Operator Overloading Basics

Fundamentals

There are many operators available that work on built-in types, like int and double such as +, -, *, /, ==, !=, <<, >> (to name a few).
Operator overloading -- is the creation of new versions of these operators for use with user-defined types.
It is not as difficult as it sounds. Some things to note:
- 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.
- C++ already does some operator overloading implicitly on built-in types. Consider the fact that the + operator already works for ints, floats, doubles, and chars. There is really a different version of the + operator for each type.
- Operator overloading is done for the purpose of using familiar operator notation on programmer-defined types (classes).

Some rules regarding operator overloading

Overloading an operator cannot change its precedence. -- a+b*c will always be equivalent to a+(b*c)
Overloading an operator cannot change its associativity. -- a+b+c will always be equivalent to (a+b)+c
Overloading an operator cannot change its "arity" (number of operands). -- the function that performs a+b will always have 2 parameters, we cannot do a+b+c in one function
It is not possible to create new operators -- only new versions of existing ones.
Operator meaning on the built-in types cannot be changed.

friend functions vs. member functions

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

Format

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==
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);
```

Example - Consider integer addition and assignment:
```
     int a;
     a = 4 + 2 + 3;
  
```
Internally, the compiler could represent this statement using the following function prototypes

     int Add(int i1, int i2);     //adds i1 and i2, returns result 
     int Equals(int& i1, int i2); //sets i1 equal to i2, returns new value of i1

and use those functions to implement the above statement as follows

     Equals(a, Add( Add(4,2), 3) ) //function calls are evaluated from most deeply nested to least

Here is an example of explicitly calling the stream insertion operator on the string class (<<).

Motivation

Example 1

Consider the arithmetic operators. These are already familiar notations from algebra and earlier.
```
   int x = 3, y = 6, z; 
   float a = 3.4, b = 2.1, c; 
   z = x + y; 
   c = a / b; 
```
Now consider this notation. This should also be familiar from high school or earlier:
```
   1/2 + 1/3	// evaluates to 5/6
   2/3 - 1/3     // evaluates to 1/3
```
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

Example 2

Consider screen output. We know that this is legal:
```
 
   int x = 5, y = 0;
   cout << x << y;
```

But how about this?

   Fraction f(3,4);
   cout << "The fraction f = " << f << '\n';

There is no insertion operator in the iostream library to handle Fraction objects. How would the library designers know what types we were going to build?
But with operator overloading, we can define a new version of the insertion operator << to make this work

Overloading the Arithmetic operators

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

Overloading as a friend function

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);
```
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
```

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 to hold the result 

     // load result Fraction with sum of adjusted 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 
   }

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);
```
Click here to see the Fraction class with operator+ added in (as a friend function)

Overloading an operator as a member function

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);

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 the calling object, n2 is the argument
```

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; 
   }

Click here to see the Fraction class with the + operator as a member function

Some other possible arithmetic operators

Prototypes only

  // multiplication overload for Fractions 
  friend Fraction operator*(Fraction f1, Fraction f2); 

  // addition operator to add a Fraction and an integer 
  friend Fraction operator+(Fraction f, int n); 

  // same as above, but this one allows the int to come first in the call 
  friend Fraction operator+(int n, Fraction f);

Note that these last two are not really needed in the Fraction class, since we have a conversion constructor! The normal operator+ function:
```
   friend Fraction operator+(const Fraction& f1, const Fraction& f2);
```
will take care of calls that involve type conversions from int to Fraction, via the conversion constructor:
```
   Fraction n1, n2, n3;
   n3 = n1 + 5;
   n3 = 10 + n2;
```
However, ask yourself... Will both of these work for the friend version and the member function version? Why or why not?

Check out this example that does not compile, and then run main2.cpp in the previous example.

Overloading comparison operators

The comparison operators can also be overloaded either as stand-alone functions or member functions

Consider the Equals function example:

   friend bool Equals(const Fraction& f1, const Fraction& f2);

We can easily write this as an operator overload:

   friend bool operator== (const Fraction& f1, const Fraction& f2);

Here are corresponding sample calls:

   Fraction n1, n2;
   if (Equals(n1, n2))
      cout << "n1 and n2 are equal";

Contrast with this:

   Fraction n1, n2;
   if (n1 == n2)
      cout << "n1 and n2 are equal";

To get a full set of comparison operations, overload all 6 of them:
- operator ==
- operator !=
- operator <
- operator >
- operator <=
- operator >=

Overloading the insertion << and extraction >> operators:

As with other operators, the << and >> operators are defined for the basic types. If you build your own class, don't expect << to automatically work with your new types of objects! If you want it to work, you have to teach the computer how to do such output. Consider the following:

 Fraction f; 
 cout << f;        // how would the machine know how to do this?

We have no reason to expect the second line to work! The insertion operator << is only pre-defined for built-in types. The iostream.h library doesn't know about the Fraction type.

The << operator is a binary operator (2 parameters, left side and right side). The first parameter is always an ostream object (we've mostly used cout, so far). Because of this, it cannot be defined as a member function (it would have to be a member of the ostream class, which we cannot change). The << and >> operators should always be defined as outside functions (usually friend functions). The second parameter is whatever new type it is being overloaded to print:

 friend ostream& operator << (ostream& s, Fraction f);

This declaration has all of the usual parts for defining a function. The name is operator<< (the keyword operator and the operator symbol). The return type is ostream&. The parameters are (ostream& s, Fraction f). When defining overloads of << and >> , always pass the stream parameters by reference. A better way to write this operator is:

 friend ostream& operator << (ostream& s, const Fraction& f);

Notice that the first one passes the Fraction by value (and makes a copy). The second passes by reference (avoiding the overhead of a copy). It is declared as a const because the Fraction does not need to change if we are just doing output.

Here is the corresponding prototype for extraction >>

 friend istream& operator >> (istream& s, Fraction& f);

Notice that the Fraction parameter for >> is also a reference parameter. This is because we are getting input into the object, so we need to work on the original, not a copy.

Remember the Show() function of the Fraction class?

 void Fraction::Show() 
 { 
   cout << numerator << '/' << denominator; 
 }

Here is how the << operator might be defined for Fraction. Notice how similar it is to the Show() function.

 ostream& operator << (ostream& s, const Fraction& f) 
 { 
   s << f.numerator << '/' << f.denominator; 
   return s; 
 }

Note the differences between this and the Show() function.

In this function, we must have a return statement, because we have a return type (as opposed to "void" in the original Show function). We return the ostream itself.
Note also that we must use "s" (not "cout") in the function body. This is the formal parameter, and a nickname for whatever was passed in (which could be cout, but also could be a different ostream).
Last, notice that this is not a member function of the Fraction class, but rather, a friend function. So, we can access the private data, but we must do it through the object:

Once this is defined, we can use a Fraction object in a cout statement:

  Fraction f1;

So now, instead of:

  cout << "Fraction f1 is "; 
  f1.Show(); 
  cout << '\n';

We can write:

  cout << "Fraction f1 is " << f1 << '\n';

Click here to see the Fraction class with the << overload used instead of Show().

Now, what would the definition of the >> overload look like for Fraction? Try it!