C++0x unique_ptr replaces scoped_ptr taking ownership?

Question

I used to write code like this:

class P {};

class Q: public P {};

class A {
    // takes ownership
    A(P* p): p_(p) {}

    scoped_ptr<P> p_;
};

A a(new Q);

With C++0x, should I rewrite class A as:

class A {
    // takes ownership
    A(unique_ptr<P>&& p): p_(p) {}

    unique_ptr<P> p_;
};

Answer 1

I've upvoted comonad's answer, but with a caveat:

Whenever you want to explicitely disallow move semantics, use a scoped_ptr const unique_ptr.

I have not come across any use cases where a const std::unique_ptr is inferior to a boost::scoped_ptr. However I'm open to education on the subject.

Edit:

Here is a use case of boost::scoped_ptr that I think should fail, but does not. It does fail for std::unique_ptr:

#include <iostream>

#ifdef USE_UNIQUEPTR

#include <memory>
typedef std::unique_ptr<int> P;

#else  // USE_UNIQUEPTR

#include <boost/scoped_ptr.hpp>
typedef boost::scoped_ptr<int> P;

#endif  // USE_UNIQUEPTR

int main()
{
    P p1(new int(1));
    {
        // new scope
#ifdef USE_UNIQUEPTR
        const P p2(new int(2));
#else  // USE_UNIQUEPTR
        P p2(new int(2));
#endif  // USE_UNIQUEPTR
        swap(p1, p2);  // should fail!
    }
    std::cout << *p1 << '\n';
}

If the promise of boost::scoped_ptr is that its resource will not escape the current scope, then it is not as good at holding that promise as a const std::unique_ptr. If we want to compare const boost::scoped_ptr to const::std::unique_ptr, I have to ask: for what purpose? They seem the same to me, except that a const std::unique_ptr allows customized construction and destruction.