Why is the destructor of the class called twice?

Question

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.

This will show you what's happening:

struct A {
  A() { cout << "contruction\n"; }
  A(A const& other) { cout << "copy construction\n"; }
  ~A() { cout << "destruction\n"; }
};

int main() {
  vector<A> t;
  t.push_back(A());
}

The destructor is called once when the temporary sent to push_back is destroyed and once when the element in t is destroyed.

There are two destructor calls because there are two objects: the argument to push_back, and the newly added element within vector t.

STL containers store copies. In your example the element added to the vector by push_back is copy constructed from the argument passed to push_back. The argument is A(), which is a temporary object, see here (variant 4).

Expanding the answer a bit, altough you havent explicitely asked for it: It might be useful to know when the temporary is destroyed. The standard (N4140) sais it pretty clearly in 12.2 p3:

... Temporary objects are destroyed as the last step in evaluating the full-expression (1.9) that (lexically) contains the point where they were created...

Side note: If you use emplace_back there is only one object. The new element in the container is directly constructed from the arguments to emplace_back. Many STL container learned an emplace variant in C++11.