I have a solid understanding of most OOP theory but the one thing that confuses me a lot is virtual destructors.
I thought that the destructor always gets called no matter what and for every object in the chain.
When are you meant to make them virtual and why?
Virtual destructors are useful when you might potentially delete an instance of a derived class through a pointer to base class:
class Base
{
// some virtual methods
};
class Derived : public Base
{
~Derived()
{
// Do some important cleanup
}
};
Here, you'll notice that I didn't declare Base's destructor to be virtual. Now, let's have a look at the following snippet:
Base *b = new Derived();
// use b
delete b; // Here's the problem!
Since Base's destructor is not virtual and b is a Base* pointing to a Derived object, delete b has undefined behaviour:
[In delete b], if the static type of the
object to be deleted is different from its dynamic type, the static
type shall be a base class of the dynamic type of the object to be
deleted and the static type shall have a virtual destructor or the
behavior is undefined.
In most implementations, the call to the destructor will be resolved like any non-virtual code, meaning that the destructor of the base class will be called but not the one of the derived class, resulting in a resources leak.
To sum up, always make base classes' destructors virtual when they're meant to be manipulated polymorphically.
If you want to prevent the deletion of an instance through a base class pointer, you can make the base class destructor protected and nonvirtual; by doing so, the compiler won't let you call delete on a base class pointer.
You can learn more about virtuality and virtual base class destructor in this article from Herb Sutter.
A virtual constructor is not possible but virtual destructor is possible.
Let us experiment.......
#include <iostream>
using namespace std;
class Base
{
public:
Base(){
cout << "Base Constructor Called\n";
}
~Base(){
cout << "Base Destructor called\n";
}
};
class Derived1: public Base
{
public:
Derived1(){
cout << "Derived constructor called\n";
}
~Derived1(){
cout << "Derived destructor called\n";
}
};
int main()
{
Base *b = new Derived1();
delete b;
}
The above code output the following:
Base Constructor Called
Derived constructor called
Base Destructor called
The construction of derived object follow the construction rule but when we delete the "b" pointer(base pointer) we have found that only the base destructor is called. But this must not happen. To do the appropriate thing, we have to make the base destructor virtual.
Now let see what happens in the following:
#include <iostream>
using namespace std;
class Base
{
public:
Base(){
cout << "Base Constructor Called\n";
}
virtual ~Base(){
cout << "Base Destructor called\n";
}
};
class Derived1: public Base
{
public:
Derived1(){
cout << "Derived constructor called\n";
}
~Derived1(){
cout << "Derived destructor called\n";
}
};
int main()
{
Base *b = new Derived1();
delete b;
}
The output changed as following:
Base Constructor Called
Derived Constructor called
Derived destructor called
Base destructor called
So the destruction of the base pointer (which takes an allocation on derived object!) follows the destruction rule, i.e first the Derived, then the Base.
On the other hand, there is nothing like a virtual constructor.
Declare destructors virtual in polymorphic base classes. This is Item 7 in Scott Meyers' Effective C++. Meyers goes on to summarize that if a class has any virtual function, it should have a virtual destructor, and that classes not designed to be base classes or not designed to be used polymorphically should not declare virtual destructors.
Make the destructor virtual whenever your class is polymorphic.
Also be aware that deleting a base class pointer when there is no virtual destructor will result in undefined behavior. Something that I learned just recently:
How should overriding delete in C++ behave?
I've been using C++ for years and I still manage to hang myself.
Calling destructor via a pointer to a base class
struct Base {
virtual void f() {}
virtual ~Base() {}
};
struct Derived : Base {
void f() override {}
~Derived() override {}
};
Base* base = new Derived;
base->f(); // calls Derived::f
base->~Base(); // calls Derived::~Derived
Virtual destructor call is no different from any other virtual function call.
For base->f(), the call will be dispatched to Derived::f(), and it's the same for base->~Base() - its overriding function - the Derived::~Derived() will be called.
Same happens when destructor is being called indirectly, e.g. delete base;. The delete statement will call base->~Base() which will be dispatched to Derived::~Derived().
Abstract class with non-virtual destructor
If you are not going to delete object through a pointer to its base class - then there is no need to have a virtual destructor. Just make it protected so that it won't be called accidentally:
// library.hpp
struct Base {
virtual void f() = 0;
protected:
~Base() = default;
};
void CallsF(Base& base);
// CallsF is not going to own "base" (i.e. call "delete &base;").
