VB.NET's Math.Round()

[DeFlipMode]

VB.NET's Math.Round() seems to be a Round_Half_Even as the data below shows: (5.5 rounds to 6.0, and 6.5 rounds to 6.0)

 

?math.round(5.2)

5.0

?math.round(5.5)

6.0

?math.round(5.8)

6.0

?math.round(6.0)

6.0

?math.round(6.5)

6.0

 

Is there an easy way to create a Round_Half_Up so that 5.5 rounds to 6.0 and 6.5 rounds to 7.0?

 

I can think of a number of ways to do this by testing, but not sure if .NET has this built in and I just can't find it in the documentaion. Any Ideas?

[Iceplug]

The only way that I can think of is this:

?Math.Floor(N + 0.5)

 

The rounding thing has something to do with statistical accuracy... The average of 5.5 and 6.5 is 6, rounding both numbers up will give you 6.5 as the average, but rounding towards even, as the .Round function, and most computer rounding algorithms, wil give you the average of 6, and similarly, you get 5 from 4.5 and 5.5 :).

[DeFlipMode]

Thanks!!

 

Looking at your method makes me think... Duh... why didn't I think of that! :rolleyes:

 

That is a lot easier than the method I was going to use, and more accurate. Glad I asked.

:D

[Nerseus]

If you need to worry about negative numbers, you'll need to use Ceiling. Here's a function I use which takes a decimal. Feel free to change it for use with double or other types.

public static decimal RoundUpToWholeNumber(decimal val)
{
if(val<decimal.Zero)
	return (decimal)Math.Ceiling((double)val - 0.5);
else
	return (decimal)Math.Floor((double)val + 0.5);

}

 

-nerseus

[DeFlipMode]

Thanks for the additional info! For the behavior I required, the Math.Floor(N+0.5) tested fine for both positive and negative numbers.... but always nice to know more than you need. :cool:

 

Later,