[Basic C++] Base Class Pointer access Derived class member

你是否有遇过这样的情况：

class Base {public:    Base(int a) : A(a) {}    virtual ~Base(){}        // print function in Base class    void printA(){ std::cout << A << endl; }private:    int A;};class Derived : public Base {public:    Derived(int a, int b) : Base(a), B(b) {}    virtual ~Derived(){}        // print function in Derived class    void printB(){ std::cout << B << endl; }private:    int B;};int main(){    Base* b0;    int a=0, b=1;    b0 = new Derived(a,b);    b0->printA();    b0->printB();    return 0;}

在执行到 b0->printB();的时候，compiler会出现以下错误：

1. static_cast<>

这种时候就直接使用大招：

static_cast<Derived*>(b0)->printB();

就可以使用了。

！！！注意！！！
Base class pointer无法存取Derived class member是正常的。任意存取Derived class member其实是非常危险的，Base class pointer不应该知道自己new了什么样的derived class。除非你真的知道你在干嘛，否则就请乖乖写virtual function并用override(c++11)方式实作这些功能。

2. virtual function

在 class Base里面加上：

class Base {public:    Base(int a) : A(a) {}    virtual ~Base(){}        // print function in Base class    void printA(){ std::cout << A << endl; }    virtual printB(){}       // virtual function to be override    // virtual printB() = 0; // = 0: pure virtual function, must be override    private:    int A;};

而在 class Derived 里面加上override保险：

class Derived : public Base {public:    Derived(int a, int b) : Base(a), B(b) {}    virtual ~Derived(){}        // print function in Derived class // override    void printB() override { std::cout << B << endl; }private:    int B;};

这样就不会有compile error了。

3. Reference

https://stackoverflow.com/questions/2436125/c-access-derived-class-member-from-base-class-pointer