// It will only call Base::f() so it doesn't need to access Base::~Base.
//-------------------
// application.cpp
struct Derived : Base {
void f() override { ... }
};
int main() {
Derived derived;
CallsF(derived);
// No need for virtual destructor here as well.
}
Virtual keyword for destructor is necessary when you want different destructors should follow proper order while objects is being deleted through base class pointer.
for example:
Base *myObj = new Derived();
// Some code which is using myObj object
myObj->fun();
//Now delete the object
delete myObj ;
If your base class destructor is virtual then objects will be destructed in a order(firstly derived object then base ). If your base class destructor is NOT virtual then only base class object will get deleted(because pointer is of base class "Base *myObj"). So there will be memory leak for derived object.
To be simple,
Virtual destructor is to destruct the resources in a proper order, when you delete a base class pointer pointing to derived class object.
#include<iostream>
using namespace std;
class B{
public:
B(){
cout<<"B()\n";
}
virtual ~B(){
cout<<"~B()\n";
}
};
class D: public B{
public:
D(){
cout<<"D()\n";
}
~D(){
cout<<"~D()\n";
}
};
int main(){
B *b = new D();
delete b;
return 0;
}
OUTPUT:
B()
D()
~D()
~B()
==============
If you don't give ~B() as virtual. then output would be
B()
D()
~B()
where destruction of ~D() is not done which leads to leak
I like to think about interfaces and implementations of interfaces. In C++ speak interface is pure virtual class. Destructor is part of the interface and expected to implemented. Therefore destructor should be pure virtual. How about constructor? Constructor is actually not part of the interface because object is always instantiated explicitly.
Virtual base class destructors are "best practice" - you should always use them to avoid (hard to detect) memory leaks. Using them, you can be sure all destructors in the inheritance chain of your classes are beeing called (in proper order). Inheriting from a base class using virtual destructor makes the destructor of the inheriting class automatically virtual, too (so you do not have to retype 'virtual' in the inheriting class destructor declaration).
I think the core of this question is about virtual methods and polymorphism, not the destructor specifically. Here is a clearer example:
class A
{
public:
A() {}
virtual void foo()
{
cout << "This is A." << endl;
}
};
class B : public A
{
public:
B() {}
void foo()
{
cout << "This is B." << endl;
}
};
int main(int argc, char* argv[])
{
A *a = new B();
a->foo();
if(a != NULL)
delete a;
return 0;
}
Will print out:
This is B.
Without virtual it will print out:
This is A.
And now you should understand when to use virtual destructors.
If you use shared_ptr(only shared_ptr, not unique_ptr), you don't have to have the base class destructor virtual:
#include <iostream>
#include <memory>
using namespace std;
class Base
{
public:
Base(){
cout << "Base Constructor Called\n";
}
~Base(){ // not virtual
cout << "Base Destructor called\n";
}
};
class Derived: public Base
{
public:
Derived(){
cout << "Derived constructor called\n";
}
~Derived(){
cout << "Derived destructor called\n";
}
};
int main()
{
shared_ptr<Base> b(new Derived());
}
output:
Base Constructor Called
Derived constructor called
Derived destructor called
Base Destructor called
What is a virtual destructor or how to use virtual destructor
A class destructor is a function with same name of the class preceding with ~ that will reallocate the memory that is allocated by the class. Why we need a virtual destructor
See the following sample with some virtual functions
The sample also tell how you can convert a letter to upper or lower
#include "stdafx.h"
#include<iostream>
using namespace std;
// program to convert the lower to upper orlower
class convertch
{
public:
//void convertch(){};
virtual char* convertChar() = 0;
~convertch(){};
};
class MakeLower :public convertch
{
public:
MakeLower(char *passLetter)
{
tolower = true;
Letter = new char[30];
strcpy(Letter, passLetter);
}
virtual ~MakeLower()
{
cout<< "called ~MakeLower()"<<"\n";
delete[] Letter;
}
char* convertChar()
{
size_t len = strlen(Letter);
for(int i= 0;i<len;i++)
Letter[i] = Letter[i] + 32;
return Letter;
}
private:
char *Letter;
bool tolower;
};
class MakeUpper : public convertch
{
public:
MakeUpper(char *passLetter)
{
Letter = new char[30];
toupper = true;
strcpy(Letter, passLetter);
}
char* convertChar()
{
size_t len = strlen(Letter);
for(int i= 0;i<len;i++)
Letter[i] = Letter[i] - 32;
return Letter;
}
virtual ~MakeUpper()
{
cout<< "called ~MakeUpper()"<<"\n";
delete Letter;
}
private:
char *Letter;
bool toupper;
};
int _tmain(int argc, _TCHAR* argv[])
{
convertch *makeupper = new MakeUpper("hai");
cout<< "Eneterd : hai = " <<makeupper->convertChar()<<" ";
delete makeupper;
convertch *makelower = new MakeLower("HAI");;
cout<<"Eneterd : HAI = " <<makelower->convertChar()<<" ";
delete makelower;
return 0;
}
From the above sample you can see that the destructor for both MakeUpper and MakeLower class is not called.
