What exactly does runtime polymorphism mean? - c++

I'm slightly confused about runtime polymorphism. Correct me if I am wrong, but to my knowledge, runtime polymorphism means that function definitions will get resolved at runtime.
Take this example:
class a
{
a();
~a();
void baseclass();
}
class b: class a
{
b();
~b();
void derivedclass1();
}
class c: class a
{
c();
~c();
void derivedclass2();
}
Calling methodology:
b derived1;
a *baseptr = &derived1; //here base pointer knows that i'm pointing to derived class b.
baseptr->derivedclass1();
In the above calling methodology, the base class knows that it's pointing to derived class b.
So where does the ambiguity exist?
In what cases will the function definitions get resolved at runtime?

This code, at run time, calls the correct version of f() depending on the type of object (A or B) that was actually created - no "ambiguity". The type cannot be known at compile-time, because it is selected randomly at run-time.
struct A {
virtual ~A() {}
virtual void f() {}
};
struct B : public A {
virtual void f() {}
};
int main() {
A * a = 0;
if ( rand() % 2 ) {
a = new A;
}
else {
a = new B;
}
a->f(); // calls correct f()
delete a;
}

There is no ambiguity exists in the example provided.
If the base class has the same function name as the derived class, and if you call in the way you specified, it will call the base class's function instead of the derived class one.
In such cases, you can use the virtual keyword, to ensure that the function gets called from the object that it is currently being pointed. It is resolved during the run time.
Here you can find more explanation..

Turn this
void baseclass();
to
virtual void baseclass();
Override this in your Derived classes b and c. Then
b *derived1 = new derived1 ();
a *baseptr = derived1; //base pointer pointing to derived class b.
baseptr->baseclass();
will invoke derived1 definition, expressing run time polymorphism. And do remember about making your destructor virtual in Base. Some basic reading material for polymorphism

Runtime means that exact method will be known only at run time. Consider this example:
class BaseClass
{
public:
virtual void method() {...};
};
class DerivedClassA : public BaseClass
{
virtual void method() {...};
};
class DerivedClassB : public BaseClass
{
virtual void method() {...};
};
void func(BaseClass* a)
{
a->method();
}
When you implement your ::func() you don't know exactly type of instance pointed by BaseClass* a. It might be DerivedClassA or DerivedClassB instance etc.
You should realize, that runtime polymorphism requires special support from language (and maybe some overhead for calling "virtual" functions). In C++ you "request" for dynamic polymorphism by declaring methods of base class "virtual" and using public inheritance.

You need to have some useful business method declared in the base and in each derived class. Then you have code such as
a->someMethod();
Now the a pointer might point to an instance of any of the derived classes, and so the type of what a is pointing to must determine which someMethod() is called.

Lets have an experiment
#include <iostream>
using namespace std;
class aBaseClass
{
public:
void testFunction(){cout<<"hello base";}///Not declared as virtual!!!!
};
class aDerivedClass:public aBaseClass
{
public:
void testFunction(){cout<<"hello derived one";}
};
class anotherDerivedClass:public aDerivedClass
{
public:
void testFunction(){cout<<"hello derived two";}
};
int main()
{
aBaseClass *aBaseClassPointer;
aBaseClassPointer=new aDerivedClass;
aBaseClassPointer->testFunction();
}
The above code does not support run time polymorphism. Lets run and analyze it.
The output is
hello base
Just change the line void testFunction(){cout<<"hello base";} to virtual void testFunction(){cout<<"hello base";} in aBaseClass. Run and analyze it. We see that runtime polymorphism is achieved. The calling of appropriate function is determined at run time.
Again change the line aBaseClassPointer=new aDerivedClass to aBaseClassPointer=new anotherDerivedClass in main function and see the output. Thus the appropriate function calling is determined at run time (when the program is running).

Related

How to defined a static interface in base class and make sure the interface must be implement in derived class?

