Question

A few hours ago I asked this question. I learned that std::vector deletes each of its elements when its destructor is called. Consider this program (a derivative of a previous example):

#include <vector>
#include <iostream>

class Bar {
    int x;
public:
    ~Bar() {
        std::cout << "~bar()" << std::endl;
    }
};

class Foo {
    std::vector<Bar*> v;
public:
    Foo() {
        this->v.push_back(new Bar());
        this->v.push_back(new Bar());
        this->v.push_back(new Bar());
    }

    ~Foo() {
    }
};

int main() {
    Foo f;

    Bar* b = new Bar();

    // Bar::~Bar() called once
    delete b;

    // Bar::~Bar() not called three times as expected

    return 0;
}

b’s destructor is called as expected; however, the destructors of the Bar* elements in f.v are not called. According to this, the destructor of each element of f.v should be called. What am I missing here?

Answer 1

The answer to this question is exactly the same as the answer to the last question. Deallocating a pointer does not call delete on it, so the thing it points to is not automatically destructed.

Put another way, Bar has a destructor, but Bar* does not. So a std::vector<Bar> would invoke the destructor of each element.