See the next sample with the virtual destructor
#include "stdafx.h"
#include<iostream>
using namespace std;
// program to convert the lower to upper orlower
class convertch
{
public:
//void convertch(){};
virtual char* convertChar() = 0;
virtual ~convertch(){}; // defined the virtual destructor
};
class MakeLower :public convertch
{
public:
MakeLower(char *passLetter)
{
tolower = true;
Letter = new char[30];
strcpy(Letter, passLetter);
}
virtual ~MakeLower()
{
cout<< "called ~MakeLower()"<<"\n";
delete[] Letter;
}
char* convertChar()
{
size_t len = strlen(Letter);
for(int i= 0;i<len;i++)
{
Letter[i] = Letter[i] + 32;
}
return Letter;
}
private:
char *Letter;
bool tolower;
};
class MakeUpper : public convertch
{
public:
MakeUpper(char *passLetter)
{
Letter = new char[30];
toupper = true;
strcpy(Letter, passLetter);
}
char* convertChar()
{
size_t len = strlen(Letter);
for(int i= 0;i<len;i++)
{
Letter[i] = Letter[i] - 32;
}
return Letter;
}
virtual ~MakeUpper()
{
cout<< "called ~MakeUpper()"<<"\n";
delete Letter;
}
private:
char *Letter;
bool toupper;
};
int _tmain(int argc, _TCHAR* argv[])
{
convertch *makeupper = new MakeUpper("hai");
cout<< "Eneterd : hai = " <<makeupper->convertChar()<<" \n";
delete makeupper;
convertch *makelower = new MakeLower("HAI");;
cout<<"Eneterd : HAI = " <<makelower->convertChar()<<"\n ";
delete makelower;
return 0;
}
The virtual destructor will call explicitly the most derived run time destructor of class so that it will be able to clear the object in a proper way.
Or visit the link
https://web.archive.org/web/20130822173509/http://www.programminggallery.com/article_details.php?article_id=138
I thought it would be beneficial to discuss the "undefined" behavior, or at least the "crash" undefined behavior that may occur when deleting through a base class(/struct) without a virtual destructor, or more precisely no vtable. The code below list a few simple structs (the same would be true for classes).
#include <iostream>
using namespace std;
struct a
{
~a() {}
unsigned long long i;
};
struct b : a
{
~b() {}
unsigned long long j;
};
struct c : b
{
~c() {}
virtual void m3() {}
unsigned long long k;
};
struct d : c
{
~d() {}
virtual void m4() {}
unsigned long long l;
};
int main()
{
cout << "sizeof(a): " << sizeof(a) << endl;
cout << "sizeof(b): " << sizeof(b) << endl;
cout << "sizeof(c): " << sizeof(c) << endl;
cout << "sizeof(d): " << sizeof(d) << endl;
// No issue.
a* a1 = new a();
cout << "a1: " << a1 << endl;
delete a1;
// No issue.
b* b1 = new b();
cout << "b1: " << b1 << endl;
cout << "(a*) b1: " << (a*) b1 << endl;
delete b1;
// No issue.
c* c1 = new c();
cout << "c1: " << c1 << endl;
cout << "(b*) c1: " << (b*) c1 << endl;
cout << "(a*) c1: " << (a*) c1 << endl;
delete c1;
// No issue.
d* d1 = new d();
cout << "d1: " << d1 << endl;
cout << "(c*) d1: " << (c*) d1 << endl;
cout << "(b*) d1: " << (b*) d1 << endl;
cout << "(a*) d1: " << (a*) d1 << endl;
delete d1;
// Doesn't crash, but may not produce the results you want.
