Re: C++0x: unique_ptr and std::move

I wanted o understand how rvalue references work, so I took GCC 4.3
with -std=c++0x flag and wrote code below.

What I don't understand is why 2nd assertion fails and move ctor is
not called. Please enlighten me :)

#include <iostream>
#include <cassert>

using std::cout;
using std::endl;
using std::move;

template<typename T>
class unique_ptr {
public:
    explicit unique_ptr(T *&&a_ptr): m_ptr(a_ptr) {
        a_ptr = NULL;
    }

    unique_ptr(unique_ptr &&p): m_ptr(p.release()) {
        cout << "Move" << endl;
    }

    T *release() {
        T *ptr = m_ptr;
        m_ptr = NULL;
        return ptr;
    }

    T *get() {
        return m_ptr;
    }

    T *operator->() {
        return m_ptr;
    }

    ~unique_ptr() {
        if(m_ptr != NULL) {
            delete m_ptr;
        }
    }

private:
    unique_ptr(const unique_ptr &);
    void operator=(const unique_ptr &);
    void operator=(unique_ptr &&p);

    T *m_ptr;

};

struct Foo
{
    Foo(int a): a(a) {
        cout << "Foo::ctor(" << a << ")" << endl;
    }

    ~Foo() {
        cout << "Foo::dtor()" << endl;
    }

    int a;
private:
    Foo(const Foo &);
    Foo(Foo &&);

};

unique_ptr<Foo> source(int a = 0) {
    return move(unique_ptr<Foo>(new Foo(a)));

}

void sink(unique_ptr<Foo> a_foo) {
    cout << a_foo->a << endl;

}

int main() {
    unique_ptr<Foo> foo( source(1) );
    unique_ptr<Foo> bar = move(foo);
    assert(foo.get() == NULL); // ok

    unique_ptr<Foo> qux( source(2) );
    sink( move(qux) );
    assert(qux.get() == NULL); // ??
}

Side question: does source() function look all right?

unique_ptr<Foo> source(int a = 0) {
    return move(unique_ptr<Foo>(new Foo(a)));