What does BOOST_SCOPE_EXIT do behind the scenes?

Trying to understand where the scope exit blocks would execute with respect to automatic variables' deallocations, I wrote this program:

#include <boost/scope_exit.hpp>
#include <iostream>

class A
{
  public:
    A()
    {
        std::cout << "A's constructor" << std::endl;
    }
    ~A()
    {
        std::cout << "A's destructor" << std::endl;
    }
};

class B
{
  public:
    B()
    {
        std::cout << "B's constructor" << std::endl;
    }
    ~B()
    {
        std::cout << "B's destructor" << std::endl;
    }
};

int main()
{
    A a;
    BOOST_SCOPE_EXIT()
    {
        std::cout << "Scope exit 1" << std::endl;
    } BOOST_SCOPE_EXIT_END
    B b;
    BOOST_SCOPE_EXIT()
    {
        std::cout << "Scope exit 2" << std::endl;
    } BOOST_SCOPE_EXIT_END
    std::cout << std::endl;
    return 0;
}

The output is the following:

A's constructor
B's constructor

Scope exit 2
B's destructor
Scope exit 1
A's destructor

There is a common order among automatic variables and scope exit blocks. I am wondering how does boost achieve this? Does the BOOST_SCOPE_EXIT macro create an object on the stack whose destructor executes the exit block?