unique_ptr as class member and move semantics fail

I can't get clang (Apple LLVM version 4.2 (clang-425.0.28)) to compile these classes:

struct A {
    int f(){return 2;}
};
class Cl{
     std::unique_ptr<A> ptr;

public:
     Cl(){ptr = std::unique_ptr<A>(new A);}

     Cl(const Cl& x) : ptr(new A(*x.ptr)) { }
     Cl(Cl&& x) : ptr(std::move(x.ptr)) { }
     Cl(std::unique_ptr<A> p) : ptr(std::move(p))  { }

    void m_ptr(std::unique_ptr<A> p){
        ptr = std::unique_ptr<A>(std::move(p));
    }
    double run(){return ptr->f();}
};

I would like to run the constructor as follows:

std::unique_ptr<A> ptrB (new A);
Cl C = Cl(ptrB);

but if I do this I get the following compiler error: ../src/C++11-2.cpp:66:10: error: call to implicitly-deleted copy constructor of 'std::unique_ptr' C.m_ptr(ptrB);

I can solve the compiler problem by running Cl(std::move(ptrB)) but this doesn't actually move the ownership of A away from ptrB: I can still run ptrB->f() without causing a run-time crash... Secondly, the constructor is not very satisfying, since I want to hide the implementation of std::move in the class interface.

Thanks in advance.

Since ptrB is passed by value to Cl's copy constructor, a call to Cl(ptrB) tries to create a copy of ptrB which in turn calls a (obviously disabled) copy constructor of unique_ptr. In order to avoid creating an extra copy of ptrB, do the following:

Cl C = Cl(std::unique_ptr<A>(new A)); //A temporary is created on initialization, no extra copy steps performed

Or:

std::unique_ptr<A> ptrB (new A);
Cl C = Cl(std::move(ptrB)); //Move semantics used. Again, no extra copy steps

Or, use pass by reference (rvalue or lvalue) in your copy constructor:

class Cl{

//...
public:
//...
     Cl(std::unique_ptr<A> &p) : ptr(std::move(p))  { }

//...

};

std::unique_ptr<A> ptrB (new A);
Cl C = Cl(ptrB);

P.S Oh and by the way: the objects stay in unspecified, but valid state after std::move(). I believe that means you can still call ptrB->f(), and it is guaranteed to return 2 :)