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Home/ Questions/Q 8512985
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Editorial Team
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Editorial Team
Asked: June 11, 20262026-06-11T04:25:48+00:00 2026-06-11T04:25:48+00:00

class A { int id; static int count; public: A() { count++; id =

  • 0
class A
{
    int id;
    static int count;
public:
    A()
    {
        count++;
        id = count;
        cout << "constructor called " << id << endl;
    }
    ~A()
    {
        //count -=2; /*Keypoint is here. */
                   /*Uncomment it later. But result doesn't change*/
        cout << "destructor called " << id << endl;
    }
};

int A::count = 0;

int main()
{
    A a[2];
    return 0;
}

The output is

constructor called 1
constructor called 2
destructor called 2
destructor called 1

The question is:
even if you uncomment the //count -=2;
the result is still the same.

Does that mean that if the constructor increments the static member by 1,then the destructor must decrement it exactly by 1 also, and you can’t change the behaviour of it?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-11T04:25:50+00:00Added an answer on June 11, 2026 at 4:25 am

    Nothing accesses count after the first destructor is invoked. The destructor does exactly what you code it to do, either modifying count or not. But you won’t see the effect unless you access count in some way.

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