Multiple template class destruction?

Question

This is a simple test:

#include <iostream>
#include <vector>

using namespace std;

class C
{
public:
    int _a;
    C(int a) { cout << "constructor C: " << (_a = a) << endl; }
    ~C()     { cout << "destructor C: "<< _a << endl; }
};

int main()
{
    vector<C> vec;

    for(int i=0; i<2; i++)
        vec.push_back(i+1);

    cout << "Go out --->\n";
    return 0;
}

Output:

constructor C: 1
destructor C: 1
constructor C: 2
destructor C: 1
destructor C: 2
Go out --->
destructor C: 1
destructor C: 2

It looks not so funny when C is a real class with responsible destructor. Any suggestions? Thank you in advance.

same issue as here -> http://stackoverflow.com/q/2745350/476681 — BЈовић, Mar 21 '14 at 12:06
You should really define (or forbid!) a full set of constructors and assignment operators for non-trivial classes like this ;-). Then you would see (and properly implement) or forbid assignments and copyings of your objects, e.g. in and out of containers. BЈовић's link illustrates the issues. — Peter - Reinstate Monica, Mar 21 '14 at 12:14
Interestingly we have three destructions for 1, but "only" two for 2. The reason is the vector growing by reallocation, thereby copying the previous elements (here 1) and destroying the old ones. As noted, the copying is not logged, but the destruction is. Ah I see, Vlad made the same observation :-) — Peter - Reinstate Monica, Mar 21 '14 at 12:39

Eric Fortin · Answer 1 · 2014-03-21T12:04:10.100

5

You get multiple class destructor logs because there is object copy when adding object in a vector. Log also in the copy contructor, you'll see it balances out. As indicated in the comments, if you are using C++11, also log in the move constructor.

That is why if object construction is expansive, you use pointers(smart_ptr) in vector. Or reserve enough space and use emplace_back(again if using C++11).

edited Mar 21 '14 at 12:04

answered Mar 21 '14 at 11:58

Eric Fortin

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1

Or possibly, the move constructor. – Puppy Mar 21 '14 at 12:02
Move constructor would be the better solution. – Ben Mar 21 '14 at 12:03
If he's using C++11 yes. Added to answer thanks. – Eric Fortin Mar 21 '14 at 12:03
1

why not just reserve + emplace_back?? – billz Mar 21 '14 at 12:04
@billz Added your suggestion. We don't know if he's using C++11 though. – Eric Fortin Mar 21 '14 at 12:06
@EricFortin Why wouldn't he be? – Ben Mar 21 '14 at 12:07
@Ben Depends on what he is working on. I can't always use it in my day to day. On PS3 or X360 for example. Also compiler don't always compile in C++11 by default. – Eric Fortin Mar 21 '14 at 12:08
I think it's pretty safe to assume he can use c++11. It looks like he's learning to me. c++11 is only going to get more support, no point giving him old solutions. – Ben Mar 21 '14 at 12:11
That's why I updated the answer base on your suggestions. Although copy constructor is still needed in case he's not using an rvalue in the call to push_back. – Eric Fortin Mar 21 '14 at 12:12
How's that not an rvalue? – Ben Mar 21 '14 at 12:14
If I have `C c(2); vec.push_back(c);` It is not safe to move right ? I might be wrong. In that case I'll remove the comment. – Eric Fortin Mar 21 '14 at 12:16
Yeah. In that case it needs to be copied. But in his example, it is an rvalue, unless I'm missing something. – Ben Mar 21 '14 at 12:18
@Ben No you're right in his example it is. But this might not be his real code so that's why I prefer to specify. – Eric Fortin Mar 21 '14 at 12:19
Yeah, that's fair enough. – Ben Mar 21 '14 at 12:21
Of course you are right, thank you. As others in comments. A silly question. Sure no questions about using pointers. – user3446122 Mar 21 '14 at 14:42

score 1 · Answer 2 · answered Mar 21 '14 at 12:18

constructor C: 1

is called for creating temporary object C( 1 ) as argument of vec.push_back(i+1); when i =0

destructor C: 1

this temporary object was copied in an object of the vector and after that it was deleted.

constructor C: 2

is called for creating temporary object C( 2 ) as argument of vec.push_back(i+1); when i =1

The vecttor need to allocate a new memory region that to accomodate the second element. So it reallocates the memory and copies the first element in the new memory.

If you would call member function reserve

vector<C> vec;
vec.reserve( 2 );

then there would not be the reallocation and you would not see the next output.

destructor C: 1

the first element of the vector in old region memory is deleted

destructor C: 2

the temporary object was copied in an object of the vector and after that it was deleted.

Multiple template class destruction?

2 Answers2