c1 = (c*) new d();
delete c1;
// Crashes due to passing an invalid address to the method which
// frees the memory.
d1 = new d();
b1 = (b*) d1;
cout << "d1: " << d1 << endl;
cout << "b1: " << b1 << endl;
delete b1;
/*
// This is similar to what's happening above in the "crash" case.
char* buf = new char[32];
cout << "buf: " << (void*) buf << endl;
buf += 8;
cout << "buf after adding 8: " << (void*) buf << endl;
delete buf;
*/
}
I'm not suggesting whether you need virtual destructors or not, though I think in general it's a good practice to have them. I'm just pointing out the reason you may end up with a crash if your base class(/struct) does not have a vtable and your derived class(/struct) does and you delete an object via a base class(/struct) pointer. In this case, the address you pass to the heap's free routine is invalid and thus the reason for the crash.
If you run the above code you'll see clearly when the issue occurs. When the this pointer of the base class(/struct) is different from the this pointer of the derived class(/struct) you're going to run into this problem. In the sample above, struct a and b don't have vtables. structs c and d do have vtables. Thus an a or b pointer to a c or d object instance will be fixed up to account for the vtable. If you pass this a or b pointer to delete it will crash due to the address being invalid to the heap's free routine.
If you plan to delete derived instances which have vtables from base class pointers, you need to ensure the base class has a vtable. One way to do that is to add a virtual destructor, which you might want anyway to properly clean up resources.
when you need to call derived class destructor from base class. you need to declare virtual base class destructor in base class.
I propose this: If a class or struct is not final, you should define virtual destructor for it.
I know this looks like an excessively vigilant overkill to become a rule of thumb. But, it is the only way to be sure that someone deriving from your class won't have UB when deleting with base pointer.
Scott Meyers' recommendation in Effective C++ quoted below is good but not enough to be sure.
if a class has any virtual function, it should have a virtual
destructor, and that classes not designed to be base classes or not
designed to be used polymorphically should not declare virtual
destructors.
For example, in the program below, base class B does not have any virtual functions, so according to Meyer, you don't need to write a virtual destructor. However, you have UB below, if you don't:
#include <iostream>
struct A
{
~A()
{
std::cout << "A::~A()" << std::endl;
}
};
struct B
{
};
struct C : public B
{
A a;
};
int main(int argc, char *argv[])
{
B *b = new C;
delete b; // UB, and won't print "A::~A()"
return 0;
}
I think most answers here miss the point, with the exception of the accepted one, which is a good thing. However, let me add one more with a different view on the issue: You need a virtual destructor if you want to polymorphically delete instances of this class.
This kind-of sidesteps the question, so let me elaborate: As many pointed out, you get undesired behaviour if you invoke delete base_ptr and the destructor is not virtual. However, there are several assumptions there that need to be make explicit:
If your class is not a baseclass, you will hopefully not write code like this. In this case I don't mean the manual memory management, which is bad in and of itself, but deriving publicly from this class. A class not designed as baseclass should not be inherited from, like e.g. std::string. C++ allows you to shoot yourself in the foot. This is your fault though, not that of the baseclass not having a virtual destructor.
If the destructor is not accessible (protected or private), this code won't compile, so the undesired behaviour can not occur. Having a protected destructor is useful, especially for mixins but also (to a lesser extent) for interfaces. You don't want to incur the overhead of virtual functions unless you actually make use of them. Making the destructor protected instead prevents undesired behaviour but doesn't restrict you otherwise.
If you actually write a class that is supposed to be derived from, you will typically have virtual functions anyways. As a user of them, you will typically only use them via a pointer to the baseclass. When this use includes disposing of them, it then needs to be polymorphic as well. This is then the case when you should make the destructor virtual.
For a similarly different view on the subject, also read When should you not use virtual destructors?
A basic definition about virtual is it determines if a member function of a class can be over-ridden in its derived classes.
A class's D-tor is called basically at the end of the scope, but there is a problem, for example when we define an instance on the Heap (dynamic allocation), we should delete it manually.
As soon as the instruction get executed, the base class destructor get called, but not for the derived one.
A Pratical example is when, in control field, you have to manipulate effectors, actuators.
At the end of the scope, if the destructor of one of the power elements (Actuator), isn't called, there will be fatal consequences.
