You need to deal with the optional Rank parameter of UBound.
Dim arr(1 To 4, 1 To 3) As Variant
Debug.Print UBound(arr, 1) '◄ returns 4
Debug.Print UBound(arr, 2) '◄ returns 3
More at: UBound Function (Visual Basic)
[This is a late answer addressing the title of the question (since that is what people would encounter when searching) rather than the specifics of OP's question which has already been answered adequately]
Ubound is a bit fragile in that it provides no way to know how many dimensions an array has. You can use error trapping to determine the full layout of an array. The following returns a collection of arrays, one for each dimension. The count property can be used to determine the number of dimensions and their lower and upper bounds can be extracted as needed:
Function Bounds(A As Variant) As Collection
Dim C As New Collection
Dim v As Variant, i As Long
On Error GoTo exit_function
i = 1
Do While True
v = Array(LBound(A, i), UBound(A, i))
C.Add v
i = i + 1
Loop
exit_function:
Set Bounds = C
End Function
Used like this:
Sub test()
Dim i As Long
Dim A(1 To 10, 1 To 5, 4 To 10) As Integer
Dim B(1 To 5) As Variant
Dim C As Variant
Dim sizes As Collection
Set sizes = Bounds(A)
Debug.Print "A has " & sizes.Count & " dimensions:"
For i = 1 To sizes.Count
Debug.Print sizes(i)(0) & " to " & sizes(i)(1)
Next i
Set sizes = Bounds(B)
Debug.Print vbCrLf & "B has " & sizes.Count & " dimensions:"
For i = 1 To sizes.Count
Debug.Print sizes(i)(0) & " to " & sizes(i)(1)
Next i
Set sizes = Bounds(C)
Debug.Print vbCrLf & "C has " & sizes.Count & " dimensions:"
For i = 1 To sizes.Count
Debug.Print sizes(i)(0) & " to " & sizes(i)(1)
Next i
End Sub
Output:
A has 3 dimensions:
1 to 10
1 to 5
4 to 10
B has 1 dimensions:
1 to 5
C has 0 dimensions:
UBound(myArray, 1) returns the number of rows in 2d arrayUBound(myArray, 2) returns the number of columns in 2d arrayHowever, let's go 1 step further and assume that you need the last row and last column of range, that has been written as a 2d array. That row (or column) should be converted to a 1d array. E.g. if our 2d array looks like this:
Then running the code below, will give you 2 1D arrays, that are the last column and last row:
Sub PrintMultidimensionalArrayExample()
Dim myRange As Range: Set myRange = Range("a1").CurrentRegion
Dim myArray As Variant: myArray = myRange
Dim lastRowArray As Variant: lastRowArray = GetRowFromMdArray(myArray, UBound(myArray, 1))
Dim lastColumnArray As Variant
lastColumnArray = GetColumnFromMdArray(myArray, UBound(myArray, 2))
End Sub
Function GetColumnFromMdArray(myArray As Variant, myCol As Long) As Variant
'returning a column from multidimensional array
'the returned array is 0-based, but the 0th element is Empty.
Dim i As Long
Dim result As Variant
Dim size As Long: size = UBound(myArray, 1)
ReDim result(size)
For i = LBound(myArray, 1) To UBound(myArray, 1)
result(i) = myArray(i, myCol)
Next
GetColumnFromMdArray = result
End Function
Function GetRowFromMdArray(myArray As Variant, myRow As Long) As Variant
'returning a row from multidimensional array
'the returned array is 0-based, but the 0th element is Empty.
Dim i As Long
Dim result As Variant
Dim size As Long: size = UBound(myArray, 2)
ReDim result(size)
For i = LBound(myArray, 2) To UBound(myArray, 2)
result(i) = myArray(myRow, i)
Next
GetRowFromMdArray = result
End Function
In addition to the already excellent answers, also consider this function to retrieve both the number of dimensions and their bounds, which is similar to John's answer, but works and looks a little differently:
Function sizeOfArray(arr As Variant) As String
Dim str As String
Dim numDim As Integer
numDim = NumberOfArrayDimensions(arr)
str = "Array"
For i = 1 To numDim
str = str & "(" & LBound(arr, i) & " To " & UBound(arr, i)
If Not i = numDim Then
str = str & ", "
Else
str = str & ")"
End If
Next i
sizeOfArray = str
End Function
Private Function NumberOfArrayDimensions(arr As Variant) As Integer
' By Chip Pearson
' http://www.cpearson.com/excel/vbaarrays.htm
Dim Ndx As Integer
Dim Res As Integer
On Error Resume Next
' Loop, increasing the dimension index Ndx, until an error occurs.
' An error will occur when Ndx exceeds the number of dimension
' in the array. Return Ndx - 1.
Do
Ndx = Ndx + 1
Res = UBound(arr, Ndx)
Loop Until Err.Number <> 0
NumberOfArrayDimensions = Ndx - 1
End Function
Example usage:
Sub arrSizeTester()
Dim arr(1 To 2, 3 To 22, 2 To 9, 12 To 18) As Variant
Debug.Print sizeOfArray(arr())
End Sub
And its output:
Array(1 To 2, 3 To 22, 2 To 9, 12 To 18)
Looping D3 ways;
Sub SearchArray()
Dim arr(3, 2) As Variant
arr(0, 0) = "A"
arr(0, 1) = "1"
arr(0, 2) = "w"
arr(1, 0) = "B"
arr(1, 1) = "2"
arr(1, 2) = "x"
arr(2, 0) = "C"
arr(2, 1) = "3"
arr(2, 2) = "y"
arr(3, 0) = "D"
arr(3, 1) = "4"
arr(3, 2) = "z"
Debug.Print "Loop Dimension 1"
For i = 0 To UBound(arr, 1)
Debug.Print "arr(" & i & ", 0) is " & arr(i, 0)
Next i
Debug.Print ""
Debug.Print "Loop Dimension 2"
For j = 0 To UBound(arr, 2)
Debug.Print "arr(0, " & j & ") is " & arr(0, j)
Next j
Debug.Print ""
Debug.Print "Loop Dimension 1 and 2"
For i = 0 To UBound(arr, 1)
For j = 0 To UBound(arr, 2)
Debug.Print "arr(" & i & ", " & j & ") is " & arr(i, j)
Next j
Next i
Debug.Print ""
End Sub
