Truncate Decimal number not Round Off

Question

Possible Duplicate:
c# - How do I round a decimal value to 2 decimal places (for output on a page)

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

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;
}

Answer 4

Here's an extension method which does not suffer from integer overflow (like some of the above answers do). It also caches some powers of 10 for efficiency.

static double[] pow10 = { 1e0, 1e1, 1e2, 1e3, 1e4, 1e5, 1e6, 1e7, 1e8, 1e9, 1e10 };
public static double Truncate(this double x, int precision)
{
    if (precision < 0)
        throw new ArgumentException();
    if (precision == 0)
        return Math.Truncate(x);
    double m = precision >= pow10.Length ? Math.Pow(10, precision) : pow10[precision];
    return Math.Truncate(x * m) / m;
}

Answer 5

This is similar to TcKs suggestion above, but using math.truncate rather than int conversions

VB: but you'll get the idea

Private Function TruncateToDecimalPlace(byval ToTruncate as decimal, byval DecimalPlaces as integer) as double 
dim power as decimal = Math.Pow(10, decimalplaces)
return math.truncate(totruncate * power) / power
end function

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

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 8

Try this

double d = 2.22912312515;
int demention = 3;
double truncate = Math.Truncate(d) + Math.Truncate((d - Math.Truncate(d)) * Math.Pow(10.0, demention)) / Math.Pow(10.0, demention);

Answer 9

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 10

Forget Everything just check out this

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