调用`std::unique_ptr` move构造函数时“使用已删除的函数”？

#include <iostream>
#include <memory>

struct CustomDeleter {
    void operator()(int* ptr) const {
        std::cout << "Deleting pointer" << std::endl;
        delete ptr;
    }

    // 确保删除器支持移动语义
    CustomDeleter(CustomDeleter&&) = default;
    CustomDeleter& operator=(CustomDeleter&&) = default;
};

int main() {
    std::unique_ptr<int, CustomDeleter> ptr1(new int(42));
    std::unique_ptr<int, CustomDeleter> ptr2(std::move(ptr1));
    return 0;
}

#include <iostream>
#include <memory>

struct NonMovable {
    NonMovable() = default;
    ~NonMovable() = default;

    // 禁止移动构造函数和移动赋值操作符
    NonMovable(NonMovable&&) = delete;
    NonMovable& operator=(NonMovable&&) = delete;
};

int main() {
    // 这将导致编译错误，因为 NonMovable 是不可移动的
    std::unique_ptr<NonMovable> ptr1(new NonMovable);
    std::unique_ptr<NonMovable> ptr2(std::move(ptr1));
    return 0;
}

#include <iostream>
#include <memory>

struct Movable {
    Movable() = default;
    ~Movable() = default;

    // 确保移动构造函数没有被删除
    Movable(Movable&&) = default;
    Movable& operator=(Movable&&) = default;
};

int main() {
    std::unique_ptr<Movable> ptr1(new Movable);
    std::unique_ptr<Movable> ptr2(std::move(ptr1));
    return 0;
}