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Home/ Questions/Q 7630719
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Editorial Team
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Editorial Team
Asked: May 31, 20262026-05-31T06:10:27+00:00 2026-05-31T06:10:27+00:00

I have a private class, User that implements an interface, IQMUser, with public static

  • 0

I have a private class, User that implements an interface, IQMUser, with

    public static bool operator ==( User a, User b ) { return Equals( a, b ); }
    public static bool operator !=( User a, User b ) { return !Equals( a, b ); }
    public override bool Equals ( object obj )
    {
        User other = obj as User;
        return m_LMUser.UserID == other.m_LMUser.UserID;
    }
    public override int GetHashCode ( )
    {
        return m_LMUser.GetHashCode ( );
    }

User is a member of a public class, UserManager, which has GetUsers( ), which returns IList<IQMUser>. In another class, I have

    IQMUser otherUser = UserManager.GetUsers( )
        .Where( u => u != CurrentUser )
        .FirstOrDefault( ); 

Now, I know for a fact that GetUsers returns an object that has the same m_LMUser.UserID as CurrentUser. However, the test u != CurrentUser is always true, even in that case.

I have set breakpoints at each of the operator overloads and the Equals override in User, but none of them hit.

I then changed the query to

    IQMUser otherUser = UserManager.GetUsers( )
        .Where( u => !u.Equals( CurrentUser ) )
        .FirstOrDefault( ); 

This hits the overridden Equals method as expected.

What is wrong with my operator overloads that Linq doesn’t even hit them?

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  1. Editorial Team
    Editorial Team
    2026-05-31T06:10:28+00:00Added an answer on May 31, 2026 at 6:10 am

    If I understand your inheritance hierarchy correctly, the type in the “Where” seems to be IQMUser (the base class), while the operator == is implemented on the subclass User.

    Since operator == is not virtual, the Linq Where expression with type IQMUser will run the IQMUser operator == instead of the one implemented on User.

    Equals on the other hand is virtual and the override works as it should, allowing it to work correctly if you call it directly instead.

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