#include <iostream>
class Mother{
public:
Mother(){
std::cout<<"Mother Ctor"<<std::endl;
}
virtual~Mother(){
std::cout<<"Mother D-tor"<<std::endl;
}
};
class Child: public Mother{
public:
Child(){
std::cout<<"Child C-tor"<<std::endl;
}
~Child(){
std::cout<<"Child D-tor"<<std::endl;
}
};
int main()
{
Mother *c = new Child();
delete c;
return 0;
}
Any class that is inherited publicly, polymorphic or not, should have a virtual destructor. To put another way, if it can be pointed to by a base class pointer, its base class should have a virtual destructor.
If virtual, the derived class destructor gets called and then the base class destructor. If not virtual, only the base class destructor gets called.
Make all destructors virtual unless you have good reason not to.
Otherwise evil like this happens:
Suppose you have an array of Fruit pointers with both Apple and Orange objects.
When you delete from the collection of Fruit objects,
~Apple() and ~Orange() fail to be called unless ~Fruit() is virtual.
Example done right:
#include <iostream>
using namespace std;
struct Fruit { // good
virtual ~Fruit() { cout << "peel or core should have been tossed" << endl; }
};
struct Apple: Fruit { virtual ~Apple() {cout << "toss core" << endl; } };
struct Orange: Fruit { virtual ~Orange() {cout << "toss peel" << endl; } };
int main() {
Fruit *basket[]={ new Apple(), new Orange() };
for (auto fruit: basket) delete fruit;
};
good output
toss core
peel or core should have been tossed
toss peel
peel or core should have been tossed
Example done wrong:
#include <iostream>
using namespace std;
struct Fruit { // bad
~Fruit() { cout << "peel or core should have been tossed" << endl; }
};
struct Apple: Fruit { virtual ~Apple() {cout << "toss core" << endl; } };
struct Orange: Fruit { virtual ~Orange() {cout << "toss peel" << endl; } };
int main() {
Fruit *basket[]={ new Apple(), new Orange() };
for (auto fruit: basket) delete fruit;
};
bad output
peel or core should have been tossed
peel or core should have been tossed
(Note: Where I used struct for brevity, normally use class and specify public)
you have to use the concept of base, derive class and use them properly. while doing it, I'm faced with some problems.
Can you help me solve the issues ?
My Code :
#include <iostream>
class Base {
public:
Base (int data) : _data(data){}
virtual void printData(){
std::cout << "Base" << _data << std::endl;
}
private:
int _data;
};
class Derived : public Base {
public:
Derived(derivdata) : Base(derivdata){}
void printData(){
cout << "Derived " << _data << endl;
}
}
int main(){
Derived *var = new Derived(5);
var-> printData();
Base* basevar = static_cast<Base>(*var);
basevar-> printData();
}
Base* basevar = static_cast<Base>(*var); is wrong,
it would be Base* basevar = static_cast<Base*>(var);
but Base* basevar = var; is sufficient.
and version without memory leak:
int main(){
Derived derived(5);
derived.printData(); // "Derived 5\n"
Base& base = *derived;
base.printData(); // "Derived 5\n" as method is virtual
}
There are several problems here, that could be typos:
a missing semicolon after the class Derived
in the constructor Derived(derivdata) : Base(derivdata){} the type of derivdata is missing. (there is no default int in c++)
you are missing the namespace std:: before cout and endl in Derived::printData
in Derived::printData the member _data is not accessible, because it is private to Base (You could make it protected or write a getter for it)
The core problem is this cast: Base* basevar = static_cast<Base>(*var); This will actually dereference the value of var, then try to cast this value to Base, and afterwards tries to assign the result (of type Base) to a variable of type Base*.
For polymorphic casting you can use dynamic_cast: Base* basevar = dynamic_cast<Base*>(var);
But don't expect basevar->printData(); to print "Base", since the pointer is still pointing to an instance of Derived and the method is declared virtual.
To access a method of the base class you need to explicitly tell the compiler to do so:
basevar->Base::printData();(prints "Base5")
In the following code, the function foo is copy constructing a Base object c from a Derived object d. My question is: are we getting an exact copy? Because I'm not getting the polymorphic behavior I'm expecting
#include<iostream>
class Base
{
public:
virtual void sayHello()
{
std::cout << "Hello Base" << std::endl ;
}
};
class Derived: public Base
{
public:
void sayHello() override
{
std::cout << "Hello Derived" << std::endl ;
}
};
void foo(Base* d)
{
Base* c = new Base(*d);
c->sayHello() ;
}
int main()
{
Derived d;
foo(&d) ; //outputs Hello Base
}
There is no virtual constructor nor copy constructor.
