“向下转型” unique_ptr<Base> 到 unique_ptr<Derived> 使用 unique_ptr<Data> 在派生

Question

“Downcasting” unique_ptr< Base > to unique_ptr< Derived > offer an elegent solution to downcasting unique_ptr. It works in most of time. But when the Derived contains unique_ptr, something go wrong:

template<typename Derived, typename Base, typename Del>
std::unique_ptr<Derived, Del> 
static_unique_ptr_cast( std::unique_ptr<Base, Del>&& p )
{
    auto d = static_cast<Derived *>(p.release());
    return std::unique_ptr<Derived, Del>(d, std::move(p.get_deleter()));
} 

struct Data
{
   int data;
};

struct Base
{
};

struct Derived : public Base
{
    Derived() 
        : data(std::make_unique<Data>())

    std::unique_ptr<Data> data;
};

int main()
{
    std::unique_ptr<Base> base = std::make_unique<Derived>();

    auto data = static_unique_ptr_case<Derived>(std::move(base))->data; // compile error

    return 0;
}

Is there a better way to fix the problem?

Eidt:

fix the typos and

@Igor Tandetnik give a solution

std::unique_ptr<Base> base = std::make_unique<Derived>();

//auto& data = static_unique_ptr_case<Derived>(std::move(base))->data; // compile error

auto derived = static_unique_ptr_case<Derived>(std::move(base));

auto& data = derived->data;

return 0;

Answer 1

The online documentation states this about unique_ptr :

The class satisfies the requirements of MoveConstructible and MoveAssignable, but not the requirements of either CopyConstructible or CopyAssignable.

So you cannot copy construct or copy assign a unique_ptr as you are trying to do in the line:

auto derived = static_unique_ptr_cast<Derived>(std::move(base))->data; // compile error