The following code is based on the example on page 298 of C++ Templates: the Complete Guide. I removed the parts not relevant to my question.
class Virtual {
public:
virtual void foo() {
}
};
class Base : private Virtual {
public:
void foo() {
std::cout << "Base::foo()" << '\n';
}
};
class Derived : public Base {
public:
void foo() {
std::cout << "Derived::foo()" << '\n';
}
};
int main()
{
Base *p = new Derived;
p->foo(); // calls Derived::foo()
}
I do not understand how the call p->foo() ends up calling Derived::foo. More concretely, the static type of p is Base*. Base::foo is non-virtual. Now, Base privately inherits from 'Virtual', which has its own foo and it looks as if Base::foo somehow gets the virtual specifier from Virtual::foo. What is actually going on here?
The virtualness of foo is inherited from the base class Virtual.
In fact, writing virtual in Derived::foo will be superfluous and many developers will omit it. You cannot undo virtualness once you've marked that function as virtual in a base class.
You can force a call of Base::foo by writing p->Base::foo(); but that's an ugly contrivance.
Related
Consider:
#include <iostream>
class Base
{
public:
virtual void foo() { std::cout << "Base::foo()\n"; };
};
class Derived : public Base
{
public:
void foo() override
{
std::cout << "Derived::foo()\n";
Base::foo();
}
};
int main()
{
Derived obj;
obj.foo();
return 0;
}
This is my code. Why can I call Base::foo() in the Derived class if I already redefined it in Derived class. Why doesn't the compiler delete Base::foo in class Derived after redefine?
"why compiler doesn't delete Base::foo in class Derived after redefine"
Because that isn't what virtual and override do. When you provide an override to a base class function, you do not replace it. You are defining a new version for that function. The base class's implementation continues to exist and to be accessible.
Consider the following code. Someone can still use a Base object, and the behaviour should not be changed because Derived exists. The output for base_obj.foo() should continue to be "Base::foo()" regardless of the existance of Derived. :
#include <iostream>
class Base
{
public:
virtual void foo() { std::cout << "Base::foo()\n"; }
};
class Derived : public Base
{
public:
void foo() override { std::cout << "Derived::foo()\n"; }
};
int main()
{
Derived obj;
obj.foo();
Base base_obj;
base_obj.foo();
return 0;
}
Also consider that multiple classes can derive from Base. I could add a class MyClass : public Base with its own version of foo(), and it should not interfere with how Base or Derived objects behave.
If overriding a member function would cause the base member function to be entirely replaced or removed, it becomes nearly impossible to reason about code without reading carefully every class that derives from it. And unless your IDE provides tools for that, it implies reading all of the code base. It would make it C++ code that uses polymorphism extremely difficult to understand.
Example code:
#include <string>
namespace vehicle
{
class Vehicle
{
public:
Vehicle(int a);
virtual ~Vehicle(); <------ not method?
protected:
int a;
};
}
Also, I don't fully get the concept of a pure virtual method where you declare a method as:
virtual method() = 0;
Why do we need this?
Let's suppose there are two classes: Base and Derived
struct Base
{
Base() {}
virtual void foo() = 0;
virtual ~Base()
{
std::cout << "Base descructor" << std::endl;
}
};
struct Derived : Base
{
int *i;
Derived(int i_): Base(), i(new int[i_]) {}
virtual void foo() override
{
std::cout << "foo" << std::endl;
}
virtual ~Derived()
{
std::cout << "Derived descructor" << std::endl;
delete [] i;
}
};
Pure virtual functions
If Derived doesn't override the foo function you can create an instance of the Derived class. When you try, your code will not compile, because foo is a pure virtual function.
error: invalid new-expression of abstract class type 'Derived'
note: because the following virtual functions are pure within 'Derived':
struct Derived : Base
Pure virtual functions are used to describe the interface.
Now about virtual destructors:
The ~(tilda) sign is used to denote the class destructor. It
is a special method that is called when the object is destroyed.
A client creates an instance of Derived:
Base *instance = new Derived;
then this variable is used somehow and when you don't need the variable you need to free the memory:
delete instance;
Let's trace the calls:
Derived descructor
Base descructor
So you can see that both base and derived destructors are called and no memory leak is possible. But if you remove this virtual keyword from the destructors
Base descructor
As you can see the descturtor of Derived is not called, so a memory leak exists (the array of Derived class don't get released).
