Is there a generic constraint I could use for the + operator?

Question

is there some 'where' type contraints in can add to make the follwing code compile ?

public class Plus<T> : BinaryOperator<T> where T : ...
{
    public override T Evaluate(IContext<T> context)
    {
        return left.Evaluate(context) + right.Evaluate(context);
    }
}

Thanks :)

Actually, you can't. See here: http://stackoverflow.com/questions/147646/solution-for-overloaded-operator-constraint-in-net-generics — dlev, May 13 '11 at 20:01
This is a frequently requested feature but it does not exist today; there's no way to genericize over the existence of a *static* method, and overloaded operators are always static. — Eric Lippert, May 13 '11 at 20:37
To add to eric's statement go down the rabbit hole, operators don't have to just take T, they don't even have to return it! As such, despite really wanting it, I'd be happy to wait for it to be done right... — ShuggyCoUk, May 15 '11 at 13:27
Possible duplicate of [Solution for overloaded operator constraint in .NET generics](https://stackoverflow.com/questions/147646/solution-for-overloaded-operator-constraint-in-net-generics) — Wai Ha Lee, Nov 14 '19 at 08:33

score 24 · Accepted Answer · edited Feb 16 '13 at 08:21

24

There are no such devices in C#. A few options are available, though:

in C# 4.0 and .NET 4.0 (or above), use dynamic, which supports + but offers no compile time checking
in .NET 3.5 (or above), MiscUtil offers an Operator class which makes operators available as methods - again, without any compile-time checking

So either:

return (dynamic)left.Evaluate(context) + (dynamic)right.Evaluate(context);

or

return Operator.Add(left.Evaluate(context), right.Evaluate(context));

edited Feb 16 '13 at 08:21

Camilo Martin

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answered May 13 '11 at 20:02

Marc Gravell

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altough this does not use a type constraints this look more like the solution i'm looking for even if there is no compile time checking ;) thanks a lot. – VANDERWEYEN Jonathan May 14 '11 at 17:05
1

If both approaches can be used, which is preferable and why? – Sam Mar 15 '13 at 08:49
Are any of these supported by Entity frameworK? – Michiel Cornille Jul 15 '13 at 16:10
1

@Mvision no, they are not – Marc Gravell Jul 15 '13 at 16:24

score 5 · Answer 2 · answered May 13 '11 at 20:04

The Type parameter constraints in C# are very limited and is listed here. So the answer is no as far as compile time check goes. If T is a type that you create and manage, one way to go about it would be to

interface IAddable 
{
   IAddable Add(IAddable foo);
}

and implement IFoo for all your types and use where T: IAddable as constraint and use Add() instead of +

score 1 · Answer 3 · answered May 13 '11 at 20:03

Using a generic constraints you can force T

to be a reference type or a value type
to inherit from a certain class
to implement certain interface
to have parameterless constructor

But that's all. You can't force the existence of the static operator+ on it.

score 1 · Answer 4 · answered Oct 15 '20 at 12:36

With C# 8 default interface methods you can achieve something similar, but it might not solve your exact use case:

You can define an interface with a default implementation of an operator:

public interface IFoo
{
    double Value { get; }

    public static IFoo operator +(IFoo a, IFoo b)
    {
        return new Foo(a.Value + b.Value);
    }
}

public class Foo : IFoo
{
    public double Value { get; }

    public Foo(double value)
    {
        Value = value;
    }
}

And consume it in a generic method/class like this:

public static class Program
{
    public static void Main()
    {
        var f1 = new Foo(1);
        var f2 = new Foo(2);
        var sum = Add(f1, f2);
        Console.WriteLine(sum.Value);
    }

    public static IFoo Add<T>(T a, T b) where T : IFoo
    {
        return a + b;
    }
}

Is there a generic constraint I could use for the + operator?

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