I have a class A and class B which inherits from class A. Class B has a variable and a function that are not available in class A. I made a pointer
A* ptr=new B();
So how can ptr access the variable and function that belongs to class B?
Class A simply cannot "see" the functions of class B. Instead, you'd have to use something like a dynamic_cast from A to B, check for null, and then proceed as you like. Here is a nice tutorial to explain this a lot better than I can. Each of the casts have their advantages and disadvantages; learn them well. Also, try to avoid C style casting.
EDIT : Seems I misread the question. The answer is still correct, though. Class A would not be able to "see" the variables of Class B. The casting would still allow you access to them.
You can force the derived class to implement a pure virtual method defined in the base class:
class A
{
public:
virtual void do_things() = 0;
};
class B : public A
{
public:
virtual void do_things()
{
//Implementation
}
};
This way you can call the method implemented by class B through a pointer A*:
A* a_ptr = new B();
//The method implemented in class B will be called
a_ptr->do_things();
You could also make the assumption that a_ptr points to an object of class B. If a_ptr doesn't point to a B the pointer returned by dynamic_cast<B*> will be a nullptr. You can use static_cast<B*> if there are no virtual methods in A.
b_ptr = dynamic_cast<B*>(a_ptr);
b_ptr->do_things;
This way you don't need the pure virtual function in class A.
In the artificial example below, if I static_cast to the base class, when I call the setSnapshot() function it still calls the actual object setSnapshot(). This is what I want to happen. My question is can I always rely on this to work?
In code I am working on, we have this class hierarchy and in the b class there are macros used which static cast to the b type. This is to downcast from a base type so that specialised function in b can be called.
#include <iostream>
class a {
};
class b: public a {
public:
virtual void setSnapshot() { std::cout << "setting b snapshot\n"; }
};
class c : public b {
public:
virtual void setSnapshot() { std::cout << "setting c snapshot\n"; }
};
int main() {
a* o = new c;
//specifically casting to b
static_cast<b*>(o)->setSnapshot(); //prints setting c snapshot - what I want to happen
delete o;
return 0;
}
The title suggests that you're misunderstanding what the case does. new c creates an object of type c, and it will remain a c until it's destructed.
If you were to cast it to an a, you'd create a copy. But yu're only casting pointers. That doesn't affect the original object. That's still a c, and that's why you end up calling c::setSnapshot().
As long as a function is virtual in the statically known type a call of it will go to the override that is most derived.
For single inheritance this can be understood as a search for an implementation up the base class chain, starting in the most derived class.
In practice, for C++, the dynamic search is not done, and the effect of the search is instead implemented as a simple table lookup.
I have the following problem. I am busy coding up an inheritance structure in C++. Briefly, this is what I am attempting to do. Class A is the base class and classes B and C are inherited class. Classes B and C each have different and unique member functions.
Now, using the boost smart pointer library I make a vector of shared pointer of class type A as follows:
class A{
A();
virtual print();
}
class B : public A{
B();
virtual print();
}
class C{
C();
virtual print();
void uniqueFunc();
}
int main(){
vector<shared_ptr<A> > myA;
shared_ptr<B> myB;
shared_ptr<C> myC;
myA.push_back(myB);
myA.push_back(myC);
}
Now I have a method for checking the type of members within the vector. This isn't the issue. My question is, how do I call uniqueFunc for a member of class type C within the vector. Would I have to downcast? Or do I have to create virtual function. I do however, need uniqueFunc to be unique to class C. I would like to avoid creating copies of members and rather directly change the member in the vector.
You either need to downcast to boost::shared_ptr<C> or create a virtual function in A.