However, it is possible to define a function that behaves like one.
In my case, it is the virtual member function copy() which I added to OP's sample:
#include <iostream>
class Base
{
public:
virtual Base* copy() const { return new Base(*this); }
virtual void sayHello()
{
std::cout << "Hello Base" << std::endl ;
}
};
class Derived: public Base
{
public:
virtual Base* copy() const override { return new Derived(*this); }
void sayHello() override
{
std::cout << "Hello Derived" << std::endl ;
}
};
void foo(Base* d)
{
Base* c = d->copy();
c->sayHello() ;
}
int main()
{
Derived d;
foo(&d) ; //outputs Hello Derived
return 0;
}
Output:
Hello Derived
Live Demo on coliru
The drawback is that every derived class of Base has to provide it to make it function properly. (I've no idea how to convince the compiler to check this for me with any trick.)
A partial solution could be to make copy() pure virtual in the class Base (assuming it is not meant to be instantiable).
you may wonna change the line of new
Base* c = new Derived(*d);
so you have the type Derived in a Base pointer. During runtime it is looked up, which type it is and you get the right output.
let me know if im wrong... just created this out of my mind on the fly.
To answer your question about whether or not this is copy constructing lets add some members. Base will have a member, m_b and Derived will inherit m_b but also have another member m_d
#include <iostream>
struct Base {
const int m_b;
Base() = delete;
Base(const int a_b) : m_b(a_b) {}
virtual void sayHello() {
std::cout << "Base " << m_b << std::endl;
}
};
struct Derived : public Base {
const int m_d;
Derived() = delete;
Derived(const int a_b, const int a_d) : Base(a_b), m_d(a_d) {}
void sayHello() override {
std::cout << "Derived " << m_b << ' ' << m_d << std::endl;
}
};
void foo(Derived* a) {
Base* b = new Base(*a);
b->sayHello(); // Output is "Base 1", 1 was copied from argument a
}
void bar(Derived* a) {
Base* d = new Derived(*a);
d->sayHello(); // Output is "Derived 1 2"
}
int main() {
Derived d(1, 2);
foo(&d);
bar(&d);
return 0;
}
The line:
Base* b = new Base(*a);
Created a Base and so sayHello calls Base's implementation which doesn't know about m_d. However this line does copy m_b from the derived class
The line:
Base* d = new Derived(*a);
Created a Derived and so sayHello calls Derived's implementation which copied both m_b and m_d
Expected polymorphic behavior will come into existence when the Base class pointer points to Derived class object. Then at run time the actual type of object pointed to by the Base class pointer will be checked and appropriate function will get called.
Base* c = new Base(*d); // <<-- case of object slicing
Here, c points to Base class object. Therefore, c->sayHello() ; is bound to call the Base::sayHello() at runtime.
are we getting an exact copy?. No since you are creating a Base object due to new Base. Due to object slicing the Base part of the *d object is passed to copy c'tor and what you get is corresponding Base object.
Base *c = new Derived(*d); will give the expected behavior.
#include <iostream>
using namespace std;
class A {
public:
void m1(){ cout << 'A'; }
virtual void m2(){ cout << 'B'; }
virtual void m3(){ cout << 'C'; }
};
class B: public A {
public:
void m1(){ cout << 'D'; }
void m2(){ cout << 'E'; }
};
class C: public B {
public:
void m3(){ cout << 'F'; }
};
int main()
{
cout << "Hello World!" << endl;
A* a = new B();
a->m1();
a->m2();
a->m3();
return 0;
}
What will be the output? I initially thought that it will be "D E C" but after running the program it was "A E C"
Could one elaborate what is going on behind this line of code:
A* a = new B();
Virtual member functions are dispatched based on the dynamic (run-time) type of the object. Non-virtual member functions are dispatched based on the static (compile-time) type of the object.
A *a = new B();
a points to an object whose dynamic type is B. The static type of a is A*, however, which means that the static type of *a is A.
Virtual functions (m2 and m3) are dispatched based on the dynamic type, so B::m2 and B::m3 are called.
