c++ - 在参数化虚拟中会发生什么？

Question

C++ 模板 - 完整指南 中的第 16.4 章参数化虚拟性，由 David Vandevoorde 和 Nicolai M. Josuttis 撰写，他说:

C++ allows us to parameterize directly three kinds of entities through templates: types, constants ("nontypes"), and templates. However, indirectly, it also allows us to parameterize other attributes such as the virtuality of a member function.

该章中说明了以下代码:

#include <iostream>

struct NotVirtual
{};

struct IsVirtual
{
    virtual void func() {;}
};

//---------------------//

template<typename T>
struct Base : T
{
    void func()
    { std::cout<< "Base::func()" <<std::endl; }
};

template<typename T>
struct Derived : Base<T>
{
    void func()
    { std::cout<< "Derived::func()" <<std::endl; }
};

//---------------------//

int main()
{
    Base<NotVirtual> *p1 = new Derived<NotVirtual>();
    p1->func();
    delete p1;

    Base<IsVirtual> *p2 = new Derived<IsVirtual>();
    p2->func();
    delete p2;
}

在线示例:https://rextester.com/OAGC66937

我理解这种技术的用法，但不幸的是，这本书没有提供任何关于这件事是如何发生的进一步细节。可以看出Base其实是一个继承自模板参数T的派生类。

问题:

1. 参数化虚拟化期间后台发生了什么？
2. 生成的 v_table 中是否发生了什么？
3. 模板参数中的virtual是如何传输过来的？

Answer 1

扩展@Quentin 的评论

Once the Base template is specialized for a specific T, everything proceeds as usual through inheritance and virtual overriding.

让我们写出一组等效的非模板类

struct BaseNV : NotVirtual
{
    void func() // first definition of func
    { std::cout<< "Base::func()" <<std::endl; }
};

struct DerivedNV : BaseNV 
{
    void func() // hides BaseNV::func
    { std::cout<< "Derived::func()" <<std::endl; }
};

struct BaseIV : IsVirtual
{
    void func() // implicitly virtual, overrides IsVirtual::func
    { std::cout<< "Base::func()" <<std::endl; }
};

struct DerivedIV : BaseIV
{
    void func() // implicitly virtual, overrides BaseIV::func (and IsVirtual::func)
    { std::cout<< "Derived::func()" <<std::endl; }
};

int main()
{
    BaseNV *p1 = new DerivedNV();
    p1->func();
    delete p1;

    BaseIV *p2 = new DerivedIV();
    p2->func();
    delete p2;
}

See it live

What happens in the background during parameterized virtuality?

Base<T>继承自 T , 并遵循继承规则

Does something happen in the resulting v_table?

只有普通virtual机制。具体来说，在 NotVirtual如果没有任何虚函数，那么可能不会有任何 vtable。

How does the virtual from the template parameter get transferred over?

按照正常的继承规则