1. Object Oriented Programming COP3330 / CGS5409

2.  Arithmetic Operator Overloading  Increment (++) / Decrement (--)  In-class exercise

3.  There are many operators available that work on built-in types, like int and double.  Operator overloading -- is the creation of new versions of these operators for use with user- defined types.  An operator in C++ is just a function that is called with special notation (usually more intuitive or familiar notation). Overloading an operator simply involves writing a function.  Operator overloading is done for the purpose of using familiar operator notation on programmer- defined types (classes).

4.  Overloading an operator cannot change its precedence.  Overloading an operator cannot change its associativity.  It is not possible to create new operators -- only new versions of existing ones.  Operator meaning on the built-in types cannot be changed.

5.  Some operators can be written as member functions of a class  Some operators can be written as stand-alone functions -- it's common to use friend on these  Some operators can be written either way (your choice)  A binary operator has two operands ◦ Written as a stand-alone function, both operands would be sent as parameters, so it would be a function with two parameters ◦ Written as a member function, the first operand would be the calling object, and the other would be sent as a parameter (i.e. a function with one parameter)  A unary operator has one operand ◦ As a stand-alone function, the operand is sent as a parameter ◦ As a member function, one calling object, no parameters

6.  An operator is just a function. This means that it must be created with a return type, a name, and a parameter list The rules above give some restrictions on the parameter list  The name of an operator is always a conjunction of the keyword operator and the operator symbol itself.  Examples: operator+ operator++ operator<< operator==

7.  So the format of an operator overload declaration is just like that of a function, with the keyword operator as part of the name: returnType operatorOperatorSymbol (parameterList);

8.  Consider the arithmetic operators. int x = 3, y = 6, z; float a = 3.4, b = 2.1, c; z = x + y; c = a / b;  Now consider this notation. 1/2 + 1/3 // evaluates to 5/6 2/3 - 1/3 // evaluates to 1/3

9.  Now, what about using arithmetic operators on our Fraction class (a user-defined type): Fraction n1, n2, n3; n3 = n1 + n2; // will the compiler accept this?  It should be clear that this would not make sense to the compiler. Fraction is a programmer-defined type. How would the computer know about common denominators, etc?  These code statements would be nice to use, however, because it's the same way we use other numeric types (like int and double).  Operator overloading makes this possible

10.  The arithmetic operators can be overloaded either as:  stand-alone functions or  member functions.

11.  To add objects, we could write a function called Add, of course. Recall this example: friend Fraction Add(Fraction f1, Fraction f2);  With this function prototype, a sample call would be: Fraction n1, n2, n3; n3 = Add(n1, n2);

12.  The + notation would certainly be more convenient.  The operator version just has a different name: friend Fraction operator+(Fraction f1, Fraction f2);  The usual function style call would look like this: n3 = operator+(n1, n2);  While this is legal, the advantage is being able to use the more common infix notation: n3 = n1 + n2; // this becomes legal

13.  Here's a full definition of the + operator for class Fraction. Note that this is also a possible definition for the Add function: Fraction operator+(Fraction f1, Fraction f2) { Fraction r; // declare a Fraction for result // load result Fraction with sum numerators r.numerator = (f1.numerator*f2.denominator) + (f2.numerator*f1.denominator); // load result with the common denominator r.denominator = f1.denominator * f2.denominator; return r; // return the result Fraction }

14.  Once this operator overload is defined, then the following is legal.  Note that cascading also works (because the operation returns a Fraction). Now we have the standard intuitive use of + Fraction n1, n2, n3, n4, n5; n5 = n1 + n2 + n3 + n4; // now it is legal!  Note: This function could also be written with const reference parameters friend Fraction operator+(const Fraction& f1, const Fraction& f2);

15.  One member function version of Add was: Fraction Add(const Fraction& f) const;  A sample call to this function: Fraction n1, n2, n3; n3 = n1.Add(n2);

16.  The corresponding operator overload. Again, we change the name to operator+ Fraction operator+(const Fraction& f) const;  Again, we could use typical function notation, and the dot- operator: n3 = n1.operator+(n2);  But the whole point is to be able to use the more familiar notation, which still works, and no dot-operator required: n3 = n1 + n2; // n1 is calling object, n2 is argument

17.  A full definition of this version of operator+ Fraction Fraction::operator+(const Fraction& f2) const { Fraction r; // result r.numerator = (numerator * f2.denominator) + (f2.numerator * denominator); r.denominator = (denominator * f2.denominator); return r; }

18.  There are two versions of ++, pre-increment and post-increment  The pre-increment operator: Has one operand, which can be the calling object (member function) or a parameter (for non-member function implementation)  Performs the increment, then returns the newly incremented object.  The post-increment operator: Has one operand, officially. But to distinguish it from the pre- increment, an extra dummy parameter (int) is included.  Performs the increment, but returns a copy of the object's previous value.

19.  operator+= It is a binary operator. In this example, it's written as a member function, so it has one parameter.  The second parameter is "added" to the calling object. Note that this is a shortcut assignment, so the left side (the calling object) is changed. The right side (the parameter) is not changed -- hence the parameter is passed as a const reference  Note that the function returns *this. The keyword this is a pointer to the calling object (from inside a member funtion). So *this is the target -- i.e. the calling object itself  Note that the return from this operator is an IntPair object, returned by reference.

20.  IntPair class IntPair class  IntPair v2 IntPair v2  Version 2 uses the +=