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Function call ()operator overloading in C++
The function call operator () can be overloaded for objects of class type. When you overload ( ), you are not creating a new way to call a function. Rather, you are creating an operator function that can be passed an arbitrary number of parameters.
Following example explain how a function call operator () can be overloaded.
#include <iostream>
using namespace std;
class Distance
{
private:
int feet; // 0 to infinite
int inches; // 0 to 12
public:
// required constructors
Distance(){
feet = 0;
inches = 0;
}
Distance(int f, int i){
feet = f;
inches = i;
}
// overload function call
Distance operator()(int a, int b, int c)
{
Distance D;
// just put random calculation
D.feet = a + c + 10;
D.inches = b + c + 100 ;
return D;
}
// method to display distance
void displayDistance()
{
cout << "F: " << feet << " I:" << inches << endl;
}
};
int main()
{
Distance D1(11, 10), D2;
cout << "First Distance : ";
D1.displayDistance();
D2 = D1(10, 10, 10); // invoke operator()
cout << "Second Distance :";
D2.displayDistance();
return 0;
}
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When the above code is compiled and executed, it produces following result:
First Distance : F: 11 I:10
Second Distance :F: 30 I:120
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