没有显式指针的C ++多态调用

本文介绍了没有显式指针的C ++多态调用的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

如果我有基础课:

struct Base { void foo() { bar(); } virtual void bar() { } };

和派生类:

struct Derived : public Base { void bar() { cerr << "Derived here\n"; } };

有可能在编写此代码时:

It is happen that when write this code:

Derived d; d.foo();

由于调用了Derived::bar，因此我将看到打印派生于此处".但是我不是通过指向base的指针来调用的，而是在这里工作的多态性.为什么?是因为Base::foo中对bar的调用实际上是对this->bar()的隐式调用，而bar是在类的vtable中找到的?

I will see printing "Derived here" - since Derived::bar was called. But I did not call via pointer to base, but polymorphism working here. Why? Is it because the call to bar in Base::foo is implicitly actually to this->bar() and bar is finding in vtable of class?

推荐答案

您的猜测是正确的(尽管请记住，C ++标准对vtable没有任何说明).

Your guess is precisely correct (although bear in mind that the C++ standard says nothing about vtables).