If I have an object of type MyBull
and a List<MyBull> orig
:
// Just an example
MyBull x = getMeTheObjectWithIdFromDB(9);
orig.add(x);
// Again same? data object
MyBull y = getMeTheObjectWithIdFromDB(9);
Why is this false then?
// This is false, eve开发者_开发问答n though all the properties
// of x and y are the same.
orig.Contains<MyBull>(y);
By default objects will expose reference based equality. If you want custom rules, such as equality based on id fields, you need to override the Equals
and GetHashCode
methods.
If you can use LINQ then you can
class Vessel
{
public int id { get; set; }
public string name { get; set; }
}
...
var vessels = new List<Vessel>() { new Vessel() { id = 4711, name = "Millennium Falcon" } };
var ship = new Vessel { id = 4711, name = "Millencolin" };
if (vessels.Any(vessel => vessel.id == ship.id))
Console.Write("There can be only one!");
This is because the MyBull instances are being compared by reference. From the point of view from .NET, x and y are both different instances and therefore not Equal.
In order to get around this you will have to override the Equals and GetHashCode methods (which means you should probably implement IEquatable<MyBull>
and override the == and != operators too).
Does your MyBull object implement IEquatable<T>.Equals
? This method will determine the equality of two objects
requested by OP
Your MyBull class would implement IEquatable
public class MyBull : IEquatable<MyBull>
and then you would need to override the Equals
method
public bool Equals(MyBull theOtherMyBull)
As David Neale mentions below, this is best used when you're comparing objects of the same type--which you are. Overriding Object.Equals and Object.GetHashCode will work too.
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