在派生的 object 上调用基础 function

Question

class base{
   public:
   virtual void foo(){
     std::cout << "base::foo was called" << std::endl;;
   }
 
   void bar(){
     foo();
     std::cout << "base::bar was called" << std::endl;;
   }
};

class derived : public base {
  public:
  void foo() {
    std::cout << "derived::foo was called" << std::endl;
  }
};

int main() {
    derived der;
    der.bar();
    // desired output = "base::foo was called" "base::bar was called"
    // actual output = "derived::foo was called" "base::bar was called"

}

我在这里遗漏了一些明显的东西吗？
Why does the bar() function when called on an object of derived class call the derived::foo function, even though the function itself only exists in the base class.

Answer 1

base::foo被声明为virtual ，因此使用虚拟调度。 如果您不希望这样，请显式调用base::foo ：

struct  base{
   virtual void foo(){
     std::cout << "base::foo was called" << std::endl;;
   }
 
   void bar(){
     base::foo();
     std::cout << "base::bar was called" << std::endl;;
   }
};

（请注意，在您的示例中，所有方法都是私有的，我想这只是一个错字）

Answer 2

When a member function in a derived class overrides a virtual member function in a base class like this, that means it entirely replaces the function definition for most purposes. 因此，在main中创建的 object der上调用 function foo通常将使用Derived::foo定义。 无论调用foo的代码是在Derived的成员中，还是在Base的成员中，或者两者都不是 class，此行为都是一致的。

两个主要的例外是：

1. 如果您使用“qualified-id” class_type::func_name语法来调用 function，这将禁用虚拟 function 逻辑，并且只调用 ZC1C425268E68385D14AB5074C17A 重载解析您命名的 ZC1C425268E68385D14AB5074C17A 重载Z9。

因此，既然您说您希望程序调用base::foo ，请将base::bar更改为：

void bar() {
    base::foo();
    std::cout << "base::bar was called" << std::endl;
}

2. 在基 class 的构造函数或析构函数期间，基 class（或其基之一）中的 function 将被调用，忽略派生类中的覆盖器。 object 被认为还不是或不再是派生类型的 object。 （如果派生的 function 被调用并使用任何尚未创建或已销毁的派生成员，这将很麻烦。）

Answer 3

When you declare a function virtual , the compiler will generate a vtable for each object created, and in your case, since the object is a derived the function vtable will always point to derived::foo() for that instance.