Why does unique_ptr<Derived> implicitly cast to unique_ptr<Base>?

Question

I wrote the following code that uses unique_ptr<Derived> where a unique_ptr<Base> is expected

class Base {
    int i;
 public:
    Base( int i ) : i(i) {}
    int getI() const { return i; }
};

class Derived : public Base {
    float f;
 public:
    Derived( int i, float f ) : Base(i), f(f) {}
    float getF() const { return f; }
};

void printBase( unique_ptr<Base> base )
{
    cout << "f: " << base->getI() << endl;
}

unique_ptr<Base> makeBase()
{
    return make_unique<Derived>( 2, 3.0f );
}

unique_ptr<Derived> makeDerived()
{
    return make_unique<Derived>( 2, 3.0f );
}

int main( int argc, char * argv [] )
{
    unique_ptr<Base> base1 = makeBase();
    unique_ptr<Base> base2 = makeDerived();
    printBase( make_unique<Derived>( 2, 3.0f ) );

    return 0;
}

and i expected this code to not compile, because according to my understanding unique_ptr<Base> and unique_ptr<Derived> are unrelated types and unique_ptr<Derived> isn't in fact derived from unique_ptr<Base> so the assignment shouldn't work.

But thanks to some magic it works, and i don't understand why, or even if it's safe to do so. Can someone explain please?

Answer 1

The bit of magic you're looking for is the converting constructor #6 here :

template<class U, class E>
unique_ptr(unique_ptr<U, E> &&u) noexcept;

It enables constructing a std::unique_ptr<T> implicitly from an expiring std::unique_ptr<U> if (glossing over deleters for clarity):

unique_ptr<U, E>::pointer is implicitly convertible to pointer

Which is to say, it mimicks implicit raw pointer conversions, including derived-to-base conversions, and does what you expect™ safely (in terms of lifetime – you still need to ensure that the base type can be deleted polymorphically).

Answer 2

Because std::unique_ptr has a converting constructor as

template< class U, class E > unique_ptr( unique_ptr<U, E>&& u ) noexcept;

and

This constructor only participates in overload resolution if all of the following is true:

a) unique_ptr<U, E>::pointer is implicitly convertible to pointer

...

A Derived* could convert to Base* implicitly, then the converting constructor could be applied for this case. Then a std::unique_ptr<Base> could be converted from a std::unique_ptr<Derived> implicitly just as the raw pointer does. (Note that the std::unique_ptr<Derived> has to be an rvalue for constructing std::unique_ptr<Base> because of the characteristic of std::unique_ptr .)

Answer 3

You can implicitly construct a std::unique_ptr<T> instance from an rvalue of std::unique_ptr<S> whenever S is convertible to T . This is due to constructor #6 here . Ownership is transferred in this case.

In your example, you have only rvalues of type std::uinque_ptr<Derived> (because the return value of std::make_unique is an rvalue), and when you use that as a std::unique_ptr<Base> , the constructor mentioned above is invoked. The std::unique_ptr<Derived> objects in question hence only live for a short amount of time, ie they are created, then ownership is passed to the std::unique_ptr<Base> object that is used further on.