c++ unique_ptr argument passing

Question

Suppose I have the following code:

class B { /* */ };

class A {
    vector<B*> vb;
public:
    void add(B* b) { vb.push_back(b); }
};


int main() {

    A a;
    B* b(new B());
    a.add(b);
}

Suppose that in this case, all raw pointers B* can be handled through unique_ptr<B>.

Surprisingly, I wasn't able to find how to convert this code using unique_ptr. After a few tries, I came up with the following code, which compiles:

class A {
    vector<unique_ptr<B>> vb;
public:
    void add(unique_ptr<B> b) { vb.push_back(move(b)); }
};


int main() {

    A a;
    unique_ptr<B> b(new B());
    a.add(move(b));
}

So my simple question: is this the way to do it and in particular, is move(b) the only way to do it? (I was thinking of rvalue references but I don't fully understand them.)

And if you have a link with complete explanations of move semantics, unique_ptr, etc. that I was not able to find, don't hesitate to share it.

EDIT According to http://thbecker.net/articles/rvalue_references/section_01.html, my code seems to be OK.

Actually, std::move is just syntactic sugar. With object x of class X, move(x) is just the same as:

static_cast <X&&>(x)

These 2 move functions are needed because casting to a rvalue reference:

1. prevents function "add" from passing by value
2. makes push_back use the default move constructor of B

Apparently, I do not need the second std::move in my main() if I change my "add" function to pass by reference (ordinary lvalue ref).

I would like some confirmation of all this, though...

Answer 1

Yes, this is how it should be done. You are explicitly transferring ownership from main to A. This is basically the same as your previous code, except it's more explicit and vastly more reliable.