, February 22, 2012 2:29 AM
- Marked as answer by Helen Zhao Friday, March 2, 2012 1:29 AM
Monday, February 20, 2012 1:52 PM

When you add copy constructor, you get the following output:

1 ctor 
2 copy ctor 
1 dtor 
2 ctor 
3 copy ctor 
2 dtor 
3 ctor 
4 copy ctor 
3 dtor 
4 ctor 
5 copy ctor 
4 dtor 
5 ctor 
6 copy ctor 
5 dtor 
4 dtor 
3 dtor 
2 dtor 
1 dtor 
0 dtor

When you add move ctor, you get the following output:

1 ctor 
2 move ctor 
1 dtor 
2 ctor 
3 move ctor 
2 dtor 
3 ctor 
4 move ctor 
3 dtor 
4 ctor 
5 move ctor 
4 dtor 
5 ctor 
6 move ctor 
5 dtor 
4 dtor 
3 dtor 
2 dtor 
1 dtor 
0 dtor

Which is what you expect probably. The final sample code is:

#include <iostream>
#include <vector>

using namespace std;

class myc
{
	static int i;
public:
	myc()
	{
		cout<<++i<<" ctor \n";
	}
	
	myc(myc const &)
	{
		cout<<++i<<" copy ctor \n";
	}
	
	myc(myc &&)
	{
		cout<<++i<<" move ctor \n";
	}
	
	~myc()
	{
		cout<<--i<<" dtor \n";
	}
};

int myc::i = 0;

int main()
{
	{
		vector<myc> vec(5);
	}
	return 0;
}

Toni Petrina
My blog: Programming ramblings
If a post answers your question, please click "Mark As Answer" on that post and "Vote as Helpful"

Proposed as answer by Helen Zhao Wednesday, February 22, 2012 2:29 AM
Marked as answer by Helen Zhao Friday, March 2, 2012 1:30 AM

Monday, February 20, 2012 1:22 PM

All replies