问题
class Material
{
public:
 void foo()
 {
  cout << "Class Material";
 }
};
class Unusual_Material : public Material
{
public:
 void foo()
 {
  cout << "Class Unusual_Material";
 }
};
int main()
{
 Material strange = Unusual_Material();
 strange.foo(); //outputs "Class Material" 
 return 0;
}
I would like for this to result in the "Class Unusual_Material" being displayed to the console. Is there a way I can achieve this? In my program I have a class Material from which other more specific materials are derived. The method Material::foo() represents a method in Material that is adequate for most materials, but occationally, another foo() needs to be defined for a material with unusual properties.
All objects in my program contain a Material field. In the event that they are assigned an unusual material, I would like the derived, unusual foo to be called.
This is probably either pretty easy, or impossible, but I can't figure it out either way.
Thanks
回答1:
What you want is polymorphism, and to enable it for a function you need to make it virtual:
class Material 
{ 
public: 
    virtual void foo() // Note virtual keyword!
    { 
        cout << "Class Material"; 
    } 
}; 
class Unusual_Material : public Material 
{ 
public: 
    void foo() // Will override foo() in the base class
    { 
        cout << "Class Unusual_Material"; 
    } 
}; 
Also, polymorphism only works for references and pointers:
int main()  
{  
    Unusual_Material unusualMaterial;
    Material& strange = unusualMaterial;
    strange.foo();  
    return 0; 
}
/* OR */
int main()  
{  
    Unusual_Material unusualMaterial;
    Material* strange = &unusualMaterial;
    strange->foo();  
    return 0; 
}
What you have in your code snippet will slice the Unusual_Material object:
int main() 
{ 
    // Unusual_Material object will be sliced!
    Material strange = Unusual_Material(); 
    strange.foo(); 
    return 0; 
} 
回答2:
Still better explanation would be..
class Base
{
public:
 void foo()     //make virtual void foo(),have derived method invoked
 {
  cout << "Class Base";
 }
};
class Derived: public Base
{
public:
 void foo()
 {
  cout << "Class Derived";
 }
};
int main()
{
 Base base1 = Derived ();
 Base1.foo(); //outputs "Class Base" 
           // Base object, calling base method
 Base *base2 = new Derived ();
 Base2->foo(); //outputs"Class Base",Again Base object calling base method
// But to have base object, calling Derived method, following are the ways
// Add virtual access modifier for base foo() method. Then do as below, to //have derived method being invoked.
//
// Base *base2 = new Derived ();
// Base2->foo();    //outputs "Class Derived" .
return 0;
}
来源:https://stackoverflow.com/questions/4209000/call-derived-class-method-from-base-class-reference