Shallow and deep copies

[Jeroen De Dauw]

Hey, I'm a little confused with the definition of a shallow copy. From the definitions I can find online, I get that reference types get copied, but not the value they refer to. In other words, they should point to the same value. However, when I test this, I can assign a new value to a reference type in either the original or the copy, and the value in it's counterpart won't change. This is the code I used to test this:

Sub Main()
    Dim list1 As New List(Of Spam)
    list1.Add(New Spam(1))
    list1.Add(New Spam(2))
    list1.Add(New Spam(3))

    Dim list2 As List(Of Spam)
    list2 = list1.GetRange(0, list1.Count) ' this should make a shallow copy
    list2.Item(1) = New Spam(5)

    For Each s As Spam In list1
        Console.Write(s.ToString & " ")
    Next
    Console.ReadKey()
End Sub

Private Class Spam
    Public Sub New(ByVal v As Integer)
        value = v
    End Sub
    Dim value As Integer
    Public Overrides Function ToString() As String
        Return value.ToString
    End Function
End Class

The output of this code is "1 2 3", and not "1 5 3", which I'd expect from the definition of a shallow copy. Please un-confuse me here :) Cheers!

Jeroen De Dauw --- Forums ; Blog ; Wiki --- 70 72 6F 67 72 61 6D 6D 69 6E 67 20 34 20 6C 69 66 65!

[Luc Pattyn]

Dag Jeroen, While you write

jeroen de dauw wrote:

I can assign a new value to a reference type

that does not really make sense (a type is a type, a type does not have a value at all); and it is not what list2.Item(1) = New Spam(5) does. Yes list2 is a reference type (it is an object, not a value type), as lists and arrays are collections holding things of a single type (either value types of reference types themselves). In this case, the list holds reference types, i.e. references to Spam instances. What your statement does is it overwrites an existing reference by a reference to a new object. So none of the existing Spam objects is being modified. As your use of the terminology isn't correct, this suggests you got a few things confused. A shallow copy copies everything: the data in value types and the data in a reference type. The trick is a reference type only holds a reference (think of it as a pointer), and that is what gets copied. The thing (object) it points to is not being copied or cloned, so the old and new pointers now point to one and the same thing. To prove, or rather disprove, your statement about shallow copies, you should add a Value property to the Spam class, then execute list2.Item(1).Value = 17 and watch how both lists reflect the change, as you now would have changed an existing object, where both lists are referring to. :)

Luc Pattyn [Forum Guidelines] [Why QA sucks] [My Articles]

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[Jeroen De Dauw]

Well, now you said it, it seems so obvious :) Thanks a lot for making me understand my error :thumbsup:

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