Hidden features of VBA
Which features of the VBA language are either poorly documented, or simply not often used?
This trick only works in Access VBA, Excel and others won't allow it. But you can make a Standard Module hidden from the object browser by prefixing the Module name with an underscore. The module will then only be visible if you change the object browser to show hidden objects.
This trick works with Enums in all vb6 based version of VBA. You can create a hidden member of an Enum by encasing it's name in brackets, then prefixing it with an underscore. Example:
Public Enum MyEnum
meDefault = 0
meThing1 = 1
meThing2 = 2
meThing3 = 3
[_Min] = meDefault
[_Max] = meThing3
End Enum
Public Function IsValidOption(ByVal myOption As MyEnum) As Boolean
If myOption >= MyEnum.[_Min] Then IsValidOption myOption <= MyEnum.[_Max]
End Function
In Excel-VBA you can reference cells by enclosing them in brackets, the brackets also function as an evaluate command allowing you to evaluate formula syntax:
Public Sub Example()
[A1] = "Foo"
MsgBox [VLOOKUP(A1,A1,1,0)]
End Sub
Also you can pass around raw data without using MemCopy (RtlMoveMemory) by combining LSet with User Defined Types of the same size:
Public Sub Example()
Dim b() As Byte
b = LongToByteArray(8675309)
MsgBox b(1)
End Sub
Private Function LongToByteArray(ByVal value As Long) As Byte()
Dim tl As TypedLong
Dim bl As ByteLong
tl.value = value
LSet bl = tl
LongToByteArray = bl.value
End Function
Octal & Hex Literals are actually unsigned types, these will both output -32768:
Public Sub Example()
Debug.Print &H8000
Debug.Print &O100000
End Sub
As mentioned, passing a variable inside parenthesis causes it to be passed ByVal:
Sub PredictTheOutput()
Dim i&, j&, k&
i = 10: j = i: k = i
MySub (i)
MySub j
MySub k + 20
MsgBox Join(Array(i, j, k), vbNewLine), vbQuestion, "Did You Get It Right?"
End Sub
Public Sub MySub(ByRef foo As Long)
foo = 5
End Sub
You can assign a string directly into a byte array and vice-versa:
Public Sub Example()
Dim myString As String
Dim myBytArr() As Byte
myBytArr = "I am a string."
myString = myBytArr
MsgBox myString
End Sub
"Mid" is also an operator. Using it you overwrite specific portions of strings without VBA's notoriously slow string concatenation:
Public Sub Example1()
''// This takes about 47% of time Example2 does:
Dim myString As String
myString = "I liek pie."
Mid(myString, 5, 2) = "ke"
Mid(myString, 11, 1) = "!"
MsgBox myString
End Sub
Public Sub Example2()
Dim myString As String
myString = "I liek pie."
myString = "I li" & "ke" & " pie" & "!"
MsgBox myString
End Sub
There is an important but almost always missed feature of the Mid() statement. That is where Mid() appears on the left hand side of an assignment as opposed to the Mid() function that appears in the right hand side or in an expression.
The rule is that if the if the target string is not a string literal, and this is the only reference to the target string, and the length of segment being inserted matches the length of the segment being replaced, then the string will be treated as mutable for the operation.
What does that mean? It means that if your building up a large report or a huge list of strings into a single string value, then exploiting this will make your string processing much faster.
Here is a simple class that benefits from this. It gives your VBA the same StringBuilder capability that .Net has.
' Class: StringBuilder
Option Explicit
Private Const initialLength As Long = 32
Private totalLength As Long ' Length of the buffer
Private curLength As Long ' Length of the string value within the buffer
Private buffer As String ' The buffer
Private Sub Class_Initialize()
' We set the buffer up to it's initial size and the string value ""
totalLength = initialLength
buffer = Space(totalLength)
curLength = 0
End Sub
Public Sub Append(Text As String)
Dim incLen As Long ' The length that the value will be increased by
Dim newLen As Long ' The length of the value after being appended
incLen = Len(Text)
newLen = curLength + incLen
' Will the new value fit in the remaining free space within the current buffer
If newLen <= totalLength Then
' Buffer has room so just insert the new value
Mid(buffer, curLength + 1, incLen) = Text
Else
' Buffer does not have enough room so
' first calculate the new buffer size by doubling until its big enough
' then build the new buffer
While totalLength < newLen
totalLength = totalLength + totalLength
Wend
buffer = Left(buffer, curLength) & Text & Space(totalLength - newLen)
End If
curLength = newLen
End Sub
Public Property Get Length() As Integer
Length = curLength
End Property
Public Property Get Text() As String
Text = Left(buffer, curLength)
End Property
Public Sub Clear()
totalLength = initialLength
buffer = Space(totalLength)
curLength = 0
End Sub
And here is an example on how to use it:
Dim i As Long
Dim sb As StringBuilder
Dim result As String
Set sb = New StringBuilder
For i = 1 to 100000
sb.Append CStr( i)
Next i
result = sb.Text