Hence virtual constructors are useful when you work with objects through pointers.
virtual ~Vehicle();
The tilde ~ denotes a destructor. It's a method, a special method for destroy the object.
Virtual destructor is used in abstract base class.
virtual method() = 0;
That's pure virtual function. It indicates that you have to provide implementation method for the implementation class implementing this abstract base class.
Consider a base class class Base which has a function virtual void foo(void). This function is implemented in Base; i.e. is not pure virtual.
Is there a pattern I can use which when inheriting from this class, i.e. class Child : public Base, compels me to override foo?
Other than making it a pure virtual function, there is no way to make the override required.
Note that the fact that a function is marked pure virtual does not mean that it cannot have an implementation in the base class - it means only that the derived class must override it.
struct Base {
virtual void foo() = 0; // foo() is pure virtual
};
struct Derived : public Base {
void foo() { // Derived overrides the pure virtual
cout << "Hello ";
Base::foo(); // Call the implementation in the base
cout << endl;
}
};
void Base::foo() {
cout << " world";
}
int main() {
Derived d;
d.foo();
return 0;
}
This prints "Hello world", with the "world" part coming from the implementation in the base class.
Demo.
C++11 introduced the override keyword to help with this:
struct Base
{
void foo();
};
struct Derived : Base
{
void foo() override; // error! Base::foo is not virtual
};
However you can not write this in Base itself to get the same effect; i.e. there is no mustoverride specifier. Ultimately, it is none of Base's business as to what derived classes do or don't override.
You can keep Base abstract whilst providing a "default" definition for your pure virtual functions:
struct Base
{
virtual void foo() = 0;
};
void Base::foo() {}
struct Derived : Base {}; // error! does not override Base::foo
struct Derived2: Base
{
virtual void foo() override
{
Base::foo(); // invokes "default" definition
}
};
This will be an acceptable solution if you are content for the entire base type to be rendered uninstantiable.
A pure-virtual member function can still have a body. The only caveat is that it must be defined outside the class definition. This is perfectly legal C++:
#include <iostream>
struct Base
{
virtual void foo() const = 0;
};
void Base::foo() const
{
std::cout << "Base!\n";
}
struct Derived : Base
{
// Uncomment following line to remove error:
//virtual void foo() const override { std::cout << "Derived\n"; Base::foo(); }
};
int main()
{
Derived d;
d.foo();
}
Live example
Notice that this makes Base an abstract class in all respects, i.e. it's impossible to instantiate Base directly.
Yes, actually there is:
#include <iostream>
class Base
{
public:
virtual void someFun() {std::cout << "Base::fun" << std::endl;}
virtual ~Base() {}
};
class AlmostBase : public Base
{
public:
virtual void someFun() = 0;
};
class Derived : public AlmostBase
{
public:
virtual void someFun() {std::cout << "Derived::fun" << std::endl;}
};
int main()
{
Derived *d = new Derived();
d->someFun();
delete d;
}
If you uncomment the someFun from Derived the compiler will complain ...
You introduce an intermediary class AlmostBase which has the function as pure virtual. This way you can have Base objects too, and the only drawback now is that all your classes will need to inherit from the intermediary base.
you can make the base method throw an exception when called, then the class must override it to avoid the parent execution.
this is used in the MFC FrameWork for example
// Derived class is responsible for implementing these handlers
// for owner/self draw controls (except for the optional DeleteItem)
void CComboBox::DrawItem(LPDRAWITEMSTRUCT)
{ ASSERT(FALSE); }
void CComboBox::MeasureItem(LPMEASUREITEMSTRUCT)
{ ASSERT(FALSE); }
int CComboBox::CompareItem(LPCOMPAREITEMSTRUCT)
{ ASSERT(FALSE); return 0; }
those methods must be inherited if the control is owner drawn it is responsible for the measuer, draw,... if you missed it while you are testing the function you will get an assert or exception with useful information thrown.
