Will the destructor of the base class called if an object throws an exception in the constructor?

Question

Will the destructor of the base class be called if an object throws an exception in the constructor?

Answer 1

Yes. The rule is that every object whose constructor has finished successfully will be destructed upon exception. E.g:

class A {
public:
    ~A() {}
};

class B : public A {
public:
    B() { throw 0; }
    ~B() {}
};

~A() is called. ~B() is not called;

EDIT: moreover, suppose you have members:

struct A {
    A(bool t) { if(t) throw 0; }
    ~A() {}
};

struct B {
    A x, y, z;
    B() : x(false), y(true), z(false) {}
};

What happens is: x is constructed, y throws, x is destructed (but neither y nor z).

Answer 2

If an exception is thrown during construction, all previously constructed sub-objects will be properly destroyed. The following program proves that the base is definitely destroyed:

struct Base
{
    ~Base()
    {
        std::cout << "destroying base\n";
    }
};

struct Derived : Base
{
    Derived()
    {
        std::cout << "throwing in derived constructor\n";
        throw "ooops...";
    }
};

int main()
{
    try
    {
        Derived x;
    }
    catch (...)
    {
        throw;
    }
}

output:

throwing in derived constructor
destroying base

(Note that the destructor of a native pointer does nothing, that's why we prefer RAII over raw pointers.)