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The %= operator is the compound remainder assignment operator. It divides the value of the left-hand operand by the value of the right-hand operand, computes the remainder, and assigns that resulting value back to the left-hand operand.

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Syntax and Equivalence

x %= y;
This expression is semantically equivalent to: 
x = (T)(x % y);
where T is the type of x. The explicit cast (T) is required in the expanded form because of C#‘s numeric promotion rules. For example, if x and y are of type byte or short, the binary operation x % y yields an int. The compound assignment operator implicitly performs the cast back to the original type. A critical distinction in the compiler’s execution is that the left-hand operand (x) is evaluated only once. If x is a property access or an indexer with side effects, those side effects occur only a single time.

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Type Support and Evaluation Rules

The %= operator is predefined for all standard numeric types in C#. Its behavior depends on the underlying types of the operands:
  • Integer Types (int, long, etc.): Computes the standard mathematical remainder. If the right-hand operand is 0, the runtime throws a DivideByZeroException.
  • Floating-Point Types (float, double): Computes the remainder using a truncating division. This explicitly differs from the IEEE 754 remainder specification, which uses a rounding division (available in C# via Math.IEEERemainder). If the right-hand operand is 0.0, the operation does not throw an exception; instead, it assigns NaN (Not a Number) to the left operand.
  • Decimal Type (decimal): Computes the remainder with higher precision. Similar to integer types, if the right-hand operand is 0m, it throws a DivideByZeroException.

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Code Visualization

// Integer evaluation
int i = 10;
i %= 3; 
// i is now 1 (10 = 3 * 3 + 1)

// Implicit casting for smaller integral types
byte b = 10;
b %= 3; 
// Equivalent to: b = (byte)(b % 3);

// Floating-point evaluation (truncating division)
double d = 5.5;
d %= 2.1; 
// d is now 1.3 (5.5 = 2.1 * 2 + 1.3)

// Division by zero behavior
int x = 5;
// x %= 0; // Throws DivideByZeroException

double y = 5.0;
y %= 0.0; 
// y is now double.NaN

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Operator Overloading

You cannot explicitly overload the %= operator in a user-defined struct or class. However, C# automatically supports compound assignment for custom types if you overload the binary remainder operator (%). 
public struct Vector
{
    public int Value;

    // Overloading the binary % operator
    public static Vector operator %(Vector left, Vector right) => 
        new Vector { Value = left.Value % right.Value };
}

// The %= operator is now implicitly supported
Vector v1 = new Vector { Value = 10 };
Vector v2 = new Vector { Value = 3 };
v1 %= v2; // v1.Value becomes 1
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