How can I call a base class method which is overridden by the derived class, from a derived class object?
class Base{
public:
void foo(){cout<<"base";}
};
class Derived:public Base{
public:
void foo(){cout<<"derived";}
}
int main(){
Derived bar;
//call Base::foo() from bar here?
return 0;
}
You can always(*) refer to a base class's function by using a qualified-id:
#include <iostream>
class Base{
public:
void foo(){std::cout<<"base";}
};
class Derived : public Base
{
public:
void foo(){std::cout<<"derived";}
};
int main()
{
Derived bar;
//call Base::foo() from bar here?
bar.Base::foo(); // using a qualified-id
return 0;
}
[Also fixed some typos of the OP.]
(*) Access restrictions still apply, and base classes can be ambiguous.
If Base::foo is not virtual, then Derived::foo does not override Base::foo. Rather, Derived::foo hides Base::foo. The difference can be seen in the following example:
struct Base {
void foo() { std::cout << "Base::foo\n"; }
virtual void bar() { std::cout << "Base::bar\n"; }
};
struct Derived : Base {
void foo() { std::cout << "Derived::foo\n"; }
virtual void bar() { std::cout << "Derived::bar\n"; }
};
int main() {
Derived d;
Base* b = &d;
b->foo(); // calls Base::foo
b->bar(); // calls Derived::bar
}
(Derived::bar is implicitly virtual even if you don't use the virtual keyword, as long as it's signature is compatible to Base::bar.)
A qualified-id is either of the form X :: Y or just :: Y. The part before the :: specifies where we want to look up the identifier Y. In the first form, we look up X, then we look up Y from within X's context. In the second form, we look up Y in the global namespace.
An unqualified-id does not contain a ::, and therefore does not (itself) specify a context where to look up the name.
In an expression b->foo, both b and foo are unqualified-ids. b is looked up in the current context (which in the example above is the main function). We find the local variable Base* b. Because b->foo has the form of a class member access, we look up foo from the context of the type of b (or rather *b). So we look up foo from the context of Base. We will find the member function void foo() declared inside Base, which I'll refer to as Base::foo.
For foo, we're done now, and call Base::foo.
For b->bar, we first find Base::bar, but it is declared virtual. Because it is virtual, we perform a virtual dispatch. This will call the final function overrider in the class hierarchy of the type of the object b points to. Because b points to an object of type Derived, the final overrider is Derived::bar.
When looking up the name foo from Derived's context, we will find Derived::foo. This is why Derived::foo is said to hide Base::foo. Expressions such as d.foo() or, inside a member function of Derived, using simply foo() or this->foo(), will look up from the context of Derived.
When using a qualified-id, we explicitly state the context of where to look up a name. The expression Base::foo states that we want to look up the name foo from the context of Base (it can find functions that Base inherited, for example). Additionally, it disables virtual dispatch.
Therefore, d.Base::foo() will find Base::foo and call it; d.Base::bar() will find Base::bar and call it.
Fun fact: Pure virtual functions can have an implementation. They cannot be called via virtual dispatch, because they need to be overridden. However, you can still call their implementation (if they have one) by using a qualified-id.
#include <iostream>
struct Base {
virtual void foo() = 0;
};
void Base::foo() { std::cout << "look ma, I'm pure virtual!\n"; }
struct Derived : Base {
virtual void foo() { std::cout << "Derived::foo\n"; }
};
int main() {
Derived d;
d.foo(); // calls Derived::foo
d.Base::foo(); // calls Base::foo
}
Note that access-specifiers both of class members and base classes have an influence on whether or not you can use a qualified-id to call a base class's function on an object of a derived type.
For example:
#include <iostream>
struct Base {
public:
void public_fun() { std::cout << "Base::public_fun\n"; }
private:
void private_fun() { std::cout << "Base::private_fun\n"; }
};
struct Public_derived : public Base {
public:
void public_fun() { std::cout << "Public_derived::public_fun\n"; }
void private_fun() { std::cout << "Public_derived::private_fun\n"; }
};
struct Private_derived : private Base {
public:
void public_fun() { std::cout << "Private_derived::public_fun\n"; }
void private_fun() { std::cout << "Private_derived::private_fun\n"; }
};
int main() {
Public_derived p;
p.public_fun(); // allowed, calls Public_derived::public_fun
p.private_fun(); // allowed, calls Public_derived::public_fun
p.Base::public_fun(); // allowed, calls Base::public_fun
p.Base::private_fun(); // NOT allowed, tries to name Base::public_fun
Private_derived r;
r.Base::public_fun(); // NOT allowed, tries to call Base::public_fun
r.Base::private_fun(); // NOT allowed, tries to name Base::private_fun
}
Accessibility is orthogonal to name lookup. So name hiding does not have an influence on it (you can leave out public_fun and private_fun in the derived classes and get the same behaviour and errors for the qualified-id calls).
