Question

Please consider the following code:

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*)! 
}  

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?

Answer 1

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.