为什么unique_ptr不会阻止自定义删除器的切片？

Question

The behavior of std::unique_ptr with custom deleter is based on the static type of the deleter. No polymorphism, no runtime behavior based on actual deleter passed in runtime, as derived deleter provided is being sliced to the static type of the declared deleter.

(It is designed this way in purpose, to allow the size of unique_ptr with default deleter or with custom deleter without any data members , to have same size as a raw pointer).

static behavior of unique_ptr with custom deleter :

class A {};

struct BaseDeleter {
    virtual void operator()(A* p) const {
        std::cout << "in BaseDeleter" << std::endl; 
        delete p;
    }
};

struct DerivedDeleter: BaseDeleter {
    void operator()(A* p) const override {
        std::cout << "in DerivedDeleter" << std::endl; 
        delete p;
    }
};

int main() {
    auto unique_var = std::unique_ptr<A, BaseDeleter>(new A);
    unique_var = std::unique_ptr<A, DerivedDeleter>(new A);
}

Output:

in BaseDeleter
in BaseDeleter

---

This is opposed to std::shared_ptr who holds its custom deleter differently and allowing dynamic behavior:

dynamic behavior of shared_ptr with custom deleter :

int main() {
    auto shared_var = std::shared_ptr<A>(new A, BaseDeleter{});
    shared_var = std::shared_ptr<A>(new A, DerivedDeleter{});
}

Output:

in BaseDeleter
in DerivedDeleter

Code: https://coliru.stacked-crooked.com/a/54a8d2fc3c95d4c1

---

The behavior of assigning std::unique_ptr with different custom deleter is actually slicing .

Why unique_ptr doesn't prevent slicing of custom deleter?

Why didn't the language block the assignment of std::unique_ptr if the assigned unique_ptr has different custom deleter , to avoid slicing?

---

This seems to be possible as presented below.

Blocking unique_ptr from slicing of custom deleter

template<typename TYPE, typename Deleter>
struct my_unique_ptr : std::unique_ptr<TYPE, Deleter> {
    using BASE = std::unique_ptr<TYPE, Deleter>;
    using std::unique_ptr<TYPE, Deleter>::unique_ptr;
    auto& operator=(std::nullptr_t) noexcept {
        return BASE::operator=(nullptr);
    }
    template<typename T, typename OtherDeleter,
      std::enable_if_t<!std::is_same<OtherDeleter, Deleter>::value>* dummy = nullptr>
    auto& operator=(std::unique_ptr<T, OtherDeleter>&& other) = delete;
};

Code: http://coliru.stacked-crooked.com/a/089cd4c7303ad63e

Answer 1

struct B {
  virtual ~B() = default;
};

struct D : B {};

std::unique_ptr<B> b;
b = std::make_unique<D>();

Here we have a classic use case. Yes, the deleter is sliced, but deletion is still well defined. Your proposal will interfere with that. And will probably be very hard to amend reliably into not interfering.

One can always specify a custom deleter like std::function<void(void*)> to get polymorphism by type erasure. Sure, it has overhead, but that's opt in.

By default unique_ptr is optimized for the more common use cases, with less common ones that require overhead being possible by request.