Answers for "xl vba dynamic string array"

xl vba dynamic string array

'Since VBA arrays can have any integer value for the base element,
'the only sure fire way to determine how many elements is in the array
'is to calculate the value:

MsgBox UBound(Arr) - LBound(Arr) + 1		'<--for elems in 1D array

MsgBox UBound(Arr, 1) - LBound(Arr, 1) + 1	'<--for rows in 2D array
MsgBox UBound(Arr, 2) - LBound(Arr, 2) + 1	'<--for cols in 2D array

xl vba dynamic string array

n = 5
 a = Array(1, 2, 3, 4, 5)
 b = Array(1, 2, 3, 4, 5)
 [a1].Resize(n) = Evaluate("{" & Join(a, ";") & "}*{" & Join(b, ";") & "}")

xl vba dynamic string array

'In VBA a dynamic string array is declared in the following ways:

Dim strArr1() As String
Dim strArr2$()

'Neither declaration style is prefered. They are exactly equivalent.

'Both ways above produce a dynamic array, which means that the exact size of
'the array (how many elements) is specified later in code:

Redim strArr1(1 to 10)

'To instead declare a static string array (an array with a fixed size), 
'specify the size within the Dim statement:

Dim strArr1(1 to 10) As String
Dim strArr2$(0 to 4)

'ReDim() cannot later be used on static arrays. Their size is fixed.

'Once a string array is declared, the array elements are assign one at a time:

strArray1(1) = "abc"

'However, the Split() function can be used to split a string and assign to and
'to also size a dynamic string array in one go:

Dim strArr3$()
strArray3 = Split("a,b,c", ",")

'strArray3 will now contain three elements, "a", "b", and "c".

'Note: String arrays can only conatin text values. When numbers are assigned
'      to string array elements, the number is first coerced to a string value.
'      Boolean and Date values are likewise coerced to text values.
'      Objects CANNOT be assinged to string array elements.
'
'      For the above reasons (and especially with Excel data) it may be better
'      to declare and use Variant arrays instead of String arrays. Variants
'      can be assigned any data type. The trade-off is that manipulating 
'      Variants is slightly slower than base data types.
'
'      Variants are the default data type in VBA so the following declares
'      an array of Variants:

Dim vArr1()

'      But the data type may also be specified explicitly:

Dim vArr1() As Variant

'      The Variant data type does not have a specific Type Declaration Character.
'      For example, the String data type has the '$' type declaration character
'      which was demonstrated at the top of this page. But since the Variant is
'      VBA's default, no type declaration character is needed. Not specifying
'      a type results in a variable's type being a Variant, by default.

'Note: it is important to realize that since a Variant can hold any other type
'      of data, it can also hold an array. It is EXTREMELY common in VBA to 
'      declare a single Variant variable that will later be assinged an array:

Dim v

'	   The above variable is a single Variant. It is not an array. Variants
'      can be assinged numbers, dates, Booleans, error values, objects, 
'      essentially anything. And in addition, variants can also
'      be assigned arrays of any type:

v = Array(1, 2, 3, "a", "b", "c", True, False, #1/1/2021#)

'      The above line creates an array of 9 Variant values and assigns the 
'      entire array to the Variant existing variable, 'v'.
'
'      Perhaps the most common way to assign an array of Variants to 
'      an existing Variant variable is to pull data from a worksheet range:

v = Sheet1.Range("A1:A10").Value

'      The above statement assigns a 2-dimension array to v, ten rows deep 
'      and one column wide. The values and data types of the 
'      ten elements will correspond to the values and types
'      in the Sheet1 range, A1:A10. It is important to realize that 
'      the VBA array created here is a copy of the data in the worksheet
'      range. Changing one does not affect the other. But it is possible
'      to write the VBA array back to Excel to the original range, or 
'      to any range in any worksheet in any open workbook.
'
'
'

xl vba dynamic string array

'VBA for passing arrays to and from JScript in the Script Control.

