Truncate Decimal number not Round Off

Question

I want to truncate the decimals like below

i.e.

• 2.22939393 -> 2.229
• 2.22977777 -> 2.229

Answer 1

You can use Math.Round:

decimal rounded = Math.Round(2.22939393, 3); //Returns 2.229

Or you can use ToString with the N3 numeric format.

string roundedNumber = number.ToString("N3");

EDIT: Since you don't want rounding, you can easily use Math.Truncate:

Math.Truncate(2.22977777 * 1000) / 1000; //Returns 2.229

Answer 2

double d = 2.22977777;
d = ( (double) ( (int) (d * 1000.0) ) ) / 1000.0 ;

Of course, this won't work if you're trying to truncate rounding error, but it should work fine with the values you give in your examples. See the first two answers to this question for details on why it won't work sometimes.

Answer 3

You can also use Math.Truncate.

decimal decimalNumber;

decimalNumber = 32.7865m;
// Displays 32
Console.WriteLine(Math.Truncate(decimalNumber));

decimalNumber = -32.9012m;
// Displays -32         
Console.WriteLine(Math.Truncate(decimalNumber));

EDIT: Never mind. Didn't notice you're truncating to a given precision.

Answer 4

What format are you wanting the output?

If you're happy with a string then consider the following C# code:

double num = 3.12345;
num.ToString("G3");

The result will be "3.12".

This link might be of use if you're using .NET. http://msdn.microsoft.com/en-us/library/dwhawy9k.aspx

I hope that helps....but unless you identify than language you are using and the format in which you want the output it is difficult to suggest an appropriate solution.

Answer 5

Try this:

decimal original = GetSomeDecimal(); // 22222.22939393
int number1 = (int)original; // contains only integer value of origina number
decimal temporary = original - number1; // contains only decimal value of original number
int decimalPlaces = GetDecimalPlaces(); // 3
temporary *= (Math.Pow(10, decimalPlaces)); // moves some decimal places to integer
temporary = (int)temporary; // removes all decimal places
temporary /= (Math.Pow(10, decimalPlaces)); // moves integer back to decimal places
decimal result = original + temporary; // add integer and decimal places together

It can be writen shorter, but this is more descriptive.

EDIT: Short way:

decimal original = GetSomeDecimal(); // 22222.22939393
int decimalPlaces = GetDecimalPlaces(); // 3
decimal result = ((int)original) + (((int)(original * Math.Pow(10, decimalPlaces)) / (Math.Pow(10, decimalPlaces));

Answer 6

Maybe another quick solution could be:

>>> float("%.1f" % 1.00001)
1.0
>>> float("%.3f" % 1.23001)
1.23
>>> float("%.5f" % 1.23001)
1.23001

Answer 7

Forget Everything just check out this

double num = 2.22939393;
num  = Convert.ToDouble(num.ToString("#0.000"));

Answer 8

A function to truncate an arbitrary number of decimals:

public decimal Truncate(decimal number, int digits)
{
  decimal stepper = (decimal)(Math.Pow(10.0, (double)digits));
  int temp = (int)(stepper * number);
  return (decimal)temp / stepper;
}