unique_ptr_U.cpp

//
// Created by light on 19-12-12.
//

#include "../RAII/shape.h"

template<typename T>
class unique_ptr {
public:
    explicit unique_ptr(
            T *ptr = nullptr) noexcept
            : ptr_(ptr) {}

    ~unique_ptr() noexcept {
        delete ptr_;
    }

    T &operator*() const noexcept { return *ptr_; }

    T *operator->() const noexcept { return ptr_; }

    operator bool() const noexcept { return ptr_; }

    T *get() const noexcept { return ptr_; }



    template<typename U>
    unique_ptr(unique_ptr<U> &&other) noexcept {
        cout << "U move ctor" << endl;
        ptr_ = other.release();
    }


    // copy and swap  始终只有一个对象有管理这块空间的权限
    unique_ptr &operator=(unique_ptr rhs) noexcept {
        rhs.swap(*this);
        return *this;
    }

    // 原来的指针释放所有权
    T *release() noexcept {
        T *ptr = ptr_;
        ptr_ = nullptr;
        return ptr;
    }

    void swap(unique_ptr &rhs) noexcept {
        using std::swap;
        swap(ptr_, rhs.ptr_);    // 转移指针所有权
    }


private:
    T *ptr_;
};

int main() {
    unique_ptr<shape> ptr1{create_shape(shape_type::circle)};
    unique_ptr<shape> ptr2{ptr1}; // 由于带模板的移动构造函数引发编译器会默认生成拷贝构造
    if (ptr1.get() != nullptr)      // bitwise copy 此时ptr1不为NULL
        ptr2.get()->print();

    unique_ptr<shape> ptr2_2{std::move(ptr1)};    // 调用的是默认的移动构造,而不是带模板的移动构造 bitwise move
    if (ptr2_2.get() != nullptr && ptr1.get() != nullptr)   // ptr1 不为空
        ptr2_2.get()->print();

    unique_ptr<shape> ptr3{create_shape(shape_type::rectangle)};
    ptr1 = ptr3;    // ok  根据形参先调用默认拷贝,再调用拷贝赋值
    ptr3 = std::move(ptr1); // ok 根据形参先调用默认移动构造，而不是带参数的移动构造,再调用移动赋值
    unique_ptr<shape> ptr4(std::move(new circle));  // ok 调用带模板的移动构造
}