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Overriding a Base's Overloaded Function in C++ [duplicate]

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Can we override already overloaded method?

So can you override an overloaded function? Yes, since the overloaded method is a completely different method in the eyes of the compiler. Overriding isn't the same thing at all.

What is overriding and overloading in C?

Overview. In C++, two or more functions can have the same name if the number or the type of parameters are different, this is known as function overloading whereas function overriding is the redefinition of base class function in its derived class with the same signature.

What is overriding & overloaded function explain with proper example?

In overriding, function signatures must be same. Scope of functions: Overridden functions are in different scopes; whereas overloaded functions are in same scope. Behavior of functions: Overriding is needed when derived class function has to do some added or different job than the base class function.


In class bar, add

using foo::a;

This is a common 'gotcha' in C++. Once a name match is found in the a class scope, it doesn't look further up the inheritance tree for overloads. By specifying the 'using' declaration, you bring all of the overloads of 'a' from 'foo' into the scope of 'bar'. Then overloading works properly.

Keep in mind that if there is existing code using the 'foo' class, its meaning could be changed by the additional overloads. Or the additional overloads could introduce ambiguity and and the code will fail to compile. This is pointed out in James Hopkin's answer.


That is the way the language used to work. Prior to the using keyword, if you overrode one overloaded function, you had to overload them all:

class bar : public foo 
{
  public:
    bar(void);
    ~bar(void);
    a(int);
    a(double d) { foo::a(d); }  // add this
}

This annoyed enough people that the language committee added the using feature, but some old habits die hard; and the habitués† have a good argument.

As James Hopkins points out, by adding using, the programmer is expressing the intention that the derived class will, without warning, add any future overrides of foo::a() to its list of acceptable signatures.

Here is an example of what he describes:

#include <iostream>
class Base {
public:
  virtual void f(double){ std::cout << "Base::Double!" << std::endl; }
  // virtual void f(int) { std::cout << "Base::Int!" << std::endl; } // (1)
  virtual ~Base() {}
};

class Derived : public Base {
public:
  // using Base::f; // (2)
  void f(double) { std::cout << "Derived::Double!" << std::endl; }
};

int main(int, char **) {
  Derived d;
  d.f(21);
  return 0;
}

The output will be "Derived::Double!" because the compiler will promote the integer argument to a double. g++ 4.0.1 -Wall will not warn that this promotion occurred.

Uncomment (1) to simulate a future change to Base adding the method Base::f(int). The code compiles, again without warning even with -Wall, and "Derived::Double!" remains the output.

Now uncomment (2) to simulate a decision by the Derived programmer to include all Base::f signatures. The code compiles (without warnings), but the output is now "Base::Int!".

† I cannot think of an English word for "those who have the habit" and "addicted" is much too strong.


It is by design. Overload resolution is restricted to a single scope. It prevents some nasty cases of valid code changing meaning when additional functions are added to a base class or to namespace scope.