A user-defined type can overload a predefined C# operator. That is, a type can provide the custom implementation of an operation in case one or both of the operands are of that type. The Overloadable operators section shows which C# operators can be overloaded.
Use the operator
keyword to declare an operator. An operator declaration must satisfy the following rules:
public
modifier.T
or T?
where T
is the type that contains the operator declaration.static
modifier, except for the compound assignment operators, such as +=
.++
) and decrement (--
) operators can be implemented as either static or instance methods.The following example defines a simplified structure to represent a rational number. The structure overloads some of the arithmetic operators:
public struct Fraction
{
private int numerator;
private int denominator;
public Fraction(int numerator, int denominator)
{
if (denominator == 0)
{
throw new ArgumentException("Denominator cannot be zero.", nameof(denominator));
}
this.numerator = numerator;
this.denominator = denominator;
}
public static Fraction operator +(Fraction operand) => operand;
public static Fraction operator -(Fraction operand) => new Fraction(-operand.numerator, operand.denominator);
public static Fraction operator +(Fraction left, Fraction right)
=> new Fraction(left.numerator * right.denominator + right.numerator * left.denominator, left.denominator * right.denominator);
public static Fraction operator -(Fraction left, Fraction right)
=> left + (-right);
public static Fraction operator *(Fraction left, Fraction right)
=> new Fraction(left.numerator * right.numerator, left.denominator * right.denominator);
public static Fraction operator /(Fraction left, Fraction right)
{
if (right.numerator == 0)
{
throw new DivideByZeroException();
}
return new Fraction(left.numerator * right.denominator, left.denominator * right.numerator);
}
// Define increment and decrement to add 1/den, rather than 1/1.
public static Fraction operator ++(Fraction operand)
=> new Fraction(operand.numerator++, operand.denominator);
public static Fraction operator --(Fraction operand) =>
new Fraction(operand.numerator--, operand.denominator);
public override string ToString() => $"{numerator} / {denominator}";
// New operators allowed in C# 14:
public void operator +=(Fraction operand) =>
(numerator, denominator ) =
(
numerator * operand.denominator + operand.numerator * denominator,
denominator * operand.denominator
);
public void operator -=(Fraction operand) =>
(numerator, denominator) =
(
numerator * operand.denominator - operand.numerator * denominator,
denominator * operand.denominator
);
public void operator *=(Fraction operand) =>
(numerator, denominator) =
(
numerator * operand.numerator,
denominator * operand.denominator
);
public void operator /=(Fraction operand)
{
if (operand.numerator == 0)
{
throw new DivideByZeroException();
}
(numerator, denominator) =
(
numerator * operand.denominator,
denominator * operand.numerator
);
}
public void operator ++() => numerator++;
public void operator --() => numerator--;
}
public static class OperatorOverloading
{
public static void Main()
{
var a = new Fraction(5, 4);
var b = new Fraction(1, 2);
Console.WriteLine(-a); // output: -5 / 4
Console.WriteLine(a + b); // output: 14 / 8
Console.WriteLine(a - b); // output: 6 / 8
Console.WriteLine(a * b); // output: 5 / 8
Console.WriteLine(a / b); // output: 10 / 4
}
}
You could extend the preceding example by defining an implicit conversion from int
to Fraction
. Then, overloaded operators would support arguments of those two types. That is, it would become possible to add an integer to a fraction and obtain a fraction as a result.
You also use the operator
keyword to define a custom type conversion. For more information, see User-defined conversion operators.
The following table shows the operators that can be overloaded:
Operators Notes+x
, -x
, !x
, ~x
, ++
, --
, true
, false
The true
and false
operators must be overloaded together. x + y
, x - y
, x * y
, x / y
, x % y
,
x & y
, x | y
, x ^ y
,
x << y
, x >> y
, x >>> y
x == y
, x != y
, x < y
, x > y
, x <= y
, x >= y
Must be overloaded in pairs as follows: ==
and !=
, <
and >
, <=
and >=
. +=
, -=
, *=
, /=
, %=
, &=
, \|=
, ^=
, <<=
, >>=
, >>>=
The compound assignment operators can be overloaded in C# 14 and later.
A compound assignment overloaded operator must follow these rules:
public
modifier.static
modifier.void
.Beginning with C# 14, the increment (++
) and decrement (--
) operators can be overloaded as instance members. Instance operators can improve performance by avoiding the creation of a new instance. An instance operator must follow these rules:
public
modifier.static
modifier.void
.The following table shows the operators that can't be overloaded:
Operators Alternativesx && y
, x || y
Overload both the true
and false
operators and the &
or |
operators. For more information, see User-defined conditional logical operators. a[i]
, a?[i]
Define an indexer. (T)x
Define custom type conversions performed by a cast expression. For more information, see User-defined conversion operators. ^x
, x = y
, x.y
, x?.y
, c ? t : f
, x ?? y
, ??= y
,
x..y
, x->y
, =>
, f(x)
, as
, await
, checked
, unchecked
, default
, delegate
, is
, nameof
, new
,
sizeof
, stackalloc
, switch
, typeof
, with
None.
Before C# 14, the compound operators can't be overloaded. Overloading the corresponding binary operator implicitly overloads the corresponding compound assignment operator.
Operator overload resolutionImportant
This section applies to C# 14 and later. Before C# 14, user-defined compound assignment operators and instance increment and decrement operators aren't allowed.
If x
is classified as a variable in a compound assignment expression such as x «op»= y
, instance operators are preferred over any static operator for «op»
. If an overloaded «op»=
operator isn't declared for the type of x
or x
isn't classified as a variable, the static operators are used.
For the postfix operator ++
, if x
isn't classified as a variable or the expression x++
is used, the instance operator++
is ignored. Otherwise, preference is given to the instance operator ++
. For example,
x++; // Instance operator++ preferred.
y = x++; // instance operator++ isn't considered.
The reason for this rule is that y
should be assigned to the value of x
before it's incremented. The compiler can't determine that for a user-defined implementation in a reference type.
For the prefix operator ++
, if x
is classified as a variable in ++x
, the instance operator is preferred over a static unary operator.
For more information, see the following sections of the C# language specification:
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