I am fairly new to VBA, so this may be a simple question but here goes.
I would like to initialize an entire array myArray
, say of integers, in VBA. I k
For VBA you need to initialise in two lines.
Sub TestArray()
Dim myArray
myArray = Array(1, 2, 4, 8)
End Sub
I want to initialize every single element of the array to some initial value. So if I have an array Dim myArray(300) As Integer of 300 integers, for example, all 300 elements would hold the same initial value (say, the number 13).
Can anyone explain how to do this, without looping? I'd like to do it in one statement if possible.
What do I win?
Sub SuperTest()
Dim myArray
myArray = Application.Transpose([index(Row(1:300),)-index(Row(1:300),)+13])
End Sub
This is easy, at least if you want a 1-based, 1D or 2D variant array:
Sub StuffVArr()
Dim v() As Variant
Dim q() As Variant
v = Evaluate("=IF(ISERROR(A1:K1), 13, 13)")
q = Evaluate("=IF(ISERROR(A1:G48), 13, 13)")
End Sub
Byte arrays also aren't too bad:
Private Declare Sub FillMemory Lib "kernel32" Alias "RtlFillMemory" _
(dest As Any, ByVal size As Long, ByVal fill As Byte)
Sub StuffBArr()
Dim i(0 To 39) As Byte
Dim j(1 To 2, 5 To 29, 2 To 6) As Byte
FillMemory i(0), 40, 13
FillMemory j(1, 5, 2), 2 * 25 * 5, 13
End Sub
You can use the same method to fill arrays of other numeric data types, but you're limited to only values which can be represented with a single repeating byte:
Sub StuffNArrs()
Dim i(0 To 4) As Long
Dim j(0 To 4) As Integer
Dim u(0 To 4) As Currency
Dim f(0 To 4) As Single
Dim g(0 To 4) As Double
FillMemory i(0), 5 * LenB(i(0)), &HFF 'gives -1
FillMemory i(0), 5 * LenB(i(0)), &H80 'gives -2139062144
FillMemory i(0), 5 * LenB(i(0)), &H7F 'gives 2139062143
FillMemory j(0), 5 * LenB(j(0)), &HFF 'gives -1
FillMemory u(0), 5 * LenB(u(0)), &HFF 'gives -0.0001
FillMemory f(0), 5 * LenB(f(0)), &HFF 'gives -1.#QNAN
FillMemory f(0), 5 * LenB(f(0)), &H80 'gives -1.18e-38
FillMemory f(0), 5 * LenB(f(0)), &H7F 'gives 3.40e+38
FillMemory g(0), 5 * LenB(g(0)), &HFF 'gives -1.#QNAN
End Sub
If you want to avoid a loop in other situations, it gets even hairier. Not really worth it unless your array is 50K entries or larger. Just set each value in a loop and you'll be fast enough, as I talked about in an earlier answer.
Fancy way to put @rdhs answer in a function:
Function arrayZero(size As Integer)
arrayZero = Evaluate("=IF(ISERROR(Transpose(A1:A" & size & ")), 0, 0)")
End Function
And use like this:
myArray = arrayZero(15)
You can initialize the array by specifying the dimensions. For example
Dim myArray(10) As Integer
Dim myArray(1 to 10) As Integer
If you are working with arrays and if this is your first time then I would recommend visiting Chip Pearson's WEBSITE.
What does this initialize to? For example, what if I want to initialize the entire array to 13?
When you want to initailize the array of 13 elements then you can do it in two ways
Dim myArray(12) As Integer
Dim myArray(1 to 13) As Integer
In the first the lower bound of the array would start with 0
so you can store 13 elements in array. For example
myArray(0) = 1
myArray(1) = 2
'
'
'
myArray(12) = 13
In the second example you have specified the lower bounds as 1
so your array starts with 1
and can again store 13 values
myArray(1) = 1
myArray(2) = 2
'
'
'
myArray(13) = 13
Wnen you initialize an array using any of the above methods, the value of each element in the array is equal to 0
. To check that try this code.
Sub Sample()
Dim myArray(12) As Integer
Dim i As Integer
For i = LBound(myArray) To UBound(myArray)
Debug.Print myArray(i)
Next i
End Sub
or
Sub Sample()
Dim myArray(1 to 13) As Integer
Dim i As Integer
For i = LBound(myArray) To UBound(myArray)
Debug.Print myArray(i)
Next i
End Sub
FOLLOWUP FROM COMMENTS
So, in this example every value would be 13. So if I had an array Dim myArray(300) As Integer, all 300 elements would hold the value 13
Like I mentioned, AFAIK, there is no direct way of achieving what you want. Having said that here is one way which uses worksheet function Rept
to create a repetitive string of 13's. Once we have that string, we can use SPLIT
using ","
as a delimiter. But note this creates a variant array but can be used in calculations.
Note also, that in the following examples myArray
will actually hold 301 values of which the last one is empty - you would have to account for that by additionally initializing this value or removing the last "," from sNum
before the Split
operation.
Sub Sample()
Dim sNum As String
Dim i As Integer
Dim myArray
'~~> Create a string with 13 three hundred times separated by comma
'~~> 13,13,13,13...13,13 (300 times)
sNum = WorksheetFunction.Rept("13,", 300)
sNum = Left(sNum, Len(sNum) - 1)
myArray = Split(sNum, ",")
For i = LBound(myArray) To UBound(myArray)
Debug.Print myArray(i)
Next i
End Sub
Using the variant array in calculations
Sub Sample()
Dim sNum As String
Dim i As Integer
Dim myArray
'~~> Create a string with 13 three hundred times separated by comma
sNum = WorksheetFunction.Rept("13,", 300)
sNum = Left(sNum, Len(sNum) - 1)
myArray = Split(sNum, ",")
For i = LBound(myArray) To UBound(myArray)
Debug.Print Val(myArray(i)) + Val(myArray(i))
Next i
End Sub
This function works with variables for size and initial value it combines tbur & Filipe responses.
Function ArrayIniValue(iSize As Integer, iValue As Integer)
Dim sIndex As String
sIndex = "INDEX(Row(1:" & iSize & "),)"
ArrayIniValue = Evaluate("=Transpose(" & sIndex & "-" & sIndex & "+" & iValue & ")")
End Function
Called this way:
myArray = ArrayIniValue(350, 13)