This question already has answers here:
Differentiate between function overloading and function overriding
(11 answers)
Overloading and overriding
(12 answers)
Closed 7 years ago.
I am studying c plus plus course in my college and i am unable to differentiate between overriding and overloading of a function can anyone please help me.
Here are two different overloads of foo:
void foo(int);
void foo(char);
Here B::bar is a function override:
class A {
public:
virtual void bar();
};
class B : public A {
public:
void bar() override;
};
overloading means functions with same name having different parameters , it does not really depend whether you are using procedural language or object oriented language you can do overloading. well as far as over riding is concerned it means we are explicitly defining a function that exist in base class in derived class . obviously you need object oriented language to perform over riding as it is done between base and derived classes.
Overloading means declaring more than one function with the same name in the same scope. They must have different parameter types, and the suitable overload is chosen at compile time based on the arguments' static types.
void f(int);
void f(double);
f(42); // selects the "int" overload
f(42.0); // selects the "double" overload
Overriding means that a derived class declares a function that matches a virtual function declared in the base class. Calling the function via a pointer or reference to the base class will select the override at run-time, based on the object's dynamic type.
struct Base {
virtual void f();
};
struct Derived : Base {
void f() override; // overrides Base::f
};
Base * b = new Base; // dynamic type is Base
Base * d = new Derived; // dynamic type is Derived
b->f(); // selects Base::f
d->f(); // selects Derived::f
Overriding means, giving a different definition of an existing function with same parameters,
and overloading means adding a different definition of an existing function with different parameters.
Example:
#include <iostream>
class base{
public:
virtual void show(){std::cout<<"I am base";} //this needs to be virtual to be overridden in derived class
void show(int x){std::cout<<"\nI am overloaded";} //this is overloaded function of the previous one
};
class derived:public base{
public:
void show(){std::cout<<"I am derived";} //the base version of this function is being overridden
};
int main(){
base* b;
derived d;
b=&d;
b->show(); //this will call the derived version
b->show(6); // this will call the base overloaded function
}
Output:
I am derived
I am overloaded
Related
Consider the following code:
#include <iostream>
#include<string>
using namespace std;
class Base
{
public:
virtual string print() const
{
return "This is Base class";
}
};
class Derived : public Base
{
public:
virtual string print() const
{
return "This is Derived class";
}
};
void describe(Base p)
{
cout << p.print() << endl;
}
int main()
{
Base b;
Derived d;
describe(b);
describe(d);
return 0;
}
On executing this code, it gives the output as
This is Base class
This is Base class
I had couple of doubts in this program:
The function print in class Base is a const member function, hence after inheritance, when we try to override the definition of print in Derived class, why doesn't the compiler give an error.
In this program, how we can we pass a derived class object (d) to the function which expects an argument of data Type Base.
Even if we could somehow pass the derived class object to the function, why does it print "This is Base class" instead of "This is Derived Class".
Does implicit type-conversion take place when I pass a derived class object to function describe?
I had be grateful if someone could explain me the working of this code.
The function print in class Base is a const member function, hence after inheritance, when we try to override the definition of print in Derived class, why doesn't the compiler give an error.
There is no error because the program is well-formed. It is unclear why you bring up the detail that the member function is const. Constness of a member function has no effect on whether the function can be overridden. Did you assume that const means final?
In this program, how we can we pass a derived class object (d) to the function which expects an argument of data Type Base.
Because that class Base is a public base of the Derived class, and thus the derived object is implicitly convertible to the base. The result of the conversion is a copy of the base class sub object. This is colloquially called "slicing".
Even if we could somehow pass the derived class object to the function, why does it print "This is Base class" instead of "This is Derived Class".
Because there is no derived class object inside the function. There is a copy of the base.
Does implicit type-conversion take place when I pass a derived class object to function describe?
Yes. When you pass an argument of a type that is different from the type of a non-reference parameter, there is always an implicit conversion.
Indirection is necessary for runtime polymorphism. Try the same using a reference to base as parameter and you'll see the overriding function called.
Your first question is why it did not give you an error because your function is const. Const function does not prohibit you from overriding it. If you want your function to be not overridden by derived classes you can declare that function final.
virtual string print() const final;
Now this will not be overridable by any other derived class. As far as you are concerned about the const. You can Overload your function on the basis of constness. For example, you can have.
virtual string print() const; // Does not modify anything
virtual string print(); Can modify `this`
For the second question can pass the Derived class object to a function witch expect the Base class object. But It is only possible if it is via reference or via a pointer.
void describe(const Base& p);
void describe(Base* p);
For the third question as you passed via copy, so your derived class object is narrowed to Base class. Which essentially loss all the additional information other than its Base class.
The fourth question has the same answer as 3.
This is fully functional code as you had expected to run.
#include <iostream>
#include<string>
using namespace std;
class Base
{
public:
virtual string print() const
{
return "This is Base class";
}
};
class Derived : public Base
{
public:
virtual string print() const override
{
return "This is Derived class";
}
};
void describe(const Base& p)
{
cout << p.print() << endl;
}
int main()
{
Base b;
Derived d;
describe(b);
describe(d);
return 0;
}
Here you can see I declared overridden function override to tell the compiler that this function is being overridden from its Base Class. Without this, it works as expected but it's like function hiding rather than overriding.
This question already has answers here:
Why does an overridden function in the derived class hide other overloads of the base class?
(4 answers)
Closed 5 years ago.
