Why is the destructor of the class called twice?

c++ destructor
dicaprio picture dicaprio · Apr 13, 2010 · Viewed 16.5k times · Source

Apologies if the question sounds silly, I was following experts in SO and trying some examples myself, and this is one of them. I did try the search option but didn't find an answer for this kind.

class A
{
    public:
         A(){cout<<"A Contruction"<<endl;}
        ~A(){cout<<"A destruction"<<endl;}
};

int main()
{
    vector<A> t;
    t.push_back(A()); // After this line, when the scope of the object is lost.
}

Why is the destructor of the class called twice ?

Answer

sharptooth picture sharptooth · Apr 13, 2010

To add the element a copy constructor is invoked on a temporary object. After the push_back() the temporary object is destroyed - that't the first destructor call. Then vector instance goes out of scope and destroys all the elements stored - that's the second destructor call.

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