vba - Looping through a Scripting.Dictionary using index/item number

Question

Similar to this issue, when using a Scripting.Dictionary object in VBA, the outcome of the code below is unexpected.

Option Explicit

Sub test()

    Dim d As Variant
    Dim i As Integer
    Dim s As String
    Set d = CreateObject("Scripting.Dictionary")

    d.Add "a", "a"
    Debug.Print d.Count ' Prints '1' as expected

    For i = 1 To d.Count
        s = d.Item(i)
        Debug.Print s ' Prints ' ' (null) instead of 'a'
    Next i

    Debug.Print d.Count ' Prints '2' instead of '1'

End Sub

Using a zero-based index, the same outcome is achieved:

For i = 0 To d.Count - 1
    s = d.Item(i)
    Debug.Print s
Next i

Watching the object, I can actually see that it has two items, the key for the newly added is 1, as added from i. If I increase this loop to a higher number, then the number of items in the dictionary is increased, once for each loop.

I have tested this in Office/VBA 2003, 2010, and 2013. All exhibit the same behavior, and I expect other versions (2007) will as well.

I can work around this with other looping methods, but this caught me off guard when I was trying to store objects and was getting an object expected error on the s = d.Item(i) line.

For the record, I know that I can do things like this:

For Each v In d.Keys
    Set o = d.item(v)
Next v

But I'm more curious about why I can't seem to iterate through the items by number.

question from:https://stackoverflow.com/questions/11296522/looping-through-a-scripting-dictionary-using-index-item-number

Answer 1

According to the documentation of the Item property:

Sets or returns an item for a specified key in a Dictionary object.

In your case, you don't have an item whose key is 1 so doing:

s = d.Item(i)

actually creates a new key / value pair in your dictionary, and the value is empty because you have not used the optional newItem argument.

The Dictionary also has the Items method which allows looping over the indices:

a = d.Items
For i = 0 To d.Count - 1
    s = a(i)
Next i