Excel's precision is only 15 decimal digits. You're using 17.
Some information he

http://www.cpearson.com/Excel/rounding.htm

--
Kind regards,

Niek Otten
Microsoft MVP - Excel

"Michiel via OfficeKB.com" <u40062@uwe wrote in message news:89e4203ddd3d3@uwe...
| Hi all,
|
| This rounding issue in VBA drive me crazy!
| Many posts report that VBA uses the bankers rounding. But I need the standard
| rounding.
|
| So I tried to make my own Mathematical Rounding function. For some
| mysterieous reasons it does not always work as expected.
|
| Anyone who likes solving strange behaviour of VBA is kindly invited to see
| why.
|
| MathRound(101.68000000000001,5) works OK -- 101.68000
| MathRound(102.63000000000001,5) does not work --- 102.63001 (102.63000
| expected)
|
| The function:
| Function MathRound(dblInput As Double, intDigits As Integer) As Double
| 'Rounds the input value (DblInput) to the number of digits specified in
| intDigits
| 'The container values dblContainer and dblContner seem to be necessary
| because
| 'without them the calculations are wrongly passed through.
| Dim dblContainer As Double
| Dim lngExpon As Long
| Dim lngLong As Long
| Dim dblContner As Double
| dblContainer = dblInput + 5 * (10 ^ -(intDigits + 1)) 'Add a bit to make
| sure truncation is done right.
| lngExpon = 10 ^ intDigits 'calculate the power of ten to be multiplied
| based on the number of digits
| dblContner = Round(dblContainer * lngExpon, 0) 'PUT value in second
| container
| lngLong = Fix(dblContner) 'truncate value
| MathRound = lngLong / lngExpon 'Devide back to the correct number of
| digits
| End Function
|
| Thanks!
|
| --
| Message posted via OfficeKB.com
| http://www.officekb.com/Uwe/Forums.a...tions/200809/1
|