No.

0 as the third argument requires an exact match - apparently you are not
making a match, so an error is returned. When you remove the zero, match
finds the closes match, so you function works but you may get unexpected
results.

--
Regards,
Tom Ogilvy

"Tom" wrote in message
...
Hi

I have a problem with the match function. This function requires three
parameters, but the following command fails:

Set wf = Application.WorksheetFunction
varMatch = wf.Match(strValue, ws.Range("B1:B1000"), 0)


but when I remove the third parameter, the function works?!? Is this a
bug?

Tom