How Can I Overload Arithmetic Operators for Generic Number Types in C#?

C# generic class arithmetic operator overloading

In C#, due to type constraints, arithmetic operator overloading of generic classes requires special consideration. When using a generic class definition like this:

public class ConstrainedNumber<T> :
    IEquatable<ConstrainedNumber<T>>,
    IEquatable<T>,
    IComparable<ConstrainedNumber<T>>,
    IComparable<T>,
    IComparable
    where T : struct, IComparable, IComparable<T>, IEquatable<T>

We discovered a limitation in using direct methods to define arithmetic operators:

public static T operator +(ConstrainedNumber<T> x, ConstrainedNumber<T> y)
{
    return x._value + y._value;
}

This code fails to compile because the ' ' operator cannot be applied to types 'T' and 'T' due to the generic nature of 'T'.

To solve this problem, we need a constraint specifically for numeric types with arithmetic operators. Unfortunately, C# does not provide specialized "IArithmetic" constraints.

An alternative is to use the 'IConvertible' constraint, which allows us to handle the arithmetic operations manually:

public static T operator +(T x, T y)
    where T : IConvertible
{
    var type = typeof(T);
    if (type == typeof(string) ||
        type == typeof(DateTime)) throw new ArgumentException(string.Format("The type {0} is not supported", type.FullName), "T");

    try { return (T)(object)(x.ToDouble(NumberFormatInfo.CurrentInfo) + y.ToDouble(NumberFormatInfo.CurrentInfo)); }
    catch (Exception ex) { throw new ApplicationException("The operation failed.", ex); }
}

This code checks for unsupported types such as 'string' and 'DateTime', then converts 'x' and 'y' to double, performs the arithmetic operation, and converts the result back to 'T'.

While this approach allows us to overload arithmetic operators for generic numeric types, it is important to note that it may not handle all potential cases gracefully.

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