c++中subtype的对象使用哪个vtable

Question

当我要执行这行代码时，它没有编译

    baseArray[1]->function5();

但是在Derived 3的vtable中，function5应该指向Derived1中的function5。所以我认为它应该输出 Derived1::function5()。但它似乎在 Base class

中找到 function5()

我只是想知道Derived3继承自Derived1，为什么它仍然检查调用的函数是否存在于Base class？

请参考以下代码。感谢您的帮助！

#include <iostream>
using namespace std;

class Base {
public:
  virtual void function1();
  virtual void function2();
  virtual void function3() = 0;
  void function4();
};

class Derived1 : public Base {
public:
  void function1();
  virtual void function5();
  void function6();
};

class Derived2 : public Base {
public:
  void function2();
  void function3();
  void function4();
  void function5();
};

class Derived3 : public Derived1 {
public:
  void function3();
};

void Base::function1(){
  cout << "Base::function1()" << endl;
}

void Base::function2(){
  cout << "Base::function2()" << endl;
}

void Base::function4(){
  cout << "Base::function4()" << endl;
}

void Derived1::function1(){
  cout << "Derived1::function1()" << endl;
}

void Derived1::function5(){
  cout << "Derived1::function5()" << endl;
}

void Derived1::function6(){
  cout << "Derived1::function6()" << endl;
}

void Derived2::function2(){
  cout << "Derived2::function2()" << endl;
}

void Derived2::function3(){
  cout << "Derived2::function3()" << endl;
}

void Derived2::function4(){
  cout << "Derived2::function4()" << endl;
}

void Derived2::function5(){
  cout << "Derived2::function5()" << endl;
}

void Derived3::function3(){
  cout << "Derived3::function3()" << endl;
}

int main(){
  Base *baseArray[] = { new Derived2, new Derived3 };
  baseArray[1]->function5(); //why the code doesn't compile when I call function5() here? 
                             //in vtable of Derived 3, function5 should points to function5 in Derived1
                             //so I think it should output Derived1::function5()

  dynamic_cast<Derived3*>(baseArray[1])->function5();
  
  baseArray[1]->function6(); //same question, why it doesn't call function6 in Derived1?
                             //which vtable does baseArray[1] use?

  dynamic_cast<Derived3*>(baseArray[1])->function6();
}

Answer 1

C++ 是一种静态类型语言。这意味着必须在编译时知道对类型、成员、变量等的引用。

baseArray[1]->function5(); 和 baseArray[1]->function6(); 不编译的原因是因为 baseArray[1] 被声明为 Base* 而 function5() 和 function6()不是 Base 的成员。就那么简单。这与 vtables 无关。这只是编译器如何选择在运行时实现虚拟方法调用调度的实现细节，但这不是 C++ 标准所要求的。

编译器不知道 Base* 指针实际上指向一个 Derived3 对象。这个事实直到运行时才确定。这是多态性的一个关键特征。

要正确调用 function5() 和 function6()，您需要将 Base* 指针类型转换为具有您要访问的成员的派生类型，因为您有已经发现了。请注意，如果转换失败，指针上的 dynamic_cast 将 return nullptr，您没有检查，例如：

// Derived1 and Derived2 are not related, but each has its own function5()...
if (Derived1 *d1 = dynamic_cast<Derived1*>(baseArray[index]))
    d1->function5();
else if (Derived2 *d2 = dynamic_cast<Derived2*>(baseArray[index]))
    d2->function5();

另一方面，如果您预先知道确切和正确的类型（如您的示例所示），则可以改用 static_cast，例如：

static_cast<Derived1*>(baseArray[index])->function5();

c++中subtype的对象使用哪个vtable

Which vtable does the object of subtype use in c++

c++

polymorphism

virtual

vtable