How to round to nearest even integer?

Question

My last goal is always to round to the nearest even integer.

For example, the number 1122.5196 I want as result 1122. I have tried this options:

Math.Round(1122.5196d, 0, MidpointRounding.ToEven);       // result 1123 Math.Round(1122.5196d, 0, MidpointRounding.AwayFromZero); // result 1123 

At the end, what I would like to get it is always the nearest even integer. For example:

• 1122.51 --> 1122
• 1122.9 --> 1122 (because the nearest int is 1123 but it is odd, and 1122 is nearer than 1124)
• 1123.0 --> 1124 (the next even value, the next higher even value)

I only work with positive numbers.

And so on.

There are some method that do that or I should to implement my own method?

Answer 1

Try this (let's use Math.Round with MidpointRounding.AwayFromZero in order to obtain "next even value" but scaled - 2 factor):

double source = 1123.0;  // 1124.0 double result = Math.Round(source / 2, MidpointRounding.AwayFromZero) * 2; 

Demo:

double[] tests = new double[] {      1.0,   1123.1,   1123.0,   1122.9,   1122.1,   1122.0,   1121.5,   1121.0, };  string report = string.Join(Environment.NewLine, tests   .Select(item => $"{item,6:F1} -> {Math.Round(item / 2, MidpointRounding.AwayFromZero) * 2}"));  Console.Write(report); 

Outcome:

   1.0 -> 2     // In case of tie, next even value 1123.1 -> 1124 1123.0 -> 1124  // In case of tie, next even value 1122.9 -> 1122 1122.1 -> 1122 1122.0 -> 1122 1121.5 -> 1122 1121.0 -> 1122  // In case of tie, next even value 

Answer 2

One liner:

double RoundToNearestEven(double value) =>     Math.Truncate(value) + Math.Truncate(value) % 2; 

Fiddle

Explanation: if we have an even number with some digits after floating point, we need to just get rid of those digits. If we have an odd number, we need to do the same and then move to the next integer that is guaranteed to be even.

P.S. Thanks to @DmitryBychenko for pointing out that casting double to long is not the brightest idea.