Excel 2011 Buggs

[pgc01]

Mike

Sorry that I did not answer before but I could not come to the board yesterday.

Using your example this is how I usually like to do it:

Sub test()
    Dim testArray() As String
 
    ReDim testArray(1 To 3)
    testArray(1) = "a"
    testArray(2) = "b"
    testArray(3) = "c"
 
    MsgBox Join(reverseArray(testArray))
End Sub
 
Function reverseArray(aArray() As String) As String()
    Dim Low As Long, High As Long, i As Long
    Dim Result() As String
 
    Low = LBound(aArray): High = UBound(aArray)
    ReDim Result(Low To High)
 
    For i = Low To High
        Result(i) = aArray(High - i + Low)
    Next i
    reverseArray = Result
End Function

Although this has worked in Windows since excel 2000, it would not work in the Mac, because it still had vba5.

I'm curious to know if this finally works in excel 2011. If you find the time to test it please post the result.

Remark: I changed the type in the arrays to String to use the Join(), that expects an array of substrings, but the function could return any typed array.

I guess I'm not sharing your dislike of variant arrays.

Well, I recognise that I'm a fan(atic) of strong typing.

I know vba needs variants sometimes but, being given the choice, I prefer to be specific about the type.

 

[mikerickson]

...I'm curious to know if this finally works in excel 2011. If you find the time to test it please post the result.
...

Yes, the code you posted works in Excel 2011.
It also works if

Redim reverseArray(1 to 4)

is added at the end.

This is good, in some situations, I prefer using String arrays so a ReDim will fill with a null stiring null rather than a 0 null.

 

[repairman615]

mikerickson,
Thanks for confirming that, and for your other helpful posts.

 

[pgc01]

Yes, the code you posted works in Excel 2011.

... I prefer using String arrays so a ReDim will fill with a null stiring null rather than a 0 null.

Great!

Good point. That's another advantaged of returning a typed array, the elements of the array are initialised with the corresponding initial value for the type.