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Home/ Questions/Q 890655
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Editorial Team
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Editorial Team
Asked: May 15, 20262026-05-15T13:46:05+00:00 2026-05-15T13:46:05+00:00

During destruction of the derived class object, i first hit the derived class destructor

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During destruction of the derived class object, i first hit the derived class destructor and then the base class destructor (which is as expected). But i was curious to find out – at what point does the functions of the derived class go out of scope (are destroyed).

Does it happen as soon as the control leaves the derived class destructor and goes toward the base? Or does it happen once we done with the base class destructor also.

Thanks

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  1. Editorial Team
    Editorial Team
    2026-05-15T13:46:06+00:00Added an answer on May 15, 2026 at 1:46 pm

    Once the destructor of the most derived class finishes, the dynamic type of the object can be considered that of the next less-derived-type. That is, a call to a virtual method in the base destructor will find that the final overrider at that point in time is at base level. (The opposite occurs during construction)

    struct base {
       base() { std::cout << type() << std::endl; }
       virtual ~base() { std::cout << type() << std::endl; }
       virtual std::string type() const {
          return "base";
       }
    };
    struct derived : base {
       virtual std::string type() const {
          return "derived";
       }
    };
    int main() {
       base *p = new derived;
       std::cout << p->type() << std::endl;
       delete p;
    }
    // output:
    // base
    // derived
    // base
    
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