Generics Operator '+' cannot be applied to operands of type 'T' and 'T'

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Solution for overloaded operator constraint in .NET generics
Implementing arithmetic in generics?

I wrote Generics class,but i am having issue as described in title.

class Program
    {
        static void Main(string[] args)
        {
            int a = 1;
            int b = 2;
            int c = 3;

            dynamic obj = new Gen<int>();
            obj.TestLine1(ref a, ref b);
            obj = new Gen<string>();
            obj.TestLine2(ref a, ref b, ref c);
            System.Console.WriteLine(a + " " + b);
            Console.ReadLine();
        }
    }

public class Gen<T>
    {
        public void TestLine1(ref T a, ref T b)
        {
            T temp;
            temp = a;
            a = b;
            b = temp;
        }
        public void TestLine2(ref T a, ref T b, ref T c)
        {
            T temp;
            temp = a;
            a = a + b;
            b = a + c;
            c = a + b;
        }
    }

Inside at method TestLine2(ref T a, ref T b, ref T c) I am getting below issue:

Operator '+' cannot be applied to operands of type 'T' and 'T'

It is a quite common problem in C#. You can't define a constraint on _T_ to specify "_T_ must implement the + operator" — Cédric Bignon, Feb 02 '13 at 19:20
You have not bounded `T` to be constrained to `+`able types, so the compiler has no way to know that `+` can be applied to `T`. — Matt Ball, Feb 02 '13 at 19:21
Duplicate of about 10,000 earlier questions: http://stackoverflow.com/search?q=%5Bc%23%5D+%5Bgenerics%5D+arithmetic Look for an answer written by Marc Gravell, he's solved this nicely — Ben Voigt, Feb 02 '13 at 19:21

score 8 · Answer 1 · answered Feb 02 '13 at 19:21

Since T can be any type, there is no guarantee that T will have a static + operator. In C# there's no way to constrain T to support static operators like +, so you'll have to pass the function to use to combine values of T to TestLine2:

public void TestLine2(ref T a, ref T b, ref T c, Func<T, T, T> op)
{
    T temp;
    temp = a;
    a = op(a, b);
    b = op(a, c);
    c = op(a, b);
}

score 1 · Answer 2 · answered Feb 02 '13 at 19:31

1

You don't know whether T implements the + operator. What if you pass object as the type parameter?

answered Feb 02 '13 at 19:31

antonijn

5,702
2
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score 0 · Answer 3 · answered Feb 02 '13 at 19:21

0

Because the type of T is not known until instantiation, there is no guarantee that the type T will support the + operator.

answered Feb 02 '13 at 19:21

Brett Wolfington

6,587
4
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1

True but it is only a diagnose rather than a solution. – Wiktor Zychla Feb 02 '13 at 19:25
@Wicktor: There was no question and no research effort. Perhaps an explanation is what the OP wanted. – Ben Voigt Feb 02 '13 at 19:26

Kevin · Answer 4 · 2013-02-08T19:53:42.583

Suppose I created an instance of your class like this: var gen = new Gen<Object>(). Now T means Object everywhere inside this instance of your class. When you call TestLine2(), the method is going to try to add to Objects, something that can't be done in C#.

More broadly, since C# doesn't know ahead of time what type arguments you're going to create a Gen object with, it restricts you to only using methods defined for all objects.

It looks to me like you really want TestLine2 to be a method for consisting strings. Why don't you just make Gen a non-generic class and tell it to use Strings everywhere instead?

Generics Operator '+' cannot be applied to operands of type 'T' and 'T'

4 Answers4