Assuming you want to downcast, you could use Boost's pointer cast functions:
boost::static_pointer_cast<C>(myA.back())->uniqueFunc();
(or use dynamic_pointer_cast<C> if you need dynamic type checking.)
conv.h
class Base
{
public:
void foo();
};
class Derived: public Base
{
public:
void bar();
};
class A {};
class B
{
public:
void koko();
};
conv.cpp
void Base::foo()
{
cout<<"stamm";
}
void Derived::bar()
{
cout<<"bar shoudn't work"<<endl;
}
void B::koko()
{
cout<<"koko shoudn't work"<<endl;
}
main.cpp
#include "conv.h"
#include <iostream>
int main()
{
Base * a = new Base;
Derived * b = static_cast<Derived*>(a);
b->bar();
Derived * c = reinterpret_cast<Derived*>(a);
c->bar();
A* s1 = new A;
B* s2 = reinterpret_cast<B*>(s1);
s2->koko();
}
output:
bar shoudn't work
bar shoudn't work
koko shoudn't work
How come the method bar is succeeded to be called in run time despite that I have created a Base class not derived?? it works even with two types of conversions (static and reinterpret cast).
same question as above but with unrelated classes (A & B) ??
Undefined behaviour can do anything, including appear to work.
It's working (read: "compiling and not crashing") 'cause you never use the this pointer in your nominally "member" functions. If you tried to print out a member variable, for example, you'd get the garbage output or crashes you expect - but these functions as they are now don't depend on anything in the classes they're supposedly part of. this could even be NULL for all they care.
The compiler knows a Derived can use member functions foo() and bar() and knows where to find them. After you did your "tricks", you had pointers to Derived.
The fact that they were pointers of type Derived -- regardless of what data was associated with those pointers -- allowed them to call the functions foo() and kook() associated with Derived.
As has been mentioned, if you had actually used the data at the pointers (i.e. read or wrote data members relative to this belonging to the Derived class (which you don't have in this case), you would have been access memory that didn't belong to your objects.
I'm having trouble understanding what the purpose of the virtual keyword in C++. I know C and Java very well but I'm new to C++
From wikipedia
In object-oriented programming, a
virtual function or virtual method is
a function or method whose behavior
can be overridden within an inheriting
class by a function with the same
signature.
However I can override a method as seen below without using the virtual keyword
#include <iostream>
using namespace std;
class A {
public:
int a();
};
int A::a() {
return 1;
}
class B : A {
public:
int a();
};
int B::a() {
return 2;
}
int main() {
B b;
cout << b.a() << endl;
return 0;
}
//output: 2
As you can see below, the function A::a is successfully overridden with B::a without requiring virtual
Compounding my confusion is this statement about virtual destructors, also from wikipedia
as illustrated in the following example,
it is important for a C++ base class
to have a virtual destructor to ensure
that the destructor from the most
derived class will always be called.
So virtual also tells the compiler to call up the parent's destructors? This seems to be very different from my original understanding of virtual as "make the function overridable"
Make the following changes and you will see why:
#include <iostream>
using namespace std;
class A {
public:
int a();
};
int A::a() {
return 1;
}
class B : public A { // Notice public added here
public:
int a();
};
int B::a() {
return 2;
}
int main() {
A* b = new B(); // Notice we are using a base class pointer here
cout << b->a() << endl; // This will print 1 instead of 2
delete b; // Added delete to free b
return 0;
}
Now, to make it work like you intended:
#include <iostream>
using namespace std;
class A {
public:
virtual int a(); // Notice virtual added here
};
int A::a() {
return 1;
}
class B : public A { // Notice public added here
public:
virtual int a(); // Notice virtual added here, but not necessary in C++
};
int B::a() {
return 2;
}
int main() {
A* b = new B(); // Notice we are using a base class pointer here
cout << b->a() << endl; // This will print 2 as intended
delete b; // Added delete to free b
return 0;
}
The note that you've included about virtual destructors is exactly right. In your sample there is nothing that needs to be cleaned-up, but say that both A and B had destructors. If they aren't marked virtual, which one is going to get called with the base class pointer? Hint: It will work exactly the same as the a() method did when it was not marked virtual.
You could think of it as follows.
All functions in Java are virtual. If you have a class with a function, and you override that function in a derived class, it will be called, no matter the declared type of the variable you use to call it.
In C++, on the other hand, it won't necessarily be called.
If you have a base class Base and a derived class Derived, and they both have a non-virtual function in them named 'foo', then
Base * base;
Derived *derived;
base->foo(); // calls Base::foo
derived->foo(); // calls Derived::foo
If foo is virtual, then both call Derived::foo.
virtual means that the actual method is determined runtime based on what class was instantiated not what type you used to declare your variable.