'JScript, running in the Microsoft Script Control, can be
'passed VBA safearrays... exactly the same as Internet Explorer can 
'be. But before JScript can do anything with the values in the 
'VBA foreign array, a new JScript array must be created based 
'on the VBA safearray:

vA = Array(1, 2, 3, 4, 5, 6, 7, 8, 9, 10)
With CreateObject("ScriptControl")
    .Language = "JScript"
    .AddCode "function jsArray(v){return new VBArray(v).toArray()}"
    .AddCode "function sum(v){for(var a=jsArray(v),r=0,t=0;t<a.length;t++)r+=a[t];return r};"
    
  	MsgBox .Run("sum", vA)				'<--displays: 55
End With

'2D arrays and higher are converted to zero-based 1D JScript arrays
'by VBArray().toArray().

'Resource: 
'	https://developer.mozilla.org/en-US/docs/Web/JavaScript/Microsoft_Extensions/VBArray

'-------------------------------------------------------------------------------------------

'Attempting to return a JScript array from the Script Control 
'results in a comma delimited string passed to the VBA side.

'To return the actual array to VBA instead of a string, the array
'must first be converted to a VBA array. This can be done by
'utilizing the .items() property of the Scripting Dictionary:

vA = Array(1, 2, 3, 4, 5, 6, 7, 8, 9, 10)
With CreateObject("ScriptControl")
    .Language = "JScript"
    .AddCode "function jsArray(v){return new VBArray(v).toArray()}"
	.AddCode "function vbArray(t){for(var r=new ActiveXObject('Scripting.Dictionary'),e=0;e<t.length;e++)r.add(e,t[e]);return r.items()}"
  	.AddCode "function roundTrip(v){return vbArray(jsArray(v))'}"
  
  	MsgBox UBound(.Run("roundTrip", vA))		'<--displays: 9  
End With

xl vba dynamic string array

arrayData = Array("A", "B", "C", "D", "E")

[a1].Resize(UBound(arrayData)) = Application.Transpose(arrayData)

xl vba dynamic string array

'VBA function to create a 2D array with every element set 
'to the same initial value:

Function Init2D(rows, cols, Optional val)
    Dim i&, j&
    ReDim v(1 To rows, 1 To cols)
    If Not IsMissing(val) Then
        For i = 1 To rows
            For j = 1 To cols
                v(i, j) = val
            Next
        Next
    End If
    Init2D = v
End Function

'This version uses no loops:
Function Init2D(rows, cols, Optional val = "")
    Const NUMER = "index(offset(a1,,,ROWS,COLS)*0+VAL,,)"
    Const ALPHA = "index(rept("""",len(offset(a1,,,ROWS,COLS))<0)&""VAL"",,)"
    DoEvents
    Init2D = Evaluate(Replace(Replace(Replace(IIf(IsNumeric(val), NUMER, ALPHA), "VAL", val), "ROWS", rows), "COLS", cols))
End Function

'--------------------------------------------------------------------

'10 rows, 1 column, all elements = 0:
a = Init2D(10, 1, 0)

'20 rows, 5 columns, all elements = "abc":
a = Init2D(20, 5, "abc")

'--------------------------------------------------------------------

'1D array version:
Function Init1D(elems, Optional val, Optional base = 1)
    Dim i&, max&
    max = base + elems - 1 
    ReDim v(base To max)
    If Not IsMissing(val) Then
        For i = base To max
            v(i) = val
        Next
    End If
    Init1D = v
End Function

xl vba dynamic string array

v = [a1:b20]

'If v is dimensioned as a variant then he above creates a 2d array, 
'20 rows high by 2 columns wide and those 40 array elements contain
'the values from the specified range.

'The square brackets are shorthand notation. Another way to code the
'same thing is:

v = Range("a1:b20")

'In both of the above examples, the Range object is returning its 
'default... the Range.Value property. But keep in mind that the
'Range.Value2 property is about 20% quicker. So it could be more
'performant to code these two examples like so:

v = [a1:b20].Value2
v = Range("a1:b20").Value2