How to deal with different classes which have the same method? - c++

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;
};

Related

Handling unique methods of Subclasses

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()
}
};

How to implement the same method in classes derived from an abstract base?

I have an abstract class Job and other classes that implement it like:
Waiter and Builder, all of them implement my function in the same way.
For example:
Waiter::changeScore()
{
score += top_score;
}
How may I prevent this kind of code duplication?
Constraints:
I want to keep Job abstract.
Each Waiter or Builder has its own top_score value (It differs between classes and objects of the same class).
Not all member functions of an abstract class need to be pure virtual (as long as at least one is). Your changeScore member is an ideal candidate as a 'real' base class function. Further, not only does it not need to be pure virtual, it doesn't even need to be virtual at all (unless you want your polymorphism to change what a pointer to a derived class will see, for that function).
As each class (or object) will have its own value of top_score (as you have stated), then that (data) member can also be part of the 'abstract' base class.
You can even add a single 'dummy' pure virtual function in your base class (which is never intended to be used, even by a derived class), just to make sure that instances aren't accidentally created. For example, your Job class could have a member:
virtual int Dummy() = 0;
Then, any derived class must have an override for that (however trivial), or the compiler won't allow you to declare an instance of that class. So, your Waiter class would need something like:
int Dummy override { return 1; }
The following code sample may help/demonstrate the idea:
#include <iostream>
#include <memory> // So we can use smart pointers
class Job {
public:
int score{ 0 }, top_score{ 0 };
public:
Job() { }
virtual ~Job() = default;
virtual void Dummy() = 0; // This is sufficient to make the class abstract!
void changeScore() {
score += top_score;
}
virtual void showName() {
std::cout << "Generic Job" << std::endl;
}
};
class Waiter : public Job {
public:
Waiter(int top = 5) { top_score = top; }
~Waiter() override = default;
void Dummy() override { } // We need this in order to use Waiter
void showName() override {
std::cout << "Waiter" << std::endl;
}
};
class Builder : public Job {
public:
Builder(int top = 10) { top_score = top; }
~Builder() override = default;
void Dummy() override { } // We need this in order to use Builder
void showName() override {
std::cout << "Builder" << std::endl;
}
};
int main()
{
Waiter w{ 6 }; // OK - uses explicit value for 'top' parameter
Builder b; // OK - uses default value for 'top' parameter
// Job j; // ERROR - Cannot instantiate abstract class
w.changeScore();
b.changeScore();
std::cout << w.score << std::endl;
std::cout << b.score << std::endl;
// Also, using pointers...
// Job* pj = new Job; // ERROR - Cannot instantiate abstract class
Job* pw = new Waiter; // OK - Now we can make use of polymorphism...
Job* pb = new Builder; // ...with either of these 2 "Job" pointers!
pw->showName();
pb->showName();
delete pw;
delete pb;
// Polymorphism also works with smart pointers (which you SHOULD be using) ...
// std::unique_ptr<Job> upj = std::make_unique<Job>(); // ERROR - Allocating an object of abstract class
std::unique_ptr<Job> upw = std::make_unique<Waiter>(15);
upw->changeScore();
std::cout << upw->score << ": ";
upw->showName();
std::unique_ptr<Job> upb = std::make_unique<Builder>(42);
upb->changeScore();
std::cout << upb->score << ": ";
upb->showName();
return 0;
}
You can define the method in the base class:
Live demo
class Job {
private:
int score;
int top_score;
protected:
//protected constructor to be inherited by derived classes
Job(int top_score) : top_score(top_score) {}
//one pure virtual method is enough to make the class abstract
virtual void some_method() = 0;
public:
void changeScore() { //single method implementation
score += top_score;
}
//to use polymorphism you must use a virtual destructor, unless you use shared_ptr
virtual ~Job(){}
};
class Waiter : public Job {
public:
Waiter(int top_score) : Job(top_score) {}
// pure virtual methods must be overridden in all derived classes
void some_method() override{}
};
class Builder : public Job {
public:
Builder(int top_score) : Job(top_score) {}
void some_method() override{}
};
changeScore() will be implemented in the abstract class and will be usable by all derived classes.
Waiter w(10); //top_score 10
Buider b(20); // top_score 20
b.changeScore();
w.changeScore();
You can make the changeScore method a pure virtual method AND provide an implementation. This would look like this:
class Job {
int score{0};
int top_score{0};
public:
virtual void changeScore() = 0;
};
void Job::changeScore()
{
score += top_score;
}
Then you can call the changeScore method of the Job base class in the child classes like this:
class Waiter : public Job {
public:
virutal void changeScore() override {
Job::changeScore();
}
};
This way if you want to change changeScore, you do not need to change all the implementations in the child classes, but you can just change the implementation in the Job class.
This way you do not need any dummy methods and the Job class remains abstract, while the override in the child classes is trivial and you have a single implementation if you ever want to change it.
EDIT:
If you are wondering where this override keyword comes from, it is introduced in C++11. Since I do not know which C++ version you are using I just wanted to point that out.
You can read about the override specifier here
EDIT II:
Regarding that ever child class has its own top_score, you should set this via the constructor of those child classes.
Like this:
class Job {
protected:
int top_score{0};
Job(top) : top_score(top) {}
...
};
class Waiter : public Job {
public:
Waiter(int top): Job(top) {}
...
};
This way each child class has its own version of top_score
EDIT III:
Putting it all together the classes would look something like this:
class Job {
protected:
int score{0};
int top_score{0};
Job(top) : top_score(top) {}
public:
virtual void changeScore() = 0;
};
void Job::changeScore()
{
score += top_score;
}
class Waiter : public Job {
public:
Waiter(int top): Job(top) {}
virutal void changeScore() override {
Job::changeScore();
}
};

