Reputation: 3458
I have the following example:
namespace ComparisonExample
{
class Program
{
static void Main(string[] args)
{
var hello1 = new Hello();
var hello2 = new Hello();
// calls Hello.Equals
var compareExplicitly = hello1.Equals(hello2);
// calls Object.Equals
var compareWithGenerics = ObjectsEqual<Hello>(hello1, hello2);
}
private static bool ObjectsEqual<TValue>(TValue value1, TValue value2)
{
return value1.Equals(value2);
}
}
class Hello : IEquatable<Hello>
{
public bool Equals(Hello other)
{
return true; // doesn't matter
}
}
}
The question is why in the second "Equals" call I'm redirected to Object.Equals instead of Hello.Equals even though I'm specifying the exact type in generic argument?
Upvotes: 6
Views: 248
Reputation: 7475
Addition from MSDN:
Unbounded Type Parameters.
Type parameters that have no constraints, such as T
in public class SampleClass<T> { }
, are called unbounded type parameters.
Unbounded type parameters have the following rules:
!=
and ==
operators cannot be used because there is no guarantee that the concrete type argument will support these operators.System.Object
or explicitly converted to any interface type.null
. If an unbounded parameter is compared to null
, the comparison will always return false
if the type argument is a value type.In this case TValue
is converted to System.Object
and Equals
method called.
Upvotes: 2
Reputation: 111840
Because you haven't told the generic method that your object implements IEquatable<T>
:
Try now with:
private static bool ObjectsEqual<TValue>(TValue value1, TValue value2)
where TValue : IEquatable<TValue> // IMPORTANT!!!
{
return value1.Equals(value2);
}
In your ObjectsEqual
method you have access only to methods/properties/fields of TValue
that are defined in the object
class plus the methods that are defined in the interfaces/base classes defined in the constraints. No constraints => you have access only to Equals(object)
, GetHashCode()
, GetType()
, (and if you have the constraint class
: operator==
, operator!=
.) Of these two are virtual (Equals(object)
, GetHashCode()
), so you'll use the "correct" version, the third isn't normally overwritten (GetType()
), so you'll probably use the "correct" version. Only the two operators ==
/!=
are often overwritten and lo and behold! In your generic method you can't use the "correct" version of the two! :-)
Upvotes: 11