Constructor/Destructor call order on stack

Question

I have the following simple code:

class A
{
    int a;
public:
    A(int a) : a(a) { cout << "Constructor a=" << a << endl; }
    ~A()            { cout << "Destructor  a=" << a << endl; }
    void print()    { cout << "Print       a=" << a << endl; }
};

void f()
{
    A a(1);
    a.print();
    a = A(2);
    a.print();
}

int main()
{
    f();
    return 0;
}

The output is:

Constructor a=1
Print       a=1
Constructor a=2
Destructor  a=2
Print       a=2
Destructor  a=2

I find that there are two destructor calls with a=2 and none with a=1 while there is one constructor call for each case. So how are contructors and destructros called in this case?

Answer 1

a = A(2);

Will use default operator= to assign new value to a, setting it's a::a member value to 2.

void f()
{
    A a(1);//a created with int constructor a.a == 1
    a.print();// print with a.a == 1
    a = A(2);//Another A created with int constructor setting a.a == 2 and immediately assigning that object to a
    //object created with A(2) deleted printing 2 it was created with
    a.print();//a.a==2 after assign
}//a deleted printing 2 it was assigned with

You probably should read about Rule of three to get better understanding what's going on.