从Derived类返回值到Base类(Return value from Derived class to Base class)

从Derived类返回值到Base类(Return value from Derived class to Base class)

我在派生类中有一个值，我希望使用基类中的函数返回，这可能吗？ 或者我必须在基类中声明变量才能这样做？

我只是在派生类中调用该函数吗？

class Base { public: int getNum() const { return number; } }; class Derived : public Base { private: int n = 50; };

I have a value in the derived class that I want to return using a function from the base class, is that possible? Or do I have to have the variable declared in the base class to do so?

Would I just call the function in the derived class?

class Base { public: int getNum() const { return number; } }; class Derived : public Base { private: int n = 50; };

最满意答案

我在派生类中有一个值，我希望使用基类中的函数返回，这可能吗？

是的，您可以使用传递给基类构造函数的引用，如下所示：

class Base { public: Base(int& number_) : number(number_) {} int getNum() const { return number; } private: int& number; }; class Derived : public Base { public: Derived() : Base(n) {} private: int n = 50; };

---

另一种方法（不使用virtual getNum() const; ）使用模板化基类（又名静态多态）：

template <typename D> class Base { public: int getNum() const { return static_cast<D*>(this)->getNum(); } }; class Derived : public Base<Derived> { public: Derived() {} int getNum() const { return n; } private: int n = 50; };

I have a value in the derived class that I want to return using a function from the base class, is that possible?

Yes, you can use a reference passed to the base classes constructor like so:

class Base { public: Base(int& number_) : number(number_) {} int getNum() const { return number; } private: int& number; }; class Derived : public Base { public: Derived() : Base(n) {} private: int n = 50; };

---

Another way to do it (without using virtual getNum() const;) is using a templated base class (aka static polymorphism):

template <typename D> class Base { public: int getNum() const { return static_cast<D*>(this)->getNum(); } }; class Derived : public Base<Derived> { public: Derived() {} int getNum() const { return n; } private: int n = 50; };