I'm normally a C# guy, and it handles this fine, largely under its handling of "best match" resolving the method to call. I'm trying to do something similar in C++ now, but am getting a compile error. Long story short, it's a combination of method overloading and overriding.
class Bar : public Foo { } //Contents don't really matter here
class Base
{
public:
virtual void Do(Foo* foo) { }
virtual void Do(Bar* bar) { }
};
class Derived : public Base
{
public:
virtual void Do(Bar* bar) { }
}
Foo* foo = new Foo();
Derived* d = new Derived();
d->Do(foo); //cannot initialize a parameter of type 'Bar*' with an rvalue of type 'Foo*'
So, it tries to resolve the method against only the methods in Derived rather than recognizing the base class implements a valid match. Again, C# finds the base method. Can C++ not do this, or am I missing something?
A function in a derived class hides any functions of the same name in a base class (even if you're overriding a virtual function). To unhide the other overloads, use a "using" statement:
class Derived : public Base
{
public:
virtual void Do(Bar* bar) { }
using Base::Do;
};
This question already has answers here:
Why do we need virtual functions in C++?
(27 answers)
Closed 6 years ago.
In c++, the virtual function in base class can be overridden in derived class.
and a member function where the specific implementation will depend on the type of the object it is called upon, at run-time.
Since virtual function has to be implemented(except for pure virtual)
Can I use regular function in base class and redefine it in derived class?
if yes.
What's the point of using virtual function?
Thanks
You can redefine it but it won't work in a polymorphic way.
so if I have a
class base
{
int foo(){ return 3; }
};
class Der : public base
{
int foo() {return 5;}
};
and then have a function that takes a base
void dostuff(base &b)
{
b.foo(); // This will call base.foo and return 3 no matter what
}
and I call it like this
Der D;
dostuff(D);
Now if I change the base to
class base
{
virtual int foo(){ return 3; }
};
void dostuff(base &b)
{
b.foo(); // This will call the most derived version of foo
//which in this case will return 5
}
so the real answer is if you want to write common code that will call the correct function from the base it needs to be virtual.
Actually, Virtual function is base of object oriented language.
In java, all functions are virtual function.
But in c++, virtual function is slower than regular function.
So if there is no need to function override, you should use regular function instead of virtual function.
This question already has answers here:
Function with same name but different signature in derived class not found
(2 answers)
Closed 5 years ago.
I need to understand why C++ don't allow to access Grandparent overloaded functions in Child if any of the overloaded function is declared in Parent. Consider the following example:
class grandparent{
public:
void foo();
void foo(int);
void test();
};
class parent : public grandparent{
public:
void foo();
};
class child : public parent{
public:
child(){
//foo(1); //not accessible
test(); //accessible
}
};
Here, two functions foo() and foo(int) are overloaded functions in Grandparent. But foo(int) is not accessible since foo() is declared in Parent (doesn't matter if it declared is public or private or protected). However, test() is accessible, which is right as per OOP.
I need to know the reason of this behavior.
The reason is method hiding.
When you declare a method with the same name in a derived class, base class methods with that name are hidden. The full signature doesn't matter (i.e. cv-qualifiers or argument list).
If you explicitly want to allow the call, you can use
using grandparent::foo;
inside parent.
Just imagine that a library has this class:
struct Base {
};
In your code you use that class as a base class:
struct Derived : Base {
void f(int);
};
and now you write:
Derived d;
d.f('a');
And now you get the shiny new version 2.0 of that library, and the base class has changed a bit:
struct Base {
void f(char);
}
If overloading applied here, your code would break.
This question already has answers here:
Closed 12 years ago.
Possible Duplicates:
C++ : implications of making a method virtual
Why is 'virtual' optional for overridden methods in derived classes?
I wonder, what is documented behavior in the following case:
You have
class A
{
virtual void A()
{
cout << "Virtual A"<<endl;
}
void test_A()
{
A();
}
}
class B: public A
{
void A()
{
cout << "Non-virtual A in derived class"<<endl;
}
void test_B()
{
A();
}
}
A a; B b;
a.test_A();
b.test_A();
b.test_B();
What it supposed to do according to C++ standard and why?
GCC works like B::A is also also virtual.
What shoudl happen in general when you override virtual method by non-virtual one in derived class?
The sub-class member function is implicitly virtual if a virtual base-class member function with the same name and signature exists.
The code should not compile as you cannot name a method with the name of the class. But regarding what I understand that is your real question:
Will making a method virtual imply that the same method in all the derived classes is virtual even if the virtual keyword is not present?
The answer is yes. Once a method is declared virtual in a class, then all overrides of that method will be virtual, and the virtual keyword is optional in derived classes (even if I recommend typing it, if only for documentation purposes). Note that for a method in a derived class to be an override it has to have the same name and signature, with only potential difference being a covariant return type:
struct A {};
struct B : A {};
struct base {
virtual A* foo();
virtual A* bar();
};
struct derived : base {
virtual B* foo(); // override, covariant return type
virtual int bar(); // not override, return type is not covariant
virtual A* bar(int); // not override, different argument list
};
This code is ill-formed. A constructor cannot have a return type (as you have done for the constructor of 'A'). Also a constructor cannot be virtual.
After fixing A's constructor, class B is ill-formed as the constructor of A is private.
So, there are many problems with this code (including missing semicolons at the class definitions).
According to standard it should be
A a; B b;
a.test_A(); //"Virtual A"
b.test_A(); //Non-virtual A in derived class
b.test_B(); //Non-virtual A in derived class