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How do you access a base class method from a derived class in Cpp?
Wiki User
∙ 11y ago
Use the scope resolution operator (::) to explicitly call the base class method. Note that the base class method must be protected or public in order for a derived class to access it. Private members are only accessible to the class itself and to friends of the class, regardless of whether the derivative uses public, protected or private inheritance.
It is quite normal for a base class to provide a "default" implementation for a virtual method for which a derived class may override. Although the derived class will generally provide its own implementation of the method, it may also call the base class method either before, during or after performing its own implementation.
The following shows minimal class declarations demonstrating a call to a protected base class method from a derived class override.
class base
{
protected:
// accessible to all instances of this class, its friends and its derivatives.
virtual void method(){
/* do something */
}
};
class derived : public base
{
public:
// full-accessible outside of class.
virtual void method(){
/* do something (or do nothing) */
base::method(); // call base class method.
/* do something else (or do nothing) */
}
};
Wiki User
∙ 11y ago
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Can a friend function of derived class access private data of base class?
In some computer languages it is possible to do so, but I would not even think or design any application in this way. A base class SHOULD NEVER know what the derived classes are. Perhaps it was created by generalizing some classes. Even at that point, this new base class should have no knowledge of the derived classes whatsoever. To do a good OO design, the base class should have a method like getPrivatePartOfDerivedClass() as abstract, then force the derived class to provide the implementation of this method.
How do you call base class method using derived class object?
If the base class method is non-private, the derived class can call the base class method implicitly. However, if the derived class overrides or overloads the method, the base class method must be called explicitly. The following demonstrates explicit calls to base class methods: #include <iostream> using namespace std; class Base { public: Base(){} void Foo(){ cout << "Base::Foo" << endl; } }; class Derived : public Base { public: Derived(){} void Foo(){ cout << "Derived::Foo" << endl; Base::Foo(); } }; int main() { Derived derived; derived.Foo(); derived.Base::Foo(); // Explicit call to base class method. return(0); } Output: Derived::Foo Base::Foo Base::Foo
How you can override base class member in derived classexplain with example?
To override a base class method you simply need to declare the base class method as being virtual. As well as creating a v-table, this also gives a visual hint to other developers that you expect the function to be overridden. The v-table ensures that all calls to the base class method are routed to the derived class method, thus ensuring objects behave polymorphically, according to their actual type, and not what we're actually pointing at. Consider the following example: #include <iostream> class base { public: virtual ~base(); virtual void PrintMe() const { std::cout << "I am a base class!" << std::endl; } }; class derived: public base { public: void PrintMe() const { std::cout << "I am a derived class!" << std::endl; } }; int main() { base b; derived d; base* pb = &d; b.PrintMe(); d.PrintMe(); pb->PrintMe(); return( 0 ); } Output: I am a base class! I am a derived class! I am a derived class! Note that although pb points to the base class instance of d, it still knows that it really is a derived class, as can be seen from the third line of output. Now try removing the virtual keyword from the base class method. The output will change as follows: Output: I am a base class! I am a derived class! I am a base class! Now your derived class thinks it is a base class. This is because the v-table no longer has no entry for that method, and therefore the call cannot be routed to the overridden derived class method. The base class method is called because that's what we're actually pointing at and the object no longer behaves polymorphically according to its actual type. Note also that if any method is declared virtual in a class, the class constructor must also be declared virtual. If you fail to do this, your classes will not be destroyed properly. The virtual keyword ensures that the most-derived class is always destroyed first, before working up the hierarchy of destructors to eventually destroy the least-derived class, the base class itself. Consider the following example without a virtual destructor: #include <iostream> class base { public: base(){ std::cout << "Base class created" << std::endl; } ~base(){ std::cout << "Base class destroyed" << std::endl; } }; class derived: public base { public: derived(){ std::cout << "Derived class created" << std::endl; } ~derived(){ std::cout << "Derived class destroyed" << std::endl; } }; int main() { derived* d = new derived(); base* b = d; delete( b ); return( 0 ); } Output: Base class created Derived class created Base class destroyed As you can see, the derived class was created but was not destroyed. We've created a memory leak: that memory cannot be recovered until the program ends. Now add the virtual keyword to the base class destructor: #include <iostream> class base { public: base(){ std::cout << "Base class created" << std::endl; } virtual ~base(){ std::cout << "Base class destroyed" << std::endl; } }; class derived: public base { public: derived(){ std::cout << "Derived class created" << std::endl; } ~derived(){ std::cout << "Derived class destroyed" << std::endl; } }; int main() { derived* d = new derived(); base* b = d; delete( b ); return( 0 ); } Output: Base class created Derived class created Derived class destroyed Base class destroyed Now we have the expected behaviour and have resolved the memory leak. Remember, if ANY method of a class is declared virtual, the destructor must also be declared virtual. Note that although derived classes need not use the virtual keyword in front of overrides (it is implied by the base class), there is no harm in explicitly declaring them as such, if only to give a visual hint that the methods are expected to be overridden by derivatives of the derivative (multi-level inheritance).
