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
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;
}
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;
}
Related
For example, I have 2 classes (in reality, it's more, that's why I'm asking this question) with the same methods:
class class1{
public:
void init(){
//something
}
void dostuff(){
//something
}
//
};
class class2{
public:
void init(){
//something
}
void dostuff(){
//something
}
//
};
And now a third one in which I want to deal with the two classes in the same manner:
class upclass{
public:
upclass(class12* argclass){
myclass=argclass;
myclass->init();
}
void domorestuff(){
myclass->dostuff();
}
private:
class12* myclass; //pointer to class 1 OR class 2
};
My question is now, do I need multiple constructors and multiple declarations to make it work or is there a way around it? Is it even possible to make "class12" a spacekeeper for these types without preprocessor-directives?
I am sorry to say, this is a wide field and there are really many many possible solution.
But I guess that we are talking about object- oriented programming, derivation and plymorphic functions. What you describe, will be typically solved with a class hierachy.
You have one base class with virtual (polymorphic) functions.
Then you derive other classes from this base class and override the virtual functions from the base class.
In a 3rd step, you create some instances of the derived classes dynamically, during runtime and you store the newly created classes (their address) in a pointer to the base class.
Later, you can call any of the virtual overriden function through the base class pointer. And mechanism behind the scenes will call the correct function for you.
Additionally. You defined some function init. Such a function name suggests the usage of a class-constructor. This will be called automatically in the correct sequence. First the base class constructor and then the derived class constructor.
Please see the below example:
#include <iostream>
#include <string>
class Base {
std::string baseName{};
public:
Base() { // Do initialization stuff
baseName = "Base";
std::cout << "\nConstructor Base\n";
}
virtual void doStuff() { // virtual function
std::cout << baseName << '\n';
}
};
class Derived1 : public Base {
std::string derivedName{};
public:
Derived1() : Base() { // Do initialization stuff
derivedName = "Derived1";
std::cout << "Constructor Derived1\n";
}
void doStuff() override { // Override virtaul function
std::cout << derivedName << '\n';
}
};
class Derived2 : public Base {
std::string derivedName{};
public:
Derived2() : Base() { // Do initialization stuff
derivedName = "Derived2";
std::cout << "Constructor Derived2\n\n";
}
void doStuff() override { // Override virtaul function
std::cout << derivedName << '\n';
}
};
int main() {
Base* base = new Base();
Base* derived1 = new Derived1(); // Store in base class pointer
Base* derived2 = new Derived2(); // Store in base class pointer
base->doStuff();
derived1->doStuff(); // Magic of polymorphism
derived2->doStuff(); // Magic of polymorphism
}
The Base class pointer will accept all classes derived from Base.
Please note. In reality you ould not use raw pointers and also to the constructor differently. This is just fotr demo.
But, you need to read several books about it to get the complete understanding.
You can explicitly write "store one of these" via std::variant and obtain the actual type (when needed) through std::visit:
#include <variant>
using class12 = std::variant<class1*, class2*>;
class upclass {
public:
upclass(class12 argclass): myclass{argclass} {
visit([](auto classn) { classn->init(); }, myclass);
}
void domorestuff() {
visit([](auto classn) { classn->dostuff(); }, myclass);
}
private:
class12 myclass;
};
If those visits get too repetitive, you might consider writing a pretty API to hide them:
class prettyclass12: public std::variant<class1*, class2*> {
private: // both g++ and clang want variant_size<>, a quick hack:
auto& upcast() { return static_cast<std::variant<class1*, class2*>&>(*this); }
public:
using std::variant<class1*, class2*>::variant;
void init() { visit([](auto classn) { classn->init(); }, upcast()); }
void dostuff() { visit([](auto classn) { classn->dostuff(); }, upcast()); }
};
class prettyupclass {
public:
prettyupclass(prettyclass12 argclass): myclass{argclass} { myclass.init(); }
void domorestuff() { myclass.dostuff(); }
private:
prettyclass12 myclass;
};
I have a component in a software that can be described by an interface / virtual class.
Which non-virtual subclass is needed is decided by a GUI selection at runtime.
Those subclasses have unique methods, for which is makes no sense to give them a shared interface (e.g. collection of different data types and hardware access).
