A Beginner's Tutorial on Operator Overloading in C#

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Introduction

This article discusses operator overloading in C#. What are the various types of operators that can be overloaded? We will create a small dummy application to see how we can overload some basic operators.

Background

In an object oriented programming language like C#, operator overloading provides a much more natural way of implementing the operations on custom types. Suppose we have a class created for Complex number and we want to perform all the arithmetic operations on this type. One way to do this is by having functions like Add, Subtract inside the class and have the functionality. Another way is to actually have the overloaded version of operators to act on this type.

Operator overloading provides a much natural abstraction for the types. When we think about possible operation on some data type, we can think of binary operators, unary operators, relational operators and perhaps some conversion operations to and from the basic types. In C#, achieving all this is possible using operator overloading.

Using the Code

Before looking at the implementation details, let's see what are the conventions that need to be followed if we want to overload an operator.

• The operator function should be a member function of the containing type.
• The operator function must be static.
• The operator function must have the keyword operator followed by the operator to be overridden.
• The arguments of the function are the operands.
• The return value of the function is the result of the operation.

So to understand the above mentioned points more clearly, we can visualize the operators getting invoked as function calls (below code will not compile, it is only for understanding purposes).

class A
{
    public static A operator+(A a1, A a2)
    {
        A temp;
        // perform actual addition
        return temp;
    }
}

// so the operations like 
A a1 = new A();
A a2 = new A();

A result = a1 + a2;

// This can actually be visualized as:
A result = A.+(a1, a2);
// Note: Only for understanding, this won't even compile

So following the above guidelines, let us see how we can implement operator overloading. We will create a small class Rational representing a rational number and will implement some basic overloaded operators in it.

class Rational
{
    // Private variables to hold the data
    int numerator;
    int denominator;

    // Constructor
    public Rational(int num, int den)
    {
        numerator = num;
        denominator = den;
    }

    public double Value
    {
        get
        {
            return ((double)numerator) / denominator;
        }
    }
}

Overloading the Binary Operators

Binary operators function will take two arguments and return a new object of the Containing type. Let us try to implement the binary operator + for our Rational class.

// Let us overload the Binary operators                
public static Rational operator +(Rational rational1, Rational rational2)
{
    int resultingDenominator = rational1.denominator * rational2.denominator;
    int resultingNumerator = rational1.numerator * rational2.denominator + 
        rational1.denominator * rational2.numerator;

    return new Rational(resultingNumerator, resultingDenominator);
}

Once we have the binary operator overloaded, we can use it simply as:

Rational r1 = new Rational(3, 2);
Rational r2 = new Rational(2, 3);

Rational result = null;

// Testing + operator on types
result = r1 + r2;

The important thing to note here is that once we have the binary operator + implemented, the compound assignment operator for that operator, i.e., += in this case also gets implemented in terms of this operator, i.e., + in this case.

The other possibility could be that we may want to have mix mode arithmetic with our type, i.e., we may want to have the arithmetic operations to work with our Rational type and some primitive type. So if we want to have the possibility of having Rational + int, then we will have to implement an overloaded function for that too.

// Binary operators for Mix mode arithmetics (Only for + and with int)
public static Rational operator +(Rational rational1, int num)
{
    return new Rational(rational1.numerator +(rational1.denominator * num), 
                        rational1.denominator);
}

public static Rational operator +(int num, Rational rational1)
{
    //let us implement this in terms of other
    return rational1 + num;
}

//Usage
static void Main(string[] args)
{
    // Testing + operator in mix mode
    result = r1 + 2;
    Console.WriteLine(result.Value.ToString());

    // Testing + operator in mix mode
    result = 3 + r1;
    Console.WriteLine(result.Value.ToString());
}

Overloading the Unary Operators

To overload the unary operators, we need a function taking only one argument of the containing type. The important thing in case of overloading unary operators is that we should not create a new object but instead change the value of the object being passed to us and return it. Let us implement the unary ++ for our Rational type now.

