多态unique_ptr的数组

Question

Using unique pointers, we can write

class A{};
class B: public A{};
class C: public A{};
std::vector<std::unique_ptr<A>> myArray;
//
f()
{
    std::unique_ptr<B> pB {new B()};
    std::unique_ptr<C> pC {new C()};
    myArray.push_back(std::move(pB));
    myArray.push_back(std::move(pC));
    // nice! now I can use the interface of A without
    // worrying about its subclasses.
}

However, if classes B and C need to have their own type-dependent custom deleter:

class A{};
class B: public A{};
class C: public A{};

template<class T>
struct custom_deleter
{ // roughly the same implementation as std::default_deleter
    void operator()(T* p)
    {
        CustomFree(p);
    }
};
//
std::vector<std::unique_ptr<A>> myArray; 
//
f()
{
    std::unique_ptr<B, custom_deleter<B>> pB {CustomAlloc(B())};
    std::unique_ptr<C, custom_deleter<C>> pC {CustomAlloc(C())};
    myArray.push_back(std::move(pB)); // error: can't convert to default deleter
    myArray.push_back(std::move(pC)); // item dito
}

Is there a way to achieve what I'm trying to do here?

Answer 1

Your first example has undefined behavior because A doesn't have a virtual destructor. unique_ptr 's default deleter, std::default_delete is stateless, so a unique_ptr<A> will always call delete p; , where the static type of p is A* .

Your code compiles because the default_delete<T> copy constructor can accept another default_delete<U> instance as long as U* is implicitly convertible to T* , which it is in this case. However, polymorphic deletion of the derived class object through base class pointer, when the base class destructor is not virtual is undefined behavior.

---

The second problem of custom deleters for each type can be solved by supplying a custom deleter that is a lambda which first casts the argument to the appropriate type before passing it on to that object's deleter.

For instance,

template<typename T>
void CustomDeleter(T *t)
{
    std::cout << __PRETTY_FUNCTION__ << std::endl;
    delete t;
}

std::vector<std::unique_ptr<A, void(*)(A *)>> myArray;

myArray.push_back(
    std::unique_ptr<A, void(*)(A *)>(new A(), [](A *a){ 
        CustomDeleter(a); }));
myArray.push_back(
    std::unique_ptr<A, void(*)(A *)>(new B(), [](A *a){ 
        CustomDeleter(static_cast<B*>(a)); }));
myArray.push_back(
    std::unique_ptr<A, void(*)(A *)>(new C(), [](A *a){ 
        CustomDeleter(static_cast<C*>(a)); }));

When the vector goes out of scope, this prints:

void CustomDeleter(T*) [with T = A]
void CustomDeleter(T*) [with T = B]
void CustomDeleter(T*) [with T = C]

Live demo

Answer 2

No, the type of the custom deleter is a template parameter to std::unique_ptr . Your easiest solution would be to add a member variable to the custom deleter:

struct custom_deleter
{
    int m_kind;
    custom_deleter(int kind)
    {
        m_kind = kind;
    }
    void operator()(A* p)
    {
        switch (m_kind)
        {
            case 0:
                CustomFree0(p);
                break;
            case 1:
                CustomFree1(p);
                break;
            //...
        }
    }
};

And then make CustomAlloc build the appropriate custom_deleter and pass it to the constructor of std::unique_ptr<A, custom_deleter> as second parameter.