基类的虚函数调用 [英] virtual function call from base class

问题描述

说我们有：

Class Base
{   
    virtual void f(){g();};
    virtual void g(){//Do some Base related code;}
};

Class Derived : public Base
{   
    virtual void f(){Base::f();};
    virtual void g(){//Do some Derived related code};
};

int main()
{
    Base *pBase = new Derived;
    pBase->f();
    return 0;  
}

其中 g code>将从 Base :: f（）调用？ Base :: g（）或 Derived :: g（）？

谢谢...

推荐答案

派生类的g将被调用。如果你想调用基类中的函数，调用

The g of the derived class will be called. If you want to call the function in the base, call

Base::g();

。如果你想调用派生，但仍然希望有基本版本被调用，安排派生版本的g在其第一个语句中调用基本版本：

instead. If you want to call the derived, but still want to have the base version be called, arrange that the derived version of g calls the base version in its first statement:

virtual void g() {
    Base::g();
    // some work related to derived
}

从基础可以调用一个虚方法和控制转移到派生类中，用于模板方法设计模式。对于C ++，它更好地称为非虚拟接口。它也广泛应用于C ++标准库（C ++流缓冲区例如具有调用执行真正工作的虚函数的函数 pub ... ，例如 pubseekoff 调用受保护的 seekoff ）。我在这个答案中写了一个例子：如何验证对象的内部状态？

The fact that a function from the base can call a virtual method and control is transferred into the derived class is used in the template method design pattern. For C++, it's better known as Non-Virtual-Interface. It's widely used also in the C++ standard library (C++ stream buffers for example have functions pub... that call virtual functions that do the real work. For example pubseekoff calls the protected seekoff). I wrote an example of that in this answer: How do you validate an object’s internal state?

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