Array blowing up

[Bill]

For some reason, when I try to run the following array, I'm getting an
error:
dim cat(0)

cat(0)="0407"
cat(1)="0102"

'I will be adding more to this array, but just trying with two for
starts.

for counter = 0 to 1

itno=cat(counter)

'I get an error when it runs through this function

Function rightvar(theVar,lengthNeeded)
Dim sResult
sResult=theVar
if Len(theVar)< lengthNeeded then sResult= String(lengthNeeded -
Len(theVar)," ")& theVar
rightvar=sResult
End Function
Can someone help me understand why?

Thanks.

[Manohar Kamath [MVP]]

You created an array to hold one item, but are filling it with two items,
perhaps that's why the error. Try

Dim Cat(1)

The "1" represents the maximum, zero-based, index you can have in the array.
In this case 1 means 0, 1 = 2 items.

--
Manohar Kamath
Editor, .netBooks
www.dotnetbooks.com
"Bill" <bi***********@gospellight.com> wrote in message
news:8d**************************@posting.google.c om...

For some reason, when I try to run the following array, I'm getting an
error:
dim cat(0)

cat(0)="0407"
cat(1)="0102"

'I will be adding more to this array, but just trying with two for
starts.

for counter = 0 to 1

itno=cat(counter)

'I get an error when it runs through this function

Function rightvar(theVar,lengthNeeded)
Dim sResult
sResult=theVar
if Len(theVar)< lengthNeeded then sResult= String(lengthNeeded -
Len(theVar)," ")& theVar
rightvar=sResult
End Function
Can someone help me understand why?

Thanks.

[Dave Anderson]

> Function rightvar(theVar,lengthNeeded)

Dim sResult
sResult=theVar
if Len(theVar)< lengthNeeded then sResult= String(lengthNeeded -
Len(theVar)," ")& theVar
rightvar=sResult
End Function

That's the oddest indentation scheme I've yet seen.

As for your errors, read up on Dim, ReDim and ReDim Preserve...
http://msdn.microsoft.com/library/en...l/vsstmdim.asp
http://msdn.microsoft.com/library/en...vsstmredim.asp

....and then tell us what kind of error you get when trying your rightvar
function.

--
Dave Anderson

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