// Let us overload the unary operator ++ now (-- will be on same lines)
public static Rational operator ++(Rational rational1)
{
    rational1 += 1;
    return rational1;
}

//Usage
static void Main(string[] args)
{
    // Post increment ++
    result = r1++;
    Console.WriteLine(result.Value.ToString());

    // Pre increment ++
    result = ++r1;
    Console.WriteLine(result.Value.ToString());
}

The thing to notice in the above code snippet is that we overloaded ++ and C# internally took care of using it in pre-increment and post-increment scenario. So unlike C++, we need not implement the separate versions for prefix and post-fix unary operators.

Overloading the Relational Operators

Relational operators like < and > can also be overloaded as functions taking two arguments. The important thing to note is that we need to overload relational operators in pairs, i.e., if I am overloading < operator, than I will have to overload > operator too. Same is true for (<=, >=) and (==, !=) operators. So let us go ahead and implement the relational operators in our Rational class.

// Let us overload the relational operator
// < and > comes in pair, if we define one we need to define other too.
// Same is applicable for <= and >=. they also come in pair.
public static bool operator < (Rational rational1, Rational rational2)
{
    return rational1.Value < rational2.Value;
}

public static bool operator >(Rational rational1, Rational rational2)
{
    return rational1.Value > rational2.Value;
}

//Usage
static void Main(string[] args)
{
    // test relational operator
    Console.WriteLine(r1 > r2);
}

Other important thing to note is while implementing the equality, i.e., == operator. If we are overloading == operator, then we need to implement != operator too (as discussed above). Also, we need to override the Equals and GetHashCode functions so that if our object returns true for ==, then it should return true for Equals function too and should return the same value from GetHashCode().

// Let is now override the equality operator
// == and != comes in pair, if we define one we need to define other too.
public static bool operator ==(Rational rational1, Rational rational2)
{
    return rational1.Value == rational2.Value;
}

public static bool operator !=(Rational rational1, Rational rational2)
{
    return rational1.Value != rational2.Value;
}

public override bool Equals(object obj)
{
    Rational r = obj as Rational;
    if (r != null)
    {
        return r == this;
    }
    return false;
}

public override int GetHashCode()
{
    return Value.GetHashCode();
}

//Usage
static void Main(string[] args)
{
    // test equality 
    Console.WriteLine(r1 == r2);
}

Overloading the Conversion Operators

We might also need to implement conversion operators sometimes so that our type can safely be converted to and from other types. We can define the conversion operators as implicit or explicit. In case of implicit conversion, the user will not have to explicitly type cast our type to target type and our conversion operation will work. In case of explicit conversion, the user will have to explicitly cast our type to target type to invoke our conversion operation. If the casting is not performed, it will give a compile time error.

Let us go ahead and define the conversion operation for our Rational type. We will implicitly convert integer types to Rational type, but we will keep the conversion from Rational to double explicit.

// Finally, let us have some conversion operators
public static implicit operator Rational(int i)
{
    // since the rational equivalent of an int has 1 as denominator
    Rational rational = new Rational(i, 1);

    return rational;
}

public static explicit operator double(Rational r)
{
    double result = ((double)r.numerator) / r.denominator;
    return result;
}

//Usage
static void Main(string[] args)
{
    // implicit conversion from int to rational
    Rational ri = 3;

    //explicit conversion from rational to double
    double value = (double)ri;
}

Now we have some basic operators overloaded/implemented for this Rational class. We have implemented some of the operators in each category of operators, rest of the operators can be overloaded on same lines.

Note: Please refer to the attached sample code to see the complete class with sample test code.

Point of Interest

This article is written as a walk-through tutorial on operator overloading in C# from a beginner's perspective. Most of the veteran programmers are already aware of this basic stuff and find the information mundane, still I hope this has been informative.

History

• 4^th September, 2012: First version