Round decimal number to smaller number

Question

With Decimal.Round, I can only choose between ToEven and AwayFromZero, now I'd like to round it always to the smaller number, that is, like truncating, remove the digits that is out of the required decimals:

public static void Main()
{
    Console.WriteLine("{0,-10} {1,-10} {2,-10}", "Value", "ToEven", "AwayFromZero");
    for (decimal value = 12.123451m; value <= 12.123459m; value += 0.000001m)
        Console.WriteLine("{0} -- {1} -- {2}", value, Math.Round(value, 5, MidpointRounding.ToEven),
                       Math.Round(value, 5, MidpointRounding.AwayFromZero));
}

// output
12.123451 -- 12.12345 -- 12.12345
12.123452 -- 12.12345 -- 12.12345
12.123453 -- 12.12345 -- 12.12345
12.123454 -- 12.12345 -- 12.12345
12.123455 -- 12.12346 -- 12.12346
12.123456 -- 12.12346 -- 12.12346
12.123457 -- 12.12346 -- 12.12346
12.123458 -- 12.12346 -- 12.12346
12.123459 -- 12.12346 -- 12.12346

I just want all those to be rounded to 12.12345, that is keep 5 decimals, and truncating remaining ones. Is there a better way to do this?

Possible duplicate of [Truncate Two decimal places without rounding](https://stackoverflow.com/questions/3143657/truncate-two-decimal-places-without-rounding) — Mighty Badaboom, Aug 07 '17 at 13:20

Fruchtzwerg · Accepted Answer · 2017-08-07T13:34:32.397

5

decimal.Truncate(value * (decimal)Math.Pow(10, 5)) / (decimal)Math.Pow(10, 5);

or simply

decimal.Truncate(value * 100000) / 100000;

should solve your problem by shifting the value 5 digits left, truncating and shifting the 5 digits back.

Example in 4 steps:

1.23456 * 100000
12345.6 decimal.Truncate
12345 / 100000
1.2345

Not as much simple as the first approach but at least twice as fast at my device is using a string and split it. Here is my implementation:

string[] splitted = value.ToString(CultureInfo.InvariantCulture).Split('.');
string newDecimal = splitted[0];
if (splitted.Length > 1)
{
    newDecimal += ".";
    newDecimal += splitted[1].Substring(0, Math.Min(splitted[1].Length, 5));
}
decimal result = Convert.ToDecimal(newDecimal, CultureInfo.InvariantCulture);

edited Aug 07 '17 at 13:34

answered Aug 07 '17 at 13:01

Fruchtzwerg

10,999
12
40
49

Thanks, this will produce the right result I want. But this involves multiple arithmetic operations, I'm doing this for a fast data stream, so is there anyway to optimize this further? – fluter Aug 07 '17 at 13:12
I've added a much faster implementation using strings to my answer @fluter – Fruchtzwerg Aug 07 '17 at 13:22
I will try this, Thanks! – fluter Aug 07 '17 at 13:25

score 1 · Answer 2 · 2017-08-07T16:38:27.697

1

You can use Math.Floor, if you modify the decimal places before you use it, and then return back to the original position like this:

public static decimal RoundDown(decimal input, int decimalPlaces)
{
    decimal power = (decimal) Math.Pow(10, decimalPlaces);
    return Math.Floor(input * power) / power;
}

Try it online!

edited Aug 07 '17 at 16:38

answered Aug 07 '17 at 13:24

score 0 · Answer 3 · edited Jun 20 '20 at 09:12

0

Are you perhaps instead looking for Math.Floor?

Floor(Decimal) Returns the largest integer less than or equal to the specified decimal number.

Floor(Double) Returns the largest integer less than or equal to the specified