It is very easy that we can make sure derived class must implement interface defined in base class.
That is pure virtual function.
For example:
class BaseClass
{
...
virtual void print()=0;
...
}
class DerivedClass :public BaseClass
{
// function must be implement, otherwise compiler will complain ...
void print()
{
}
};
Can we defined a static interface in base class and make sure the interface must be implement in derivate class?
I want something like this
class BaseClass
{
...
static void print(); // base class only define static interface
...
}
class DerivedClass :public BaseClass
{
// derived class must implement interface, otherwise compiler will complain ...
static void print()
{
}
};
I have no idea about this.
Thanks for your time.
It is not possible to make a virtual static function. For the simple reason that when calling a static function, you always know the class that defines that function in compile time. Unlike virtual functions, where you don't know the type of the object whose method you're calling.
For example:
class A
{
public:
virtual void f() {printf("A");}
};
class B : public A
{
virtual void f() override {printf("B");}
};
void g(A& a)
{
a.f();
}
int main()
{
B b;
g(b);
return 0;
}
In the above example, inside the function g, the correct function is invoked (B::f). Even though while compiling the function it is not known what the type of its argument is (it could be A or any class derived from A).
Without making f() virtual, you would have overloaded the method f, rather than overridden it. Which means that in the following example, the output would be "A", even though you might expect it to be "B":
class A
{
public:
void f() {printf("A");}
};
class B : public A
{
void f() {printf("B");}
};
void g(A& a)
{
a.f();
}
int main()
{
B b;
g(b);
return 0;
}
This may cause serious bugs, and it is suggested to never overload base class methods, and to always use the override keyword when overriding a virtual method to escape those bugs.
When making a static function, you can simply overload it, it would not create a compilation error. However, you probably never should overload it, because it may hide a bug that is very difficult to track (you are certain that B::f() is being called while actually A::f() is being called).
Furthermore, it is not possible to 'force' the derived class to implement a static interface, because there is no such thing as a static interface. Because you have no virtual static functions, you may not pass a reference or pointer to the interface that would implement this function.

Explicitly stop late binding

If we create an object pointer to base class class which points to its child class object then we use virtual key word for late binding
So.,in case of late binding,, our code goes like this :-
#include<iostream>
using namespace std;
struct A{
virtual void print() {
cout<<"function1";
}
};
struct B : public A{
void print(){
cout<<"function2";
}
};
struct C : public B{
void print(){
cout<<"function3";
}
};
int main(){
A* a = new C();
A* p = new B();
a->print();
p->print();
}
Now my question is : when we use virtual keyword in base class, all the functions of derived classes created in base class will become virtual.
In multilevel inheritance, is there any way so that we can stop the function of class c from being virtual??
Any way to break this chain of virtual functions ?
Sorry for any mistakes in question but i tried my best.. ☺️☺️
It is not possible to make a function that is already virtual be not virtual in a child class.
Virtuality is a property of the top-level base class, not the child classes. Once a method has been marked as virtual, and pointer to that class must use dynamic dispatch for that function when calling because the pointer could be pointing to a child class that has overridden the behaviour.
Consider this code:
A a;
B b;
C c;
A * ap = &a;
A * bp = &b;
A * cp = &c;
ap->print(); // 'function1'
bp->print(); // 'function2'
cp->print(); // 'function3'
Here, the calls to print cannot tell which function to call at compile time, they absolutely must use dynamic dispatch.
However, you can make C::print behave like A::print
struct C : public B {
void print() {
A::print();
}
};
Which results in:
ap->print(); // 'function1'
bp->print(); // 'function2'
cp->print(); // 'function1'
And if the behaviour of A::print() changes, C::print() mirrors those changes.
This will still be overridable though, unless you use the final keyword as outlined below.
Original answer:
I believe you are looking for the final specifier.
It's only available as of C++11 though.
To quote en.cppreference's page about the final specifier:
When used in a virtual function declaration or definition, final
ensures that the function is virtual and specifies that it may not be
overridden by derived classes. The program is ill-formed (a
compile-time error is generated) otherwise.
And a variation of the example they give, demonstrating the solution to your problem:
struct A
{
virtual void foo();
};
struct B : A
{
void foo() final; // B::foo is overridden and it is the final override
};
struct C : B
{
void foo() override; // Error: foo cannot be overridden as it's final in B
};