The error in p.Base::private_fun() is different from the error in r.Base::public_fun() by the way: The first one already fails to refer to the name Base::private_fun (because it's a private name). The second one fails to convert r from Private_derived& to Base& for the this-pointer (essentially). This is why the second one works from within Private_derived or a friend of Private_derived.
First of all Derived should inherit from Base.
class Derived : public Base{
That said
First of you can just not have foo in Derived
class Base{
public:
void foo(){cout<<"base";}
};
class Derived : public Base{
}
int main(){
Derived bar;
bar.foo() // calls Base::foo()
return 0;
}
Second you can make Derived::foo call Base::foo.
class Base{
public:
void foo(){cout<<"base";}
};
class Derived : public Base{
public:
void foo(){ Base::foo(); }
^^^^^^^^^^
}
int main(){
Derived bar;
bar.foo() // calls Base::foo()
return 0;
}
Third you can use qualified id of Base::foo
int main(){
Derived bar;
bar.Base::foo(); // calls Base::foo()
return 0;
}
Consider making foo() virtual in the first place.
class Base {
public:
virtual ~Base() = default;
virtual void foo() { … }
};
class Derived : public Base {
public:
virtual void foo() override { … }
};
However, this does the job:
int main() {
Derived bar;
bar.Base::foo();
return 0;
}
An important [additional] note: you will still have compilation errors if Name Hiding occurs.
In this case, either utilize the using keyword, or use the qualifer. Additionally, see this answer as well.
#include <iostream>
class Base{
public:
void foo(bool bOne, bool bTwo){std::cout<<"base"<<bOne<<bTwo;}
};
class Derived : public Base
{
public:
void foo(bool bOne){std::cout<<"derived"<<bOne;}
};
int main()
{
Derived bar;
//bar.foo(true,true); // error: derived func attempted
bar.foo(true); // no error: derived func
bar.Base::foo(true,true); // no error: base func, qualified
return 0;
}
Say we have the following two classes, A is the base class with virtual destructor and B is the derived class whose destructor doesn't have 'virtual' qualifier. My question is, if I going to derive more classes from B, will B's destructor automatically inherit the virtualness or I need to explicitly put 'virtual' before '~B() {...}'
class A
{
public:
A() { std::cout << "create A" << std::endl;};
virtual ~A() { std::cout << "destroy A" << std::endl;};
};
class B: A
{
public:
B() { std::cout << "create B" << std::endl;};
~B() { std::cout << "destroy B" << std::endl;};
};
From C++ standard (section 10.3):
If a virtual member function vf is declared in a class Base and in
a class Derived, derived directly or indirectly from Base, [...]
then Derived::vf is also virtual (whether or not it is so declared).
So yes.
If base class method is virtual then all the subsequent derived class methods will become virtual. However, IMO it's a good programming practice to put virtual ahead of the method; just to indicate the reader the nature of the function.
Also note that there are some corner case where you might get unexpected results:
struct A {
virtual void foo(int i, float f) {}
};
sturct B : A {
void foo(int i, int f) {}
};
Here actually, B::foo() is not overriding A::foo() with virtual mechanism; rather it's hiding it. So irrespective of you make B::foo() virtual, there is no advantage.
In C++0x, you have override keyword, which overcomes such problems.
Virtualness is inherited all the way down. You only need to specify it in the top base class.
This is true for destructors as well as normal member functions.
Example:
class Base { virtual void foo() { std::cout << "Base\n"; } };
class Derived1 : public Base { void foo() { std::cout << "Derived1\n"; } };
class Dervied2 : public Derived1 { void foo() { std::cout << "Derived2\n"; } };
int main()
{
Base* b = new Base;
Base* d1 = new Derived1;
Base* d2 = new Derived2;
Derived1* d3 = new Derived2;
b->foo(); // Base
d1->foo(); // Derived1
d2->foo(); // Derived2
d3->foo(); // Derived2
}
or I need to explicitly put 'virtual' before '~B() {...}'
No, you need not, although you can put virtual here to make code more clear for the reader. This applies not only for destructors but for all member functions.