编译器如何知道 vtable 中的哪个条目对应于调用的虚函数？

Question

首先，这个问题听起来很重复，但我通过谷歌探索了这个问题，但作为外行没有得到满意的答案

问题： 假设我们在父类和派生类中有 3 个虚函数，如下例所示

class Base1 {
    public :
    virtual void Print1() {
       cout << "Print1 Base1" << endl;
    }
    virtual void Print2() {
       cout << "Print1 Base1" << endl;
    }
    virtual void Print3() {
       cout << "Print1 Base1" << endl;
    }
};

class Derived : public Base1 {
public:  
    void Print1() {
       cout << "Print1 Derived1" << endl;
    }
    void Print2() {
       cout << "Print1 Derived2" << endl;
    }
    void Print3() {
       cout << "Print1 Derived3" << endl;
    }
};

int main()
{
    Derived d;
    Base1 *bptr = &d;  
    bptr->Print2(); // -> Here Derived::Print2() shall be called
    return 0;
 }

如上例所示，在main()中调用Derived::Print2()

这里，基类和派生类vTable应由3个虚函数的3个条目组成

 [0] -> Address of Print1()
 [1] -> Address of Print2()
 [2] -> Address of Print3()

我想知道， 在运行时编译器如何在 Derived vTable 中搜索 Print2() ？

Answer 1

它不搜索。 “Print2 is the second entry in the vtable”的对应关系是在编译时建立的，当编译器查看Base1类的定义时。

Answer 2

没有搜索 - 表不是按名称而是按偏移量（即它是一个数组）。

在实践中，它的工作原理通常很像这个非常简化的类 C 实现：

// Simplified type; the actual implementation doesn't need to care about the type.
typedef void(*function)(void*);

// Base class
struct Base1 {
    function* vtable;
};

// Derived class
struct Derived
{
    Base1 parent; // The base class subobject    
};

// Virtual members
void Base1_Print1(void*) {
    cout << "Print1 Base1" << endl;
}

void Base1_Print2(void*) {
    cout << "Print2 Base1" << endl;
}

// v-table
function Base1_vtable[2] = { &Base1_Print1, &Base1_Print2 };

// Constructor
void Base1_create(Base1* self)
{
    self->vtable = Base1_vtable;
}

// Overridden members
void Derived_Print2(void*) {
    cout << "Print2 Derived" << endl;
}

// v-table
function Derived_vtable[2] = { &Base_Print1, &Derived_Print2 };

// Constructor
void Derived_create(Derived* self)
{
    Base1_create(&self->parent); // Construct the base subobject
    self->parent.vtable = Derived_vtable; // Adjust the vtable
}

int main()
{
    Derived d;
    Derived_create(&d); // Construct d
    Base1 *bptr = &d.parent; // Implicit conversion made explicit.
    (bptr->vtable[0])(bptr); // Virtual call to Base_Print1
    (bptr->vtable[1])(bptr); // Virtual call to Derived_Print2
 }