Can you access in drive class from base class private data?
Base class should no knowledge about derived classes. The "private" modifier on a data member means private to the class which defined it. Base class cannot directly reference/access the private data member of the derived class, and the derived classes cannot access the private data member defined in the base class. Either way the accessing the private data member should be done via properties or getters
Difference between Method overloading and method overriding in C plus plus?
Yes. Any base class method that is declared virtual can be overridden by a derived class. Overriding a method that is not declared virtual can still be called, but will not be called polymorphically. That is, if you call the base class method, the base class method will execute, not the override. To call a non-virtual override you must call it explicitly.
What is overriding in c?
Overriding relates to derived classes, where the derived class provides a new implementation for a method declared in the base class. The override is said to be a more-specialised implementation of the base class method, which is itself described as being a generic method. However, the derived class method can still call the base class method, if required.When the designer of a class can predict that their class will be derived from, they will normally provide virtual methods. These methods are expected to be overridden by the derived class. Overriding a non-virtual method can have side effects if the method is also overloaded. Overriding just one overloaded method will effectively hide all the other overloads in the base class, which may be undesirable.
Can a friend function of derived class access private data of base class?
In some computer languages it is possible to do so, but I would not even think or design any application in this way. A base class SHOULD NEVER know what the derived classes are. Perhaps it was created by generalizing some classes. Even at that point, this new base class should have no knowledge of the derived classes whatsoever. To do a good OO design, the base class should have a method like getPrivatePartOfDerivedClass() as abstract, then force the derived class to provide the implementation of this method.
How do you call base class method using derived class object?
If the base class method is non-private, the derived class can call the base class method implicitly. However, if the derived class overrides or overloads the method, the base class method must be called explicitly. The following demonstrates explicit calls to base class methods: #include <iostream> using namespace std; class Base { public: Base(){} void Foo(){ cout << "Base::Foo" << endl; } }; class Derived : public Base { public: Derived(){} void Foo(){ cout << "Derived::Foo" << endl; Base::Foo(); } }; int main() { Derived derived; derived.Foo(); derived.Base::Foo(); // Explicit call to base class method. return(0); } Output: Derived::Foo Base::Foo Base::Foo
When the concept of method overriding is necessary?
concept of overriding is very important as due to overriding the derived class can use the function of the base class! when the function has same name and prototype in both the classes(base and derived) then the derived class can use the funtion of base class!
How you can override base class member in derived classexplain with example?
To override a base class method you simply need to declare the base class method as being virtual. As well as creating a v-table, this also gives a visual hint to other developers that you expect the function to be overridden. The v-table ensures that all calls to the base class method are routed to the derived class method, thus ensuring objects behave polymorphically, according to their actual type, and not what we're actually pointing at. Consider the following example: #include <iostream> class base { public: virtual ~base(); virtual void PrintMe() const { std::cout << "I am a base class!" << std::endl; } }; class derived: public base { public: void PrintMe() const { std::cout << "I am a derived class!" << std::endl; } }; int main() { base b; derived d; base* pb = &d; b.PrintMe(); d.PrintMe(); pb->PrintMe(); return( 0 ); } Output: I am a base class! I am a derived class! I am a derived class! Note that although pb points to the base class instance of d, it still knows that it really is a derived class, as can be seen from the third line of output. Now try removing the virtual keyword from the base class method. The output will change as follows: Output: I am a base class! I am a derived class! I am a base class! Now your derived class thinks it is a base class. This is because the v-table no longer has no entry for that method, and therefore the call cannot be routed to the overridden derived class method. The base class method is called because that's what we're actually pointing at and the object no longer behaves polymorphically according to its actual type. Note also that if any method is declared virtual in a class, the class constructor must also be declared virtual. If you fail to do this, your classes will not be destroyed properly. The virtual keyword ensures that the most-derived class is always destroyed first, before working up the hierarchy of destructors to eventually destroy the least-derived class, the base class itself. Consider the following example without a virtual destructor: #include <iostream> class base { public: base(){ std::cout << "Base class created" << std::endl; } ~base(){ std::cout << "Base class destroyed" << std::endl; } }; class derived: public base { public: derived(){ std::cout << "Derived class created" << std::endl; } ~derived(){ std::cout << "Derived class destroyed" << std::endl; } }; int main() { derived* d = new derived(); base* b = d; delete( b ); return( 0 ); } Output: Base class created Derived class created Base class destroyed As you can see, the derived class was created but was not destroyed. We've created a memory leak: that memory cannot be recovered until the program ends. Now add the virtual keyword to the base class destructor: #include <iostream> class base { public: base(){ std::cout << "Base class created" << std::endl; } virtual ~base(){ std::cout << "Base class destroyed" << std::endl; } }; class derived: public base { public: derived(){ std::cout << "Derived class created" << std::endl; } ~derived(){ std::cout << "Derived class destroyed" << std::endl; } }; int main() { derived* d = new derived(); base* b = d; delete( b ); return( 0 ); } Output: Base class created Derived class created Derived class destroyed Base class destroyed Now we have the expected behaviour and have resolved the memory leak. Remember, if ANY method of a class is declared virtual, the destructor must also be declared virtual. Note that although derived classes need not use the virtual keyword in front of overrides (it is implied by the base class), there is no harm in explicitly declaring them as such, if only to give a visual hint that the methods are expected to be overridden by derivatives of the derivative (multi-level inheritance).
