Why doesn't the destructor get called? - c++

Question

I am using RAII and using try/catch to not leak memory for example. Here is an implementation in C++:
#include <iostream>
using namespace std;
void g(){
throw 5;
}
class Rand{
public:
~Rand(){
cout << "Randomm Destructor called" << endl;
}
int a = 17;
};
void f(){
auto p = std::make_unique<Rand>(); //Should call dtor
Rand* r = new Rand(); //Shouldnt call dtor
cout << p->a << endl; //Prints 17
g();
cout << "This never executes" << endl;
}
int main(){
f();
}
Due to stackunwinding and using RAII with the std::unique_ptr, shouldn't the destructors for stack allocated objects be called as a basic guarantee to throw/try since an exception is being thrown?

Answer 1

From throw:
Stack unwinding
As the control flow moves up the call stack, destructors are invoked
for all objects with automatic storage duration constructed, but not
yet destroyed, since the corresponding try-block was entered, in
reverse order of completion of their constructors.
There is no corresponding try-block in your code, so no destructors are called and the program is terminated.
If you change the program as:
try
{
auto p = std::make_unique<Rand>(); //Should call dtor
Rand* r = new Rand(); //Shouldnt call dtor
cout << p->a << endl; //Prints 17
g();
cout << "This never executes" << endl;
}
catch (int) {}
you will see, that the destructor for the object which is wrapped into unique_ptr is called.