Hosam Aly Hosam Aly - 3 months ago 10
C# Question

Can't operator == be applied to generic types in C#?

According to the documentation of the

==
operator in MSDN,


For predefined value types, the
equality operator (==) returns true if
the values of its operands are equal,
false otherwise. For reference types
other than string, == returns true if
its two operands refer to the same
object. For the string type, ==
compares the values of the strings.
User-defined value types can overload
the == operator (see operator). So can
user-defined reference types, although
by default == behaves as described
above for both predefined and
user-defined reference types.



So why does this code snippet fail to compile?

void Compare<T>(T x, T y) { return x == y; }


I get the error Operator '==' cannot be applied to operands of type 'T' and 'T'. I wonder why, since as far as I understand the
==
operator is predefined for all types?

Edit: Thanks everybody. I didn't notice at first that the statement was about reference types only. I also thought that bit-by-bit comparison is provided for all value types, which I now know is not correct.

But, in case I'm using a reference type, would the the
==
operator use the predefined reference comparison, or would it use the overloaded version of the operator if a type defined one?

Edit 2: Through trial and error, we learned that the
==
operator will use the predefined reference comparison when using an unrestricted generic type. Actually, the compiler will use the best method it can find for the restricted type argument, but will look no further. For example, the code below will always print
true
, even when
Test.test<B>(new B(), new B())
is called:

class A { public static bool operator==(A x, A y) { return true; } }
class B : A { public static bool operator==(B x, B y) { return false; } }
class Test { void test<T>(T a, T b) where T : A { Console.WriteLine(a == b); } }

Answer

"...by default == behaves as described above for both predefined and user-defined reference types."

Type T is not necessarily a reference type, so the compiler can't make that assumption.

However, this will compile because it is more explicit:

    bool Compare<T>(T x, T y) where T : class
    {
        return x == y;
    }

Follow up to additional question, "But, in case I'm using a reference type, would the the == operator use the predefined reference comparison, or would it use the overloaded version of the operator if a type defined one?"

I would have thought that == on the Generics would use the overloaded version, but the following test demonstrates otherwise. Interesting... I'd love to know why! If someone knows please share.

namespace TestProject
{
 class Program
 {
    static void Main(string[] args)
    {
        Test a = new Test();
        Test b = new Test();

        Console.WriteLine("Inline:");
        bool x = a == b;
        Console.WriteLine("Generic:");
        Compare<Test>(a, b);

    }


    static bool Compare<T>(T x, T y) where T : class
    {
        return x == y;
    }
 }

 class Test
 {
    public static bool operator ==(Test a, Test b)
    {
        Console.WriteLine("Overloaded == called");
        return a.Equals(b);
    }

    public static bool operator !=(Test a, Test b)
    {
        Console.WriteLine("Overloaded != called");
        return a.Equals(b);
    }
  }
}

Output

Inline: Overloaded == called

Generic:

Press any key to continue . . .

Follow Up 2

I do want to point out that changing my compare method to

    static bool Compare<T>(T x, T y) where T : Test
    {
        return x == y;
    }

causes the overloaded == operator to be called. I guess without specifying the type (as a where), the compiler can't infer that it should use the overloaded operator... though I'd think that it would have enough information to make that decision even without specifying the type.