Dynamic Array of dynamic arrays

Question

I can't seem to find this problem addressed anywhere.

I need to declare a bunch of dynamic arrays as follow:

Dim list1 () as variant
Dim list2() as variant
Dim list3() as variant
...
Dim listN() as Variant

Each list is a one-dimensional dynamic array. However, I wouldn't know what "N" will be during the program. I want to make these "N" lists dynamic as well. I have tried two-dimensional dynamic arrays. But the "redim" statement requires both dimensions to be declared at the same time. In particular, I do this:

Dim BigList() as variant
...
Redim BigList(listNum, listLength)

To access/pass into a sub "list1", "list2" , "list3"..., calling "BigList(1)", "BigList(2)" gives me error. In particular, somewhere in my code, there is this portion:

sub ProcessList(byref listToProcess() as variant)
...
end sub

sub main()
...
call ProcessList(list1)
call ProcessList(list2)
...
call ProcessList(listN)

end sub

Now I can do a loop:

for i = 1 to N
    Call ProcessList(list"i")
next i

This requires list"i" to be a one-dimensional dynamic array. So, after a redim BigList(listNum,listLength) and I do this:

for i = 1 to N
    Call ProcessList(BigList(i))    'i refers to listNum
next i

This gives me error "Incompatible type".

BigList(i)(j) would be item j of array i. Also -- if you want dynamic arrays -- just use variants rather than variants() — John Coleman
– John Coleman, Commented Jan 13, 2016 at 18:55
What determines N? You are implying that you don't know what N will be until you get there, but perhaps you could do an initial loop to check the size of the first dimension, and then knowing that, redim to the appropriate size? — Grade 'Eh' Bacon
– Grade 'Eh' Bacon, Commented Jan 13, 2016 at 18:55
Thanks Grade 'Eh' Bacon. I meant "I wouldn't know what N will be". So, in advance, I don't know either the number of lists or the length of each list. "listNum" and "listLength" are external variables for me to take in. But there are subs in which I need to work on each list individually. Hence, the need to call "BigList(1)", "BigList(2)". — user5700405
– user5700405, Commented Jan 13, 2016 at 19:00
You want BigList as an array of one-dimensional arrays? Or as an array of Multi-dimensional arrays? — David Zemens
– David Zemens, Commented Jan 13, 2016 at 19:02
Thanks, John Coleman. But how do I pass each individual list as an argument into a sub. As you pointed out, I have to access BigList elements by providing two indices "i" and "j". Is it possible to access the 2-dimensional array with just the front index "i"? — user5700405
– user5700405, Commented Jan 13, 2016 at 19:02

David Zemens · Accepted Answer · 2016-01-13 19:21:02Z

1

Here is one example of creating a Dictionary which is keyed to integer values (i.e., the N) and each value is initially an empty array.

You can then use something like the ExtendList function to resize those empty arrays as needed.

Sub foo()
Dim BigList As Object
Dim N As Long
Dim v as Variant

'Create an empty dictionary object
Set BigList = CreateObject("Scripting.Dictionary")

'Add N empty array to the dictionary:
N = 3
For i = 1 To N
    BigList(i) = Array()
Next

'Resize one of the items in your BigList
BigList(2) = ExtendList(BigList(2), 1, 10)

v = BigList(2)  'Here you can examine v in the Locals window and see it is an array, of dimensions 1 x 10
End Sub
Function ExtendList(lst, a As Long, b As Long)

    ReDim lst(a, b)
    ExtendList = lst

End Function

On review of your edited question, I think you merely misunderstood how the ReDim statement works:

Redim BigList(listNum, listLength)

This re-dimensions the BigList based on the parameters listNum and listLength. It does not (as it seems you may have expected) create a list of arrays within BigList.

I think this might also work (untested, and remember arrays are zero-index):

ReDim Preserve BigList(listNum)
BigList(listNum) = Array()
ReDim BigList(listNum)(listSize)

edited Jan 13, 2016 at 19:21

answered Jan 13, 2016 at 19:15

David Zemens

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2 Comments

user5700405 Over a year ago

Thank you, David Zemens. This Dictionary thing seems to solve my problem. I will dig more on this.

David Zemens Over a year ago

Cheers @user5700405 I added a brief note explaining what I think was your original problem, based on your edited Q and code samples.

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