C# – Limiting double to 3 decimal places

cdoubletruncate

This i what I am trying to achieve:

If a double has more than 3 decimal places, I want to truncate any decimal places beyond the third. (do not round.)

Eg.: 12.878999 -> 12.878

If a double has less than 3 decimals, leave unchanged

Eg.:   125   -> 125
       89.24 -> 89.24

I came across this command:

double example = 12.34567;
double output = Math.Round(example, 3);

But I do not want to round. According to the command posted above,
12.34567 -> 12.346

I want to truncate the value so that it becomes: 12.345

Best Answer

Doubles don't have decimal places - they're not based on decimal digits to start with. You could get "the closest double to the current value when truncated to three decimal digits", but it still wouldn't be exactly the same. You'd be better off using decimal.

Having said that, if it's only the way that rounding happens that's a problem, you can use Math.Truncate(value * 1000) / 1000; which may do what you want. (You don't want rounding at all, by the sounds of it.) It's still potentially "dodgy" though, as the result still won't really just have three decimal places. If you did the same thing with a decimal value, however, it would work:

decimal m = 12.878999m;
m = Math.Truncate(m * 1000m) / 1000m;
Console.WriteLine(m); // 12.878

EDIT: As LBushkin pointed out, you should be clear between truncating for display purposes (which can usually be done in a format specifier) and truncating for further calculations (in which case the above should work).

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