assigning multidimensional arrays and then using Ubound.........i am using it wrong.

[welchs101]

I have the following code:

ReDim stdy_file_array1(UBound(file_array1), 3)

For x = 1 To UBound(file_array1)
stdy_file_array1(x, 1) = Left(file_array1(x), InStr(1, file_array1(x), "_") - 1)
stdy_file_array1(x, 2) = path1
stdy_file_array1(x, 3) = file_array1(x)
Next x

But for some reason when i try to find the ubound using the following .....i get an error.

MsgBox ("ubound of stdy_file_array1 is   " & UBound(stdy_file_array1, 1))
MsgBox ("ubound of stdy_file_array1 is   " & UBound(stdy_file_array1, 2))
MsgBox ("ubound of stdy_file_array1 is   " & UBound(stdy_file_array1, 3))

So i think i am confused about something or not doing something correct. I view the array above as having 3 cols by x-number of rows.

[nfuids]

Your array only has 2 dimensions.. and you try to get the UBound of the 3rd dimension...

UBound(stdy_file_array1, 3) means "get the upper bound of the 3rd dimension", but in your redim, you only have 2 dimensions.

[nfuids]

and also, be advised you can only redim the last dimension if you want to preserve data.

So sometime, you need to add columns instead of rows (for a 2 dimensions array). making each column represent a record. Once you've done, you can transpose it to make it a row per record, as we are more used to use...

[welchs101]

for some reason i thought that by having the "stdy_array1(14,3)" that this ment i had 3 dimensions......but your right.......now that i think about it its only 2 dimensions.

Makes sense

[welchs101]

you read my mind.............so if i used


lets assume i had the following:

redim stdy_array1(10,3)


then i do somethings in my code


now i want to add items to my array

so i do the following

redim preserve stdy_array(100,3)


this wont work? is that what your saying? the first 10 items in my array wont be preserved?

[shg]

When you dimension an array, you should explicitly define the lower boundary of each dimension:

dim myArray(1 to 10, -3 to 6, 0 to 2) as whatever

Otherwise, the lower boundary defaults to zero, unless you use Option Base 1, which is not great practice.

[welchs101]

not to get off topic......but what is wrong with arrays starting at 1...........what is the advantage of starting them at "0"......i thought there was no advantage either way just a preference thing.

[shg]

It doesn't matter what they start with -- it's a matter of convenience. But it does matter how many elements you have, especially if you're writing an array to a worksheet:

Sub x()
    Dim i           As Long
    Dim ad1(1 To 3) As Double
    Dim ad2(3)      As Double

    For i = 1 To 3
        ad1(i) = i ^ 2
        ad2(i) = i ^ 2
    Next i

    Range("A1:C1").Value = ad1
    Range("A2:D2").Value = ad2
End Sub

[nfuids]

Exactly, you can't redim preserve an array if you change the first dimension.

That's why it would be better to do this

redim stdy_array1(3,10)

and later do

redim preserve stdy_array1(3, 100)

that way, the 3 headers you had when you declared it (10,3) will now be the first column instead of the first row. and you can add as many column as you want.