Unexpected behaviour when comparing GUIDs in .NET

Question

I have attempted to create an extension method that looks like this...

public static IEnumerable<T> Distinct<T>(this IEnumerable<T> value, IEnumerable<T> compareTo, Func<T, object> compareFieldPredicate)
{
    return value.Where(o => !compareTo.Exists(p => compareFieldPredicate.Invoke(p) == compareFieldPredicate.Invoke(o)));
}

The idea is that I would be able to do something like this...

IEnumerable<MyCollection> distinctValues = MyCollection.Distinct(MyOtherCollection, o => o.ID); //Note that o.ID is a guid

Now at this point I would have expected to have only my distinct items returned to me but what I found is that this was never the case.

Upon further research breaking down this method using the following code.

Guid guid1 = Guid.NewGuid();
Guid guid2 = new Guid(guid1.ToString());

Func<MyObject, object> myFunction = o => o.ID;
Func<MyObject, object> myFunction1 = o => o.ID;

bool result = myFunction(MyObject) == myFunction1(MyObject);
//result = false

I have found that infact even if the Guids are the same the comparison will always return false.

What is the cause of this?

Answer 1

Your problem is that you're boxing the Guids into Objects before you compare them. Consider this code:

Guid g1 = Guid.NewGuid();
var g2 = g1;

Console.WriteLine(g1 == g2);

object o1 = g1;
object o2 = g2;

Console.WriteLine(o1 == o2);

That actually outputs:

true
false

Since "o1" and "o2", while equal to the same Guid, are not the same object.

If you truly want your "Distinct" extension method to not be tied to a specific type (like Guid), you can do this:

public static IEnumerable<TItem> Distinct<TItem, TProp>(this IEnumerable<TItem> value, IEnumerable<TItem> compareTo, Func<TItem, TProp> compareFieldPredicate)
    where TProp : IEquatable<TProp>
{
    return value.Where(o => !compareTo.Any(p => compareFieldPredicate(p).Equals(compareFieldPredicate(o))));
} 

Answer 2

bool result = (guid1==guid2); //result -> true

You can try to change return type Object to GUID in myfunction and myfunction1

Func<MyObject, Guid> myFunction = o => o.ID;
Func<MyObject, Guid> myFunction1 = o => o.ID;

Otherwise, the return value (true) is boxed to Object, and Reference equality is checked, which is false.