In your case this is a static override it will go for the method defined for class B no matter what was the actual type of the object created
So virtual also tells the compiler to call up the parent's destructors? This seems to be very different from my original understanding of virtual as "make the function overridable"
Your original and your new understanding are both wrong.
Methods (you call them functions) are always overridable. No matter if virtual, pure, nonvirtual or something.
Parent destructors are always called. As are the constructors.
"Virtual" does only make a difference if you call a method trough a pointer of type pointer-to-baseclass. Since in your example you don't use pointers at all, virtual doesn't make a difference at all.
If you use a variable a of type pointer-to-A, that is A* a;, you can not only assign other variables of type pointer-to-A to it, but also variables of type pointer-to-B, because B is derived from A.
A* a;
B* b;
b = new B(); // create a object of type B.
a = b; // this is valid code. a has still the type pointer-to-A,
// but the value it holds is b, a pointer to a B object.
a.a(); // now here is the difference. If a() is non-virtual, A::a()
// will be called, because a is of type pointer-to-A.
// Whether the object it points to is of type A, B or
// something entirely different doesn't matter, what gets called
// is determined during compile time from the type of a.
a.a(); // now if a() is virtual, B::a() will be called, the compiler
// looks during runtime at the value of a, sees that it points
// to a B object and uses B::a(). What gets called is determined
// from the type of the __value__ of a.
As you can see below, the function A::a is successfully overridden with B::a without requiring virtual
It may, or it may not work. In your example it works, but it's because you create and use an B object directly, and not through pointer to A. See C++ FAQ Lite, 20.3.
So virtual also tells the compiler to call up the parent's destructors?
A virtual destructor is needed if you delete a pointer of base class pointing to an object of derived class, and expect both base and derived destructors to run. See C++ FAQ Lite, 20.7.
You need the virtual if you use a base class pointer as consultutah (and others while I'm typing ;) ) says it.
The lack of virtuals allows to save a check to know wich method it need to call (the one of the base class or of some derived). However, at this point don't worry about performances, just on correct behaviour.
The virtual destructor is particulary important because derived classes might declare other variables on the heap (i.e. using the keyword 'new') and you need to be able to delete it.
However, you might notice, that in C++, you tend to use less deriving than in java for example (you often use templates for a similar use), and maybe you don't even need to bother about that. Also, if you never declare your objects on the heap ("A a;" instead of "A * a = new A();") then you don't need to worry about it either. Of course, this will heavily depend on what/how you develop and if you plan that someone else will derive your class or not.
Try ((A*)&b).a() and see what gets called then.
The virtual keyword lets you treat an object in an abstract way (I.E. through a base class pointer) and yet still call descendant code...
Put another way, the virtual keyword "lets old code call new code". You may have written code to operate on A's, but through virtual functions, that code can call B's newer a().
Say you instantiated B but held it as an instance of an A:
A *a = new B();
and called function a() whose implementation of a() will be called?
If a() isn't virtual A's will be called. If a() was virtual the instantiated sub class version of a() would be called regardless of how you're holding it.
If B's constructor allocated tons of memory for arrays or opened files, calling
delete a;
would ensure B's destructor was called regardless as to how it was being held, be it by a base class or interface or whatever.
Good question by the way.
I always think about it like chess pieces (my first experiment with OO).
A chessboard holds pointers to all the pieces. Empty squares are NULL pointers. But all it knows is that each pointer points a a chess piece. The board does not need to know more information. But when a piece is moved the board does not know it is a valid move as each pice has different characteristica about how it moves. So the board needs to check with the piece if the move is valid.
Piece* board[8][8];
CheckMove(Point const& from,Point const& too)
{
Piece* piece = board[from.x][from.y];
if (piece != NULL)
{
if (!piece->checkValidMove(from,too))
{ throw std::exception("Bad Move");
}
// Other checks.
}
}
class Piece
{
virtual bool checkValidMove(Point const& from,Point const& too) = 0;
};
class Queen: public Piece
{
virtual bool checkValidMove(Point const& from,Point const& too)
{
if (CheckHorizontalMove(from,too) || CheckVerticalMoce(from,too) || CheckDiagonalMove(from,too))
{
.....
}
}
}