Truncating a number to specified decimal places
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
Using decimal.ToString('0.##')
also imposes rounding:
1.119M.ToString("0.##") // -> 1.12
(Yeah, likely should be a comment, but it's hard to format well as such.)
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;
}
You've answered the question yourself; it seems you just misunderstood what division by zero means. The correct way to do this is to multiply, truncate, then devide, like this:
decimal TruncateTo100ths(decimal d)
{
return Math.Truncate(d* 100) / 100;
}
TruncateTo100ths(0m); // 0
TruncateTo100ths(2.919m); // 2.91
TruncateTo100ths(2.91111m); // 2.91
TruncateTo100ths(2.1345m); // 2.13
There is no division by zero here, there is only division by 100, which is perfectly safe.