Truncate Decimal number not Round Off [duplicate]

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 

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.

readonly 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 ? pow10[precision] : Math.Pow(10, precision); return Math.Truncate(x * m) / m; } 

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; } 

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 

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.

Maybe another quick solution could be:

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

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); 

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));