Truncating a number to specified decimal places

Question

I need to truncate a number to 2 decimal places, which basically means chopping off the extra digits.

Eg:

2.919     ->      2.91

2.91111   ->     2.91

Why? This is what SQL server is doing when storing a number of a particular precision. Eg, if a column is Decimal(8,2), and you try to insert/update a number of 9.1234, the 3 and 4 will be chopped off.

I need to do exactly the same thing in c# code.

The only possible ways that I can think of doing it are either:

1. Using the stringformatter to "print" it out only two decimal places, and then converting it to a decimal, eg:

     decimal tooManyDigits = 2.1345
   
   decimal ShorterDigits = Convert.ToDecimal(tooManyDigits.ToString("0.##"));
   
   // ShorterDigits is now 2.13
   

   I'm not happy with this because it involves a to-string and then another string to decimal conversion which seems a bit mad.

2. Using Math.Truncate (which only accepts an integer), so I can multiply it by 100, truncate it, then divide by 100. eg:

   decimal tooLongDecimal = 2.1235;
   
   tooLongDecimal = Math.Truncate(tooLongDecimal * 100) / 100;
   

   I'm also not happy with this because if tooLongDecimal is 0, I'll get a divide by 0 error.

Surely there's a better + easier way! Any suggestions?

Answer 1

The previously offered mathematical solutions are vulnerable to overflow with large numbers and/or a large number of decimal places. Consider instead the following extension method:

public static decimal TruncateDecimal(this decimal d, int decimals)
{
    if (decimals < 0)
        throw new ArgumentOutOfRangeException("decimals", "Value must be in range 0-28."); 
    else if (decimals > 28)
        throw new ArgumentOutOfRangeException("decimals", "Value must be in range 0-28.");
    else if (decimals == 0)
        return Math.Truncate(d);
    else
    {
        decimal integerPart = Math.Truncate(d);
        decimal scalingFactor = d - integerPart;
        decimal multiplier = (decimal) Math.Pow(10, decimals);

        scalingFactor = Math.Truncate(scalingFactor * multiplier) / multiplier;

        return integerPart + scalingFactor;
    }
}

Usage:

decimal value = 18446744073709551615.262626263m;
value = value.TruncateDecimal(6); // Result: 18446744073709551615.262626

Answer 2

I agree with p.s.w.g. I had the similar requirement and here is my experience and a more generalized function for truncating.

http://snathani.blogspot.com/2014/05/truncating-number-to-specificnumber-of.html

public static decimal Truncate(decimal value, int decimals)
{
    decimal factor = (decimal)Math.Pow(10, decimals);
    decimal result = Math.Truncate(factor * value) / factor;
    return result;
}