如何通过编译器解决函数？

Question

如何确定以下调用是在编译时还是在运行时绑定？

object.member_fn;//object is either base class or derived class object
p->member_fn;//p is either base class or derived class pointer

编辑:

#include <iostream>
using namespace std;
class Base
{
       public:
          Base(){ cout<<"Constructor: Base"<<endl;}
          ~Base(){ cout<<"Destructor : Base"<<endl;}
};
class Derived: public Base
{
     //Doing a lot of jobs by extending the functionality
       public:
           Derived(){ cout<<"Constructor: Derived"<<endl;}
           ~Derived(){ cout<<"Destructor : Derived"<<endl;}
 };
void foo()
{
    Base & Var = Derived();
    Base*pVar = new Derived;
    delete pVar;
}
void main()
{
    foo();
        std::cin.get();
}


out put:
Constructor: Base
Constructor: Derived
Constructor: Base
Constructor: Derived
Destructor : Base // the Derived is not called,
                  // the PVal is of type Base* and the fn is not virtual 
                  //so compile time binding
Destructor : Derived
Destructor : Base

Answer 1

如果该方法不是虚拟的,则两个调用都将在编译时解析.如果该方法是虚拟的,那么您的question(obj.method())中的第一个调用将在编译时针对对象解析,但在运行时将用于引用.第二个call(objp->method())将在运行时解析.您还可以在编译时强制调用方法的非派生版本.

struct base {
   void f();
   virtual void v();
};
struct derived : public base {
   void f();
   void v(); // intentionally left virtual out, it does not really matter
};
int main() {
   derived d;
   base & b = d;
   base * bp = &d;

   // compile time:
   d.f();   // derived::f
   d.v();   // derived::v
   b.f();   // base::f   -- non-virtual
   bp->f(); // base::f   -- non-virtual

   // runtime:
   b.v();   // derived::v
   bp->v(); // derived::v

   // compile time (user forced):
   b.base::v();   // base::v
   bp->base::v(); // base::v
}