overriding C++ not working as expected [duplicate]

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";
};
};

C++ Overriding method not working

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";
};
};

Avoid new when storing base-typed member variable that could be initialized with different derived types

My code structure is like below where multiple classes implement Interface. In Example class I store a pointer to the Interface and new() it in the constructor appropriately (depending on constructor parameters not shown here). I'm looking for ways to avoid using new() in this scenario but haven't got a solution yet. What's the best practice for something like this?
class Interface
{
virtual void Foo() = 0;
};
class A : public Interface
{
void Foo() { ... }
};
class B : public Interface
{
void Foo() { ... }
};
class Example
{
private:
Interface* m_bar;
public:
Example()
{
m_bar = new A(); // deleted in destructor
}
};
There are two ways this is typically done, each with their own merits.
If A is truely defined at compile time, than a typical way to handle this is to simply use a template type:
template <typename T>
class TemplateExample
{
T m_bar;
public:
TemplateExample() : m_bar() {};
}
This has some downsides. TemplateExample<A> becomes unrelated to TemplateExample<B>, the error messages when T doesn't follow the correct interface are pretty obtuse, ect. The upside is this may use duck typing rather than interface typing, and m_bar is a concrete instance.
The other (arguable more common) way is to do the following
class UniquePtrExample
{
std::unique_ptr<Interface> m_bar;
public:
UniquePtrExample() : m_bar(new A()){}
};
This has the benefit of being able to be run time configuratble if you follow a cloable pattern:
class Interface
{
public:
virtual void Foo() = 0;
virtual Interface* clone() const = 0;
};
template <typename T>
class CloneHelper : public Interface
{
public:
virtual Interface* clone() const { return new T(static_cast<const T&>(*this));}
};
class A : public CloneHelper<A>
{
virtual void Foo() { std::cout << 'A' << std::endl; }
};
class B : public CloneHelper<B>
{
virtual void Foo() { std::cout << 'B' << std::endl; }
};
class UniquePtrExample
{
std::unique_ptr<Interface> m_bar;
public:
UniquePtrExample() : m_bar(new A()){}
UniquePtrExample(const Interface& i) : m_bar(i.clone());
};
Note you can further extend the above to have a move variant of the clone function.