重载虚拟函数调用解析

问题描述

请考虑以下代码：

class Abase{};
class A1:public Abase{};
class A2:public A1{};
//etc

class Bbase{
public:
    virtual void f(Abase* a);
    virtual void f(A1* a);
    virtual void f(A2* a);
};

class B1:public Bbase{
public:
    void f(A1* a);
};

class B2:public Bbase{
public:
    void f(A2* a);
};

int main(){
    A1* a1=new A1();
    A2* a2=new A2();
    Bbase* b1=new B1();
    Bbase* b2=new B2();
    b1->f(a1); // calls B1::f(A1*), ok
    b2->f(a2); // calls B2::f(A2*), ok
    b2->f(a1); // calls Bbase::f(A1*), ok
    b1->f(a2); // calls Bbase::f(A2*), no- want B1::f(A1*)!
}

$ b $ p我有兴趣知道为什么C ++选择解析函数调用最后一行通过将对象的这个指针向上转换到基类，而不是上传 f（）的参数 ？

I'm interested to know why C++ chooses to resolve the function call on the last line by upcasting the this pointer of the object to the base class, rather than upcasting the argument of f()? Is there any way that I can get the behaviour I want?

推荐答案

选择哪个版本的 f 通过查看参数的编译时类型进行调用。此名称解析不考虑运行时类型。由于 b1 的类型为 Bbase * ，所有 Bbase 考虑成员;一个 A2 * 是最好的匹配，所以这是一个被调用。

The choice of which version of f to call is made by looking at the compile-time type of the parameter. The run-time type isn't considered for this name resolution. Since b1 is of type Bbase*, all of Bbase's members are considered; the one that takes an A2* is the best match, so that's the one that gets called.

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