Math.Round for decimal in C #

Question

Possible duplicate:
Why does .NET use banker rounding by default?

Here is a sample code

decimal num1=390, num2=60, result;
result=num1/num2; // here I get 6.5
result=Math.Round(result,0);

the final value of the result should be 7 , but I get 6 . Why is this behavior?

Answer 1

Check out the third MidpointRounding parameter .

The default is MidpointRounding.ToEven, so

Math.Round(result,0); // 6.0 
//or
Math.Round(result,0, MidpointRounding.ToEven); // 6.0 

//But:
Math.Round(result,0, MidpointRounding.AwayFromZero); // 7.0 

Answer 2

This rounding is sometimes called rounding to the nearest or rounding of the banker. It minimizes rounding errors due to constant rounding of the average value in one direction.

http://msdn.microsoft.com/en-us/library/3s2d3xkk.aspx

:

//       11.1 --> 11
//       11.2 --> 11
//       11.3 --> 11
//       11.4 --> 11
//       11.5 --> 11
//       11.6 --> 12

Answer 3

MSDN:

d , , - , .

Answer 4

decimal.Math.Round MidpointRounding.ToEven .

, , . . IEEE 754, 4. . , - .

Answer 5

,

      Math.Round(6.5, 0);

6, 7. MSDN ,

  Console.WriteLine(Math.Round(3.45, 1)); //Returns 3.4.
  Console.WriteLine(Math.Round(4.35, 1)); // Returns 4.4

to another MDSN document - these are states

The nearest integer parameter d. If the fractional component d is halfway between two integers, one of which is even and the other is odd, an even number is returned. Note that this method returns a decimal number instead of an integer type.

Answer 6

Use the Math.Ceiling () method. decimal num1 = 390, num2 = 60, result;
Result = Math.Ceiling (num1 / pit2);