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Home/ Questions/Q 5849011
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Editorial Team
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Editorial Team
Asked: May 22, 20262026-05-22T12:56:23+00:00 2026-05-22T12:56:23+00:00

I tried the following program on Visual Studio 2010. #include <iostream> using namespace std;

  • 0

I tried the following program on Visual Studio 2010.

#include <iostream>
using namespace std;

class A {
public:
        int p;

        /*A(){
            cout << "Constructor A" << endl;
        }*/

        ~A(){
            cout << "Destructor in A" << endl;
        }
};

class D: public A
{
public: 

        /*D(){
            cout << "Constructor D" << endl;
        }*/

        ~D(){
            cout << "Destructor in D" << endl;
        }
};

int main()
{
    D d =  D();
    cout << "Exiting main" << endl;
}

The output that I got was –

Destructor in D
Destructor in A
Exiting main
Destructor in D
Destructor in A

I am not able to understand why the destructor of class D and A are being called
before “Exiting main” statement is executed?

I tried another thing – I uncommented the Class D constructor in the code above, then the output was as I
expected –

Constructor D
Exiting main
Destructor in D
Destructor in A

What am I missing here?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-22T12:56:24+00:00Added an answer on May 22, 2026 at 12:56 pm

    The line

    D d =  D();
    

    first creates a temporary, unnamed object, which then is copied to d. What you see is the temporary object being destroyed when the statement is ended. The named object d is destroyed when it goes out of scope, after main() is completed.

    If you add a copy constructor to D you’ll see that it is invoked.

    When commenting out the constructor I think that you see the expected behaviour because the compiler can do some optimizations.

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