I have defined the following interface
public interface IHaveAProblem
{
string Issue { get; set; }
}
And here is the implementation of the IHaveAProblem
public class SomeProblem : IHaveAProblem
{
public string Issue { get; set; }
public override bool Equals(object obj)
{
SomeProblem otherObj = obj as SomeProblem;
if (otherObj == null)
{
return false;
}
return this.Issue == otherObj.Issue;
}
public override int GetHashCode()
{
return base.GetHashCode();
}
public static bool operator ==(SomeProblem rhs, SomeProblem lhs)
{
// Null check
if (Object.ReferenceEquals(rhs, null) || Object.ReferenceEquals(lhs, null))
{
if (Object.ReferenceEquals(rhs, null) && Object.ReferenceEquals(lhs, null))
{
// Both are null. They do equal each other
return true;
}
// Only 1 is null the other is not so they do not equal
return false;
}
return rhs.Equals(lhs);
}
public static bool operator !=(SomeProblem rhs, SomeProblem lhs)
{
// Null check
if (Object.ReferenceEquals(rhs, null) || Object.ReferenceEquals(lhs, null))
{
if (Object.ReferenceEquals(rhs, null) && Object.ReferenceEquals(lhs, null))
{
// Both are null. They do equal each other
return false;
}
// Only 1 is null the other is not so they do not equal
return true;
}
return !rhs.Equals(lhs);
}
}
When I use the object I can get the correct results for the == compare.
SomeProblem firstTest = new SomeProblem()
{
Issue = "Hello World"
};
SomeProblem secondTest = new SomeProblem()
{
Issue = "Hello World"
};
// This is true
bool result = firstTest == secondTest;
However when I try to compare the interfaces it is doing a memory compare rather than the operator == on SomeProblem
IHaveAProblem firstProblem = new SomeProblem()
{
Issue = "Hello World"
};
IHaveAProblem secondProblem = new SomeProblem()
{
Issue = "Hello World"
};
Is it possible have the interface use the == on SomeProblem rather than a memory compare?
I know I can do a firstProblem.Equals(secondProblem)
and get the proper result, however I am creating a frame work and I will not know how it is used in the end. I thought == would work correctly.