A minimal code example looks like this:
#include <iostream>
#include <memory>
using namespace std;
// interface base class
class Base
{
public:
virtual void shared()=0;
};
// some subclasses with shared and unique methods
class A : public Base
{
public:
void shared()
{
cout << "do A stuff\n";
}
void methodUniqueToA()
{
cout << "stuff unique to A\n";
}
};
class B : public Base
{
public:
void shared()
{
cout << "do B stuff\n";
}
void methodUniqueToB()
{
cout << "stuff unique to B\n";
}
};
// main
int main()
{
// it is not known at compile time, which subtype will be needed. Therefore: pointer has base class type:
shared_ptr<Base> basePtr;
// choose which object subtype is needed by GUI - in this case e.g. now A is required. Could also have been B!
basePtr = make_shared<A>();
// do some stuff which needs interface functionality... so far so good
basePtr->shared();
// now I want to do methodUniqueToA() only if basePtr contains type A object
// this won't compile obviously:
basePtr->methodUniqueToA(); // COMPILE ERROR
// I could check the type using dynamic_pointer_cast, however this ist not very elegant!
if(dynamic_pointer_cast<A>(basePtr))
{
dynamic_pointer_cast<A>(basePtr)->methodUniqueToA();
}
else
if(dynamic_pointer_cast<B>(basePtr))
{
dynamic_pointer_cast<B>(basePtr)->methodUniqueToB();
}
else
{
// throw some exception
}
return 0;
}
Methods methodUniqueTo*() could have different argument lists and return data which is omitted here for clarity.
I suspect that this problem isn't a rare case. E.g. for accessing different hardware by the different subclasses while also needing the polymorphic functionality of their container.
How does one generally do this?
For the sake of completeness: the output (with compiler error fixed):
do A stuff
stuff unique to A
You can have an enum which will represent the derived class. For example this:
#include <iostream>
#include <memory>
using namespace std;
enum class DerivedType
{
NONE = 0,
AType,
BType
};
class Base
{
public:
Base()
{
mType = DerivedType::NONE;
}
virtual ~Base() = default; //You should have a virtual destructor :)
virtual void shared() = 0;
DerivedType GetType() const { return mType; };
protected:
DerivedType mType;
};
// some subclasses with shared and unique methods
class A : public Base
{
public:
A()
{
mType = DerivedType::AType;
}
void shared()
{
cout << "do A stuff\n";
}
void methodUniqueToA()
{
cout << "stuff unique to A\n";
}
};
class B : public Base
{
public:
B()
{
mType = DerivedType::BType;
}
void shared()
{
cout << "do B stuff\n";
}
void methodUniqueToB()
{
cout << "stuff unique to B\n";
}
};
// main
int main()
{
shared_ptr<Base> basePtr;
basePtr = make_shared<B>();
basePtr->shared();
// Here :)
if(basePtr->GetType() == DerivedType::AType)
static_cast<A*>(basePtr.get())->methodUniqueToA();
else if(basePtr->GetType() == DerivedType::BType)
static_cast<B*>(basePtr.get())->methodUniqueToB();
return 0;
}
You can store an enum and initialize it at the constructor. Then have a Getter for that, which will give you the Type. Then a simple static cast after getting the type would do your job!
The goal of using polymorphism for the client is to control different objects with a single way. In other words, the client do not have to pay any attention to the difference of each object. That way, checking the type of each object violates the basic goal.
To achieve the goal, you will have to :
write the concrete method(methodUniqueToX()).
write a wrapper of the concrete method.
name the wrapper method abstract.
make the method public and interface/abstract.
class Base
{
public:
virtual void shared()=0;
virtual void onEvent1()=0;
virtual void onEvent2()=0;
};
// some subclasses with shared and unique methods
class A : public Base
{
private:
void methodUniqueToA()
{
cout << "stuff unique to A\n";
}
public:
void shared()
{
cout << "do A stuff\n";
}
void onEvent1()
{
this.methodUniqueToA()
}
void onEvent2()
{
}
};
class B : public Base
{
private:
void methodUniqueToB()
{
cout << "stuff unique to B\n";
}
public:
void shared()
{
cout << "do B stuff\n";
}
void onEvent1()
{
}
void onEvent2()
{
methodUniqueToB()
}
};
I am expecting "My Game" to print out but I am getting "Base"
This only happens when using methods internally inside the class.
#include <iostream>
namespace Monster { class App {
public:
App(){}
~App(){}
void run(){
this->speak();
}
void speak(){
std::cout << "Base" << "\n";
};
};}; // class / namespace
class MyGame : public Monster::App {
public:
MyGame(){}
~MyGame(){}
void speak(){
std::cout << "My Game" << "\n";
};
};
int main(){
MyGame *child = new MyGame;
child->run();
return 0;
}
In C++ you need to specifically declare a function to be virtual:
class BaseClass {
virtual void speak () {
...
}
};
In C++ a method can only be overridden if it was marked virtual. You can think of virtual as a synonym for "overridable".
The virtual keyword has to appear in the base class. It may also appear optionally in the subclasses at the point of override, but it does not have to.
If you are using a compiler that supports C++11 (and you should if you are learning C++), I recommend that you always use the new override keyword when you mean to override:
class Base {
public:
virtual void speak() {
std::cout << "Base";
}
};
class Derived : public Base {
public:
void speak() override { // <---
std::cout << "Derived";
}
};
If the method isn't actually an override, the compiler will tell you so by giving an error.
