Function Call Operator() Overloading in C++

Using the concept of Operator Overloading we can overload the function call operator () , and when we say we can overload the function call operator () we do not mean we can redefine a new way to call a function, rather in function call operator overloading we redefine the parameters accepted by the function and after function call operator overloading we can pass an arbitrary number of parameters in a class member function. Example

#include <iostream>
using namespace std;

class Displacement {
   private:
      int meters;  
   public:
      // required constructors

      Displacement() 
      {
         meters = 0;
      }

      Displacement(int m)
      {
         meters=m;
      }  

      // overload function call

      Displacement operator()(int a, int b, int c) 
       {
         Displacement D;
         D.meters = a+b+c;
         return D;
       } 

      // it show the displacement
      void show_displacement() 
        {
          cout  << meters << " Meters " << endl;
        }  
};

int main() {
   Displacement D1(100), D2;
   cout << "D1 Displacement : ";
   D1.show_displacement();
   D2 = D1(200, 200, 200); // invoke operator() overloading method
   cout << "D2 Displacement:";
   D2.show_displacement();
   return 0;
}

Output D1 Displacement: 100 Meters D2 Displacement: 600 Meters

Conclusion

So, this was all about function call operator() overloading in C++. When you overload the function call operator(), it does not mean that you create another way to call a function. Instead, you create an operator function that can accept an arbitrary number of parameters. We hope the example mentioned in this tutorial helped you develop a better understanding of how function call operator() overloading works in C++. Also, if you have any queries or suggestions, feel free to share them in the comments section below. People are also reading:

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