Polymorphism in C++ why is this isn't working?

class Base {
public:
virtual void f();
void f(int);
virtual ~Base();
};
class Derived : public Base {
public:
void f();
};
int main()
{
Derived *ptr = new Derived;
ptr->f(1);
delete ptr;
return 0;
}
ptr->f(1); is showing the following error: "too many arguments in function call".
Why is this isn't possible? isn't derived inherited all the functions form base and is free to use any of them?
I could call it explicitly and it would work but why isn't this allowed?
What you are seeing is called hiding.
When you override the function void f() in the Derived class, you hide all other variants of the f function in the Base class.
You can solve this with the using keyword:
class Derived : public Base {
public:
using Base::f; // Pull all `f` symbols from the base class into the scope of this class
void f() override; // Override the non-argument version
};
As mentioned by #Some Programming Dude : it is because of Hiding.
To understand hiding in relatively simpler language
Inheritance is meant to bring Baseclass variables / functions in Derived class.
But, on 1 condition : "If its not already available in Derived Class"
Since f() is already available in Derived, it doesn't make sense to look at Base class from compiler perspective.
That's the precise reason why you need to scope clarify while calling this function
void main()
{
Derived *ptr = new Derived;
ptr->Base::f(1);
delete ptr;
}
Suppose for time being that Derived do have the access to the function void Base::f(int);. Then it will be a case of function overloading. But, this is invalid case of function overloading since one function f(int);is in Base and other function f(); is in Derived. Function Overloading happens inside a single class. Function Overriding happens across classes. The example you posted is a case of Name Hiding in Inheritance
"Derived *ptr" This definition will only allow "ptr" to access all the member functions which are defined via Derived class or its child class. But, it will not allow u to access the member functions which are coming in Derived class because of inheritance.
If u want to access base class version of function "f" then use "Base *ptr" and it will choose the correct version of function automatically as shown :)
class Base {
public:
virtual void f()
{
cout<<"Here 2"<<endl;
}
void f(int x)
{
cout<<"Here 1"<<endl;
}
virtual ~Base() {}
};
class Derived : public Base {
public:
void f()
{
cout<<"Here 3"<<endl;
}
virtual ~Derived() {}
};
int main()
{
Base *ptr = new Derived;
ptr->f(1);
delete ptr;
return 0;
}
output is
Here 1

How to make an inherited virtual function not usable?

I have an abstract base class which declares a pure virtual function (virtual method() = 0;). Some of the inherited classes specialize and use this method but there's one of those inherited classes in which I don't want to make this method usable. How do I do it? Is making it private the only choice?
Well, you could throw that will make tacking where it is called easier.
void method() override { throw /* whatever */ ; }
Dynamic polymorphism is a runtime property. Hence a runtime error. If you look after something that will trigger at compile time, you need static polymorphism.
template<typename Child>
struct Parent {
void callMe() {
static_cast<Child*>(this)->callMeImpl();
}
};
struct SomeChild : Parent<SomeChild> {
};
Now, if you try to call callMe form the parent that is extended by SomeChild, it will be a compile time error.
You can also hold pointer to the parent just like dynamic polymorphism, as the parent will call the child function
Is making it private the only choice?
No, that's not a choice at all since you can still access the method if it's public or protected in the base classes.
Other than implementing the method in the class and resorting to run-time failures, there's not a lot you can do. You could port the whole thing to templates and use static polymorphism which, with further trickey, you could contrive a compile-time failure in certain instances, but that could be design overkill.
I guess you could make it a normal virtual function instead of a pure virtual function like this:
virtual void method() { /* code */ }
If this function is not being used in another class, you will be able to catch that. For example you could warn yourself:
virtual void method() { error = true; } //or whatever
As others have said there is no way of enforcing this at compile time. If you are referring to a pointer to a base class there is no way the compiler can know if that pointer is referring to one of the derived classes that does implement this method or one that doesn't.
So the case will have to be handled at runtime. One option is to just throw an exception. Another option is to introduce a level of indirection so that you can ask your base class if it implements a certain function before you call it.
Say you have a Base class with three methods foo, bar and doit and some derived classes do not want to implement foo then you could split up the Base class into two base classes:
class Base1 {
public:
virtual void foo() = 0;
};
class Base2 {
public:
virtual void bar() = 0;
virtual void doit() = 0;
};
Then in places where you are currently using Base you instead use a BaseSource:
class BaseSource {
public:
virtual Base1* getBase1() = 0;
virtual Base2* getBase2() = 0;
};
where getBase1 and getBase2 can return nullptr if a BaseSource does not offer that interface:
class Derived : public BaseSource, public Base2 {
public:
// don't implement foo();
// Implemementation of Base2
void bar() override;
void doit() override;
Base1* getBase1() override { return nullptr; } // Doesn't implement Base1
Base2* getBase2() override { return this; }
};
int main() {
std::vector<std::unique_ptr<BaseSource>> objects;
objects.push_back(std::make_unique<Derived>());
for (auto& o : objects) {
auto b1 = o->getBase1();
if (b1)
b1->foo();
auto b2 = o->getBase2();
if (b2)
b2->bar();
}
}
Live demo.

