c++11 - unique_ptr::operator= 在释放方面做了什么

Question

我无法完全理解 unique_ptr 的赋值运算符。我知道我们只能移动它们，因为复制构造函数和赋值运算符被删除了，但是如果已经包含分配的 unique_ptr 被移动操作覆盖了吗？之前存储在智能指针中的内容是否被释放？

#include <iostream>
#include <memory>

class A{
public:
    A() = default;
    virtual void act() const {
        std::cout << "act from A" << std::endl;
    }
    virtual ~A() {
        std::cout << "destroyed A" << std::endl;
    }
};
class B : public A {
public:
    B() : A{} {}
    void act() const override {
        std::cout << "act from B" << std::endl;
    }

    ~B() override {
        std::cout << "destroyed from B " << std::endl;
    }
};
int main() {
    auto pP{std::make_unique<A>()};
    pP->act();

    ==================== ! =======================
    pP = std::make_unique<B>(); // || std::move(std::make_unique<B>())
    ==================== ! =======================

    pP->act();
    return 0;
}

当我做的时候

pP = std::make_unique<B>();

这是否意味着在第一行分配给 pP (new A()) 的内容会自动销毁？或者我应该选择:

pP.reset();
pP = std::make_unique<B>();

Answer 1

是的，请参阅 C++11 draft standard 的第 20.9.1 节第 4 段

Additionally, u can, upon request, transfer ownership to another unique pointer u2. Upon completion of such a transfer, the following postconditions hold:

• u2.p is equal to the pre-transfer u.p,
• u.p is equal to nullptr, and
• if the pre-transfer u.d maintained state, such state has been transferred to u2.d.

As in the case of a reset, u2 must properly dispose of its pre-transfer owned object via the pre-transfer associated deleter before the ownership transfer is considered complete

换句话说，它会像您期望的那样在分配后自行清理。