It is not always obvious on the first read whether a method is an override. For example the following is correct thanks to return type covariance:
class A {};
class B : public A {};
class Base {
public:
virtual A* foo() {
return nullptr;
}
};
class Derived : public Base {
public:
B* foo() override {
return nullptr;
}
};
This might not be useful very often, but override makes it clear in case someone has to read it.
Also, if you have at least one virtual method in your class, also make its destructor virtual. This will assure that all the destructors will run when needed and things get cleaned up properly:
class App {
public:
App() {}
virtual ~App() {} // <---
void run() {
this->speak();
}
virtual void speak() {
std::cout << "Base\n";
};
};
I am expecting "My Game" to print out but I am getting "Base"
This only happens when using methods internally inside the class.
#include <iostream>
namespace Monster { class App {
public:
App(){}
~App(){}
void run(){
this->speak();
}
void speak(){
std::cout << "Base" << "\n";
};
};}; // class / namespace
class MyGame : public Monster::App {
public:
MyGame(){}
~MyGame(){}
void speak(){
std::cout << "My Game" << "\n";
};
};
int main(){
MyGame *child = new MyGame;
child->run();
return 0;
}
In C++ you need to specifically declare a function to be virtual:
class BaseClass {
virtual void speak () {
...
}
};
In C++ a method can only be overridden if it was marked virtual. You can think of virtual as a synonym for "overridable".
The virtual keyword has to appear in the base class. It may also appear optionally in the subclasses at the point of override, but it does not have to.
If you are using a compiler that supports C++11 (and you should if you are learning C++), I recommend that you always use the new override keyword when you mean to override:
class Base {
public:
virtual void speak() {
std::cout << "Base";
}
};
class Derived : public Base {
public:
void speak() override { // <---
std::cout << "Derived";
}
};
If the method isn't actually an override, the compiler will tell you so by giving an error.
It is not always obvious on the first read whether a method is an override. For example the following is correct thanks to return type covariance:
class A {};
class B : public A {};
class Base {
public:
virtual A* foo() {
return nullptr;
}
};
class Derived : public Base {
public:
B* foo() override {
return nullptr;
}
};
This might not be useful very often, but override makes it clear in case someone has to read it.
Also, if you have at least one virtual method in your class, also make its destructor virtual. This will assure that all the destructors will run when needed and things get cleaned up properly:
class App {
public:
App() {}
virtual ~App() {} // <---
void run() {
this->speak();
}
virtual void speak() {
std::cout << "Base\n";
};
};
This is part 2 to a problem I previously asked: Is it possible to have polymorphic member overloading in C++?
Using the Wiki example I created this example.
http://en.wikipedia.org/wiki/Double_dispatch
My problem is that the compiled code never looks up the vtable, and always uses the base instead of the inherited class. Here is my code:
#include <iostream>
class xEntity;
class xVehicle;
class xMapObject
{
public:
virtual void Bump(xMapObject&) { std::cout << "MapObject Bump MapObject\n"; };
virtual void Bump(xEntity&) { std::cout << "MapObject Bump Entity\n"; };
virtual void Bump(xVehicle&) { std::cout << "MapObject Bump Vehicle\n"; };
};
class xEntity : public xMapObject
{
public:
virtual void Bump(xMapObject&) { std::cout << "Entity Bump MapObject\n"; };
virtual void Bump(xEntity&) { std::cout << "Entity Bump Entity\n"; };
virtual void Bump(xVehicle&) { std::cout << "Entity Bump Vehicle\n"; };
};
class xVehicle : public xEntity
{
public:
virtual void Bump(xMapObject&) { std::cout << "Vehicle Bump MapObject\n"; };
virtual void Bump(xEntity&) { std::cout << "Vehicle Bump Entity\n"; };
virtual void Bump(xVehicle&) { std::cout << "Vehicle Bump Vehicle\n"; };
};
int main(int argv, char **argc)
{
xEntity Entity;
xVehicle Vechile;
xMapObject &EntityRef = Entity;
xMapObject &VehicleRef = Vechile;
VehicleRef.Bump(EntityRef);
return 0;
}
However, the output is always:
Vehicle Bump MapObject
Any help on this mystery is greatly appreciated.
You only did single dispatch, not double dispatch. The idea is that in xVehicle, taking an xMapObject&, you would call ref.bump(*this); which is double dispatch.
It is using the vtable; that's why it's calling xVechicle::Bump()! The vtable isn't used on arguments, that doesn't make sense (in C++, at least).
The typical solution is to have e.g. Bump(xMapObject& obj) call obj.Bump(*this);.
Is a bad design the xEntity::Bump(xVehicle&) because you are using as parameter in a base class a derived class.
and at least that your contract change, you don't need redefine base Bump methods.
the problem is that you are creating the xMapRef vars, that convert your derived class to a base class.
if you want that appropriated method is called, just call with the derived class object