Can you access in drive class from base class private data?
Base class should no knowledge about derived classes. The "private" modifier on a data member means private to the class which defined it. Base class cannot directly reference/access the private data member of the derived class, and the derived classes cannot access the private data member defined in the base class. Either way the accessing the private data member should be done via properties or getters
Difference between Method overloading and method overriding in C plus plus?
Yes. Any base class method that is declared virtual can be overridden by a derived class. Overriding a method that is not declared virtual can still be called, but will not be called polymorphically. That is, if you call the base class method, the base class method will execute, not the override. To call a non-virtual override you must call it explicitly.
What is a public class?
A public class is a base class declared with public inheritance: class base { // ... }; class derived : public base { // ... }; In the above example, base is a public class of derived, thus derived is regarded as being a type of base. The derived class inherits all the public and protected methods of its base. Protected methods are accessible to the derived class, its derivatives and their friends. If base were declared protected, its public methods become protected methods of derived. The base class is then an implementation detail of derived; only members of derived, its derivatives and their friends can treat derived as being a type of base. If declared private, the public and protected methods of base become private methods of derived. The base class is then an implementation detail of derived; only members of derived and its friends can treat derived as a type of base.
Can additional methods of base class be hidden from the derived class in object oriented software?
Yes. During inheritance only the public members of the parent class are visible to the child class. If you declare a method or a variable as private, the child class cannot access it. In other words, the methods and variables are hidden from the derived class.
How can the base class data be accessed from database directly using derived class?
In order to access a base class member the base class member must be declared protected or public. Private data is only accessible to members of the same class and to friends of the class. I'm not entirely sure what you mean by accessing from a database. A derived class does not require a database in order to access its base class members, it only requires protected access at the very least. Note that although you can declare derived classes to be friends of their base classes and thus allow private access, it would be an unusual design since base classes should never know anything about any of their derivatives.
How can a base class protect derived classes so that changes to the base class will not affect them explain with example?
In order for a base class to protect derived classes, so that changes to the base class do not affect the derived classes, you must make sure that the public interface exposed by the base class does not change when the implementation of those public methods do change. You can also prevent inadvertant access to the base class attributes from the derived class, by making them private. class base { public: base(...) {...}; /* constructor */ base~(...) {...}; /* destructor */ method(...) {...}; /* other public methods */ private: ... etc. } class child : base { public: ... private: ... } So long as the calling sequence and functionality (interface) of the base class public methods do not change, the implementation of those public methods can change, and the private methods and attributes can change, without impacting any child class.
What do you mean by protected derivation of a sub class from base class?
When you derive a class (the sub-class) from a base class using protected access, all public members of the base class become protected members of the derived class, while protected members of the base class will remain protected. Private members are never inherited so they remain private to the base class. By contrast, if you use public inheritance, the public members of the base class remain public to the derived class, while protected members of the base class remain protected in the derived class. If you use private inheritance, both the public and protected members of the base class become private to the derived class. Note that accessibility cannot be increased, only reduced or left the same. That is, a protected member of a base class cannot be inherited as a public member of a derived class -- it can only be declared private or remain protected. Note also that accessibility is viewed from outside of the derived class. That is, all members of a base class other than the private members are inherited by the derived class and are therefore fully accessible to the derived class. But from outside of the derived class, all base class accessibility is determined by the access specified by the type of inheritance.