Non-virtual functions are dispatched based on the static type. The static type of *a is A, so A::m1 is called.
What exactly is going on in the new line? A new object of type B is created dynamically, and the new expression returns a pointer to that object (of type B*). Then, a derived-to-base conversion is applied to that pointer to convert it to an A*, which is used to initialise the variable a.
In pseudo-code, showing intermediary steps:
B *tmp_b = new B(); // allocate and initialise B object
A *tmp_a = convert_derived_to_base(tmp_b);
A *a = tmp_a;
m1 method is not virtual in class A, it can not be override .m2 is virtual so it override with class B method .
Here Class A is your parent class, so to override parent class's method you need to make it virtual in class A. So that Class B's method will be called. Please see more information on Virtual Functions
The output will be
AEC
The reason is that A::m1 was not declared virtual. In this line
A* a = new B();
You are declaring an A* variable named a, which is actually an instance of the derived class B. This is allowed because "B is a type of A".
When a function is declared virtual, if a derived class defines that function then the one in the base class is overridden with the version in the derived class. So the following calls will be made
a->m1(); // A::m1
a->m2(); // B::m2
a->m3(); // A::m3
void m1(){ cout << 'A'; }
virtual void m2(){ cout << 'B'; }
virtual void m3(){ cout << 'C'; }
The output is totally fine as per your programs code.
void m1(){ cout << 'A'; is not a virtual function.
So a->m1(); calls the base version of m1 and gives the o/p as A.
m2() is a virtual function and it has a new implementation inside class B.
void m2(){ cout << 'E'; }
So a->m2(); calls the class B version of the m2 function and gives the output E.
There is no new implementation of m3 inside class B.
So a->m3(); calls m3 which is inherited from class A.
To get the o/p as DEC you have to just change the m1 function to a virtual function like below
class A {
public:
virtual void m1(){ cout << 'A'; }
virtual void m2(){ cout << 'B'; }
virtual void m3(){ cout << 'C'; }
};
What's the difference between redefining and using a virtual function? Don't they serve the same purpose? You are allowing objects of a derived class to call their own versions of a function by the same name in both cases..so where is the distinction?
An example says it best:
#include <iostream>
using namespace std;
class A {
public:
virtual void f1() { cout << "Class A" << endl; }
void f2() { cout << "Class A" << endl; }
virtual ~A(){}
};
class B : public A {
public:
virtual void f1() { cout << "Class B" << endl; }
void f2() { cout << "Class B" << endl; }
virtual ~B(){}
};
int main()
{
A *a = new B;
a->f1();
a->f2();
}
...
$ ./override
Class B
Class A
You can see that when we reference the instance of B, f1() still calls B's version, but f2() calls A's.
When you declare a function virtual, you are saying that when we call it we should use the vtable to look up the right version of the function to call, and so you will always get the most derived version of the function, even if you reference it as an ancestor type. Without virtual, it will simply use the definition in the type you are referencing it as.
The distinction is when you have a reference or pointer to the base class. A call to a virtual function will call the most-derived version while a call to a normal function will call the base-class version.
If you're using a variable directly or a reference or pointer to the most-derived class, there is no practical difference.
TL;DR
The only way to take advantage of the polymorphism in C++ is via virtual functions and pointers (and references). The virtual keyword will tell the compiler to allocate a virtual function table where to look when deciding what version of the virtual function to call when dealing with pointer to the base class with a dynamic type that is of a child class.
How does polymorphism work in C++
Let's take a simple example:
class A { public: virtual void eat() { std::cout << "Class A" << std::endl; }
class B : public A {};
class C : public B { virtual void eat() { std::cout << "Class C" << std::endl; }
Note: the virtual keyword could be omitted after the first function/method definition.
The following:
A a; B b; C c;
A* ptrA = &a; A* ptrB = &b; A* ptrC = &c;
ptrA->eat();
ptrB->eat();
ptrC->eat();
will print:
Class A
Class A
Class C
If we wouldn't have declared the function eat virtual, the output would have simply been:
Class A
Class A
Class A
But why
A* ptrA = &a; A* ptrB = &b; A* ptrC = &c;
Instead of
A* ptrA = &a; B* ptrB = &b; C* ptrC = &c;
With the second version, the eat() method can simply be redefined for each derived class, without the need for virtual, right?