Why does Math.Round(2.5) return 2 instead of 3?

Question

In C#, the result of Math.Round(2.5) is 2.

It is supposed to be 3, isn\'t it? Why is it 2 instead in C#?

Answer 1

I had this problem where my SQL server rounds up 0.5 to 1 while my C# application didn't. So you would see two different results.

Here's an implementation with int/long. This is how Java rounds.

int roundedNumber = (int)Math.Floor(d + 0.5);

It's probably the most efficient method you could think of as well.

If you want to keep it a double and use decimal precision , then it's really just a matter of using exponents of 10 based on how many decimal places.

public double getRounding(double number, int decimalPoints)
{
    double decimalPowerOfTen = Math.Pow(10, decimalPoints);
    return Math.Floor(number * decimalPowerOfTen + 0.5)/ decimalPowerOfTen;
}

You can input a negative decimal for decimal points and it's word fine as well.

getRounding(239, -2) = 200