How might I convert a double to the nearest integer value?

Question

double d = 1.234;
int i = Convert.ToInt32(d);

Reference

Handles rounding like so:

rounded to the nearest 32-bit signed integer. If value is halfway between two whole numbers, the even number is returned; that is, 4.5 is converted to 4, and 5.5 is converted to 6.

Use Math.round(), possibly in conjunction with MidpointRounding.AwayFromZero

eg:

Math.Round(1.2) ==> 1
Math.Round(1.5) ==> 2
Math.Round(2.5) ==> 2
Math.Round(2.5, MidpointRounding.AwayFromZero) ==> 3

You can also use function:

//Works with negative numbers now
static int MyRound(double d) {
  if (d < 0) {
    return (int)(d - 0.5);
  }
  return (int)(d + 0.5);
}

Depending on the architecture it is several times faster.

double d;
int rounded = (int)Math.Round(d);

I know this question is old, but I came across it in my search for the answer to my similar question. I thought I would share the very useful tip that I have been given.

When converting to int, simply add .5 to your value before downcasting. As downcasting to int always drops to the lower number (e.g. (int)1.7 == 1), if your number is .5 or higher, adding .5 will bring it up into the next number and your downcast to int should return the correct value. (e.g. (int)(1.8 + .5) == 2)