When does the program not call the destructor in c++?

Question

So here's the code:

#include <iostream>

using namespace std;

class C {
public:
    C() {i = 6; cout << "A:" << i << endl;}

    C(int i0) {i = i0; cout << "B:" << i << endl;}

    ~C() {cout << "C:" << i << endl;}
private:
    int i;
};

int main(int argc, char* argv[]) {
    cout << "X" << endl;
    C *c = new C;
    cout << "Y" << endl;
}

For some reason the output for that code is

X
A:6
Y

And for some reason the destructor (C:6) is never called once you reach the end of the code. Why is that? Also this code does call the destructor:

#include <iostream>

using namespace std;

class C {
public:
    C() {i = 0; cout << "A:" << i << endl;}

    C(int i0) {i = i0; cout << "B:" << i << endl;}

    ~C() {cout << "C:" << i << endl;}
private:
    int i;
};

int main(int argc, char* argv[]) {
    cout << "X" << endl;
    C c;
    cout << "Y" << endl;
}

Answer 1

Because you forgot to write

delete c;

If you just go on in your program without deleting a variable instantiated with new you will cause a memory leak.

Edit, since you edited your question:

If you write something like

C c;
C c{1};
C c = C{1};

you create a variable with automatic storage duration. It will run out of scope once the function it is declared in exits (or more precise: once the block it is declared in exits). In this case the constructor is called automatically.

If you write

C* c = new C{};

you create a pointer to a (new) C. The pointer itself has automatic storage duration, which means c will run out of scope as well. But the pointer only holds the adress of the object of type C. And this object is only deleted if you call delete c;. If you don't call delete, your program "forgets" the address of the object but it does not free the memory or destroy the object. That's a memory leak.
However once your program ends, all memory is freed (without calling destructors), so in your small example you wont notice.