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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 explains 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; }
When the above code is compiled and executed, it produces the following result −
First Distance : F: 11 I:10 Second Distance :F: 30 I:120
cpp_overloading.htm
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