Hiding of all Overloaded Methods with Same Name in Base Class in C++
In C++, function overloading is possible i.e., two or more functions from the same class can have the same name but different parameters. However, if a derived class redefines the base class member method then all the base class methods with the same name become hidden in the derived class.
For example, the following program doesn’t compile. Here, Derived redefines Base’s method fun() and this makes fun(int i) hidden.
CPP
#include <iostream>
using namespace std;
class Base {
public :
int fun() { cout << "Base::fun() called" ; }
int fun( int i) { cout << "Base::fun(int i) called" ; }
};
class Derived : public Base {
public :
int fun() { cout << "Derived::fun() called" ; }
};
int main()
{
Derived d;
d.fun(5);
return 0;
}
|
Output
prog.cpp: In function ‘int main()’:
prog.cpp:20:12: error: no matching function for call to ‘Derived::fun(int)’
d.fun(5); // Compiler Error
^
prog.cpp:13:9: note: candidate: int Derived::fun()
int fun() { cout << "Derived::fun() called"; }
^
prog.cpp:13:9: note: candidate expects 0 arguments, 1 provided
Even if the signature of the derived class method is different, all the overloaded methods in the base class become hidden. For example, in the following program, Derived::fun(char ) makes both Base::fun() and Base::fun(int ) hidden.
CPP
#include <iostream>
using namespace std;
class Base {
public :
int fun() { cout << "Base::fun() called" ; }
int fun( int i) { cout << "Base::fun(int i) called" ; }
};
class Derived : public Base {
public :
int fun( char c)
{
cout << "Derived::fun(char c) called" ;
}
};
int main()
{
Derived d;
d.fun( 'e' );
return 0;
}
|
Output
Derived::fun(char c) called
Note: The above facts are true for both static and non static methods.
There is a way to mitigate this kind of issue. If we want to overload a function of a base class, it is possible to unhide it by using the ‘using’ keyword. This keyword brings a base class method ?or variable into the current class’s scope.
C++
#include <iostream>
using namespace std;
class Base {
public :
int fun() { cout << "Base::fun() called" ; }
};
class Derived : public Base {
public :
using Base::fun;
int fun( char c)
{
cout << "Derived::fun(char c) called" ;
}
};
int main()
{
Derived d;
d.fun();
return 0;
}
|
Output
Base::fun() called
Last Updated :
04 Jan, 2022
Like Article
Save Article
Share your thoughts in the comments
Please Login to comment...