腌制在装饰器中定义其类的对象时发生错误

Question

我正在尝试腌制一个用类包装器装饰的对象

BaseObj

但是当我用pickle / dill / joblib编写Can't pickle <class '{package}.{file}.Wrapper'>: it's not found as {package}.{file}.Wrapper 的实例时，我得到了

template<typename T>
class Numeric
{
public:
    Numeric() : val(T()) { cout << "ctor default\n"; }
    explicit Numeric(const T& v) : val(v) { cout << "ctor value\n"; }
    Numeric(const Numeric& v) : val(v.val) { cout << "copy ctor\n"; }
    Numeric(Numeric&& v) { val = v.val; cout << "cmove\n"; v.val = 0; }
    Numeric& operator=(const Numeric& v) { val = v.val; cout << "copy assignment\n"; return *this; }
    Numeric& operator=(Numeric&& v) { val = v.val;cout << "amove\n"; return *this; }
    ~Numeric() { cout << "dtor\n"; };

private:
    T val;

};

// ----------- main ------
Numeric<int> c1(Numeric<int>(2)); // calls the normal constructor instead of copy constructor

任何人都有想法/建议。实际位置不是{package}。{file} .Wrapper，因为它是在装饰器中定义的？

谢谢！