problem with vb's round function

Hello experts !

Recently i detected a bug in the round function. In the immediate window of the vb-ide i get the following response:

? round(96.645,2)
 96,64

The return value should be 96.65 by definition. Format works properly:

? format(96.645,"0.00")
96,65

Can i get a fix (i have vb6 SP5)or is there an other elegant function. I don't like using format numbers which aren't been displayed or printed.

Also i know that i can use a function like this for example:

Public Function RoundIt(Number As Variant, NumDigitsAfterDecimal As Long) As Variant
    If Not IsNumeric(Number) Then
        RoundIt = Number
    Else
        RoundIt = Int(0.5 + Number * (10 ^ NumDigitsAfterDecimal)) / (10 ^ NumDigitsAfterDecimal)
    End If
End Function

... but this is perhaps a slow and inelegant way like format. Any suggestions ?

Regards

V.K.
LVL 6
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TimCotteeConnect With a Mentor Commented:
This is the way that the round function works. There are a whole host of questions here and answers in MSDN which explain why. You have the two best options available: format or a custom function so I would suggest you choose one and live with it. Anyway here is the reason why:

http://support.microsoft.com/support/kb/articles/Q196/6/52.ASP

And

http://support.microsoft.com/support/kb/articles/Q194/9/83.ASP 
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Guy Hengel [angelIII / a3]Billing EngineerCommented:
i have vb6 sp3:

? round(96.645,2)
 96.64

Cheers
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VKAuthor Commented:
Hello angel !

...

? round(96.645,2)
96.64

This is not correct as said before. if you have to round with 2 digits after decimal point you have to evaluate the 3rd digit after decimal:

>=5: round up
<5:  round down

Hope now the problem is clear.

V.K.
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Guy Hengel [angelIII / a3]Billing EngineerCommented:
Sorry, I wanted to post more saying that it won't be the sp that fixes it. Like TimCottee indicates, that's the way round function works...
CHeers
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VKAuthor Commented:
Hello Experts !

Thanks to tim (although i have to live with this fact :-)).
I have to do some benchmarking to find out a fast and bulletproof function.

V.K.
0
 
VKAuthor Commented:
Hello Experts !

Thanks to tim (although i have to live with this fact :-)).
I have to do some benchmarking to find out a fast and bulletproof function.

V.K.
0
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