Virtual Function During Construction Workaround

I've got a base class that has a virtual function. I want to call that class during the construction because I want the function called for each of the derived classes. I know I can't call a virtual function during construction, but I can't think of an elegant (i.e., avoid repeating code) solution.
What are some work arounds to calling a virtual function during construction?
The reason I want to avoid this is because I don't want to have to create constructors that just call the base class.
class A {
public:
A() {
read();
}
// This never needs to be called
virtual void read() = 0;
}
class B:A {
public:
B():A() { };
read() { /*Do something special for B here.*/ }
}
class C:A {
public:
C():A() { };
read() { /*Do something special for C here.*/ }
}
PS: The Python way of doing this is simply to raise NotImplementedError in A::read(). I'm returning to C++ and I'm more rusty than I thought.
The FAQ perspective.
This is a Frequently Asked Question.
See the C++ FAQ item titled “Okay, but is there a way to simulate that behavior as if dynamic binding worked on the this object within my base class's constructor?”.
It’s very often a good idea to check the FAQ (and generally, googling or altavista’ing) before asking.
The question as “Derived class specific base initialization”.
To be clear, while the literal question above is
“What are some work arounds to calling a virtual function during construction?”
it is evident that what’s meant is
“How can a base class B be designed so that each derived class can specify part of what goes on during B construction?”
A major example is where C style GUI functionality is wrapped by C++ classes. Then a general Widget constructor might need to instantiate an API-level widget which, depending on the most derived class, should be a button widget or a listbox widget or whatever. So the most derived class must somehow influence what goes on up in Widget’s constructor.
In other words, we’re talking about derived class specific base construction.
Marshall Cline called that “Dynamic Binding During Construction”, and it’s problematic in C++ because in C++ the dynamic type of an object during class T construction and destruction, is T. This helps with type safety, in that a virtual member function is not called on a derived class sub-object before that sub-object has been initialized, or its initialization has started. But a major cost is that DBDI (apparently) can’t be done in a way that is both simple and safe.
Where the derived class specific init can be performed.
In the question the derived class specific action is called read. Here I call it derived_action. There are 3 main possibilities for where the derived_action is invoked:
Invoked by instantiation code, called two-phase construction.
This essentially implies the possibility of having a mostly unusuable not fully initialized object at hand, a zombie object. However, with C++11 move semantics that has become more common and accepted (and anyway it can be mitigated to some extent by using factories). A main problem is that during the second phase of construction the ordinary C++ protection against virtual calls on uninitialized sub-objects, due to dynamic type changes during construction, is not present.
Invoked by Derived constructor.
For example, derived_action can be invoked as an argument expression for the Base constructor. A not totally uncommon technique is to use a class template to generate most derived classes that e.g. supply calls of derived_action.
Invoked by Base constructor.
This implies that knowledge of derived_action must be passed up to the constructor, dynamically or statically. A nice way is to use a defaulted constructor argument. This leads to the notion of a parallel class hierarchy, a hierarchy of derived class actions.
This list is in order of increasing sophistication and type safety, and also, to the best of my knowledge, reflects the historical use of the various techniques.
E.g. in Microsoft’s MFC and Borland’s ObjectWindows GUI early 1990’ libraries two-phase construction was common, and that kind of design is now, as of 2014, regarded as very ungood.
