C++ 中的原型设计模式示例

问：

我正在学习原型设计模式并实现演示，我将深度复制构造函数和克隆函数添加到这些类中，但我不确定我是否正确实现了它们。我在otherNode中做了dynamic_cast转换，这有点奇怪。有什么建议吗？

class base {
public:
    explicit base() {}
    virtual ~base() = default;
    virtual base* clone() const = 0;
};

class ttNode : public base {
public:
    explicit ttNode() : base() {}
    ~ttNode() = default;
};

class SonNode : public ttNode {
public:
    SonNode() : ttNode() {}
    ~SonNode() = default;

    base* clone() const override {
        return new SonNode(*this);
    }
private:
    int c_{0};
    int a_{0};
};

class FatherNode : public ttNode {
public:
    FatherNode() : ttNode() {}
    ~FatherNode() = default;

    FatherNode(const FatherNode& node) : ttNode(node) {
        for (const auto &son : node.sons_) {
            sons_.emplace_back(new SonNode(*son));
        }
    }
    base* clone() const override {
        return new FatherNode(*this);
    }

private:
    std::vector<SonNode*> sons_;
};

class otherNode : public base {
public:
    otherNode() : base() {}
    ~otherNode() = default;
    base* clone() const override { return new otherNode(*this); }
    otherNode(const otherNode& node) : base(node) {
        ttt = dynamic_cast<ttNode*>(node.ttt->clone());
    }
private:
    ttNode* ttt;
};


TEST_F(tt, base1) {
    base *o = new otherNode();
    base *f = new FatherNode();
    base *o1 = o->clone();
    base *f1 = f->clone();
    delete o1;
    delete f1;
}

我不能使用，因为ttNode是抽象类。ttt = new ttNode(*node.ttt);