This is the factory method approach, putting the factory into the base class:
class A {
public:
virtual void read() = 0;
template<class X> static X* create() {X* r = new X;X->read();return X;}
virtual A* clone() const = 0;
};
class B : public A {
B():A() { };
friend class A;
public:
void read() { /*Do something special for B here.*/ }
B* clone() const {return new B(*this);}
};
class C : public A {
C():A() { };
friend class A;
public:
void read() { /*Do something special for C here.*/ }
C* clone() const {return new C(*this);}
};
Added a clone-method with covariant return type as a bonus.
Using CRTP:
class A {
public:
// This never needs to be called
virtual void read() = 0;
virtual A* clone() const = 0;
};
template<class D, class B> struct CRTP : B {
D* clone() {return new D(*this);}
static D* create() {return new D();}
};
class B : public CRTP<B, A> {
B() { };
public:
void read() { /*Do something special for B here.*/ }
};
class C : public CRTP<C, A> {
C() { };
public:
void read() { /*Do something special for C here.*/ }
};
One way to achieve this, would be simply to delegate it to another class (that is perhaps a friend) and can be sure to be called when fully constructed.
class A
{
friend class C;
private:
C& _c; // this is the actual class!
public:
A(C& c) : _c(c) { };
virtual ~A() { };
virtual void read() = 0;
};
class B : public A
{
public:
B(C& c) : A(c) { };
virtual ~B() { };
virtual void read() {
// actual implementation
};
};
class C
{
private:
std::unique_ptr<A> _a;
public:
C() : _a(new B(*this)) { // looks dangerous? not at this point...
_a->read(); // safe now
};
};
In this example, I just create a B, but how you do that can depend on what you want to achieve and use templates on C if necessary, e.g:
template<typename VIRTUAL>
class C
{
private:
using Ptr = std::unique_ptr<VIRTUAL>;
Ptr _ptr;
public:
C() : _ptr(new VIRTUAL(*this)) {
_ptr->read();
};
}; // eo class C
The workaround is to call the virtual function after construction. You can then couple the two operations (construction + virtual call) in factory function. Here is the basic idea:
class FactoryA
{
public:
A *MakeA() const
{
A *ptr = CreateA();
ptr->read();
return ptr;
}
virtual ~FactoryA() {}
private:
virtual A *CreateA() const = 0;
};
class FactoryB : public FactoryA
{
private:
virtual A *CreateA() const { return new B; }
};
// client code:
void f(FactoryA &factory)
{
A *ptr = factory.MakeA();
}
As mentioned by Benjamin Bannier, you can use CRTP (a template which defines the actual read() function.) One problem with that method is that templates have to always be written inline. That can at times be problematic, especially if you are to write really large functions.
Another way is to pass a function pointer to the constructor. Although, in a way, it is similar to calling the function in your constructor, it forces you to pass a pointer (although in C++ you could always pass nullptr.)
class A
{
public:
A(func_t f)
{
// if(!f) throw ...;
(*f)();
}
};
class B : A
{
public:
B() : A(read) {}
void read() { ... }
};
Obviously, you have the "can't call other virtual functions" problem within the read() function and any function it calls. Plus, variable members of B are NOT yet initialized. That is probably a much worst problem in this case...
For that reason, writing it this way is safer:
B() : A()
{
read();
}
However, in cases like that, that may be the time when you an some for of init() function. That init() function can be implemented in A() (if you make it accessible: i.e. use public A when deriving) and that function can call all the virtual functions as expected:
class A
{
public:
void init()
{
read();
}
};
class B : public A
{
public:
...
};
I know a lot of people say that an init() function is evil because people who create a B object now need to know to call it... but there isn't much else you can do. That being said, you could have a form of factory, and that factory can do the init() call as required.
class B : public A
{
public:
static B *create() { B *b(new B); b->init(); return b; }
private:
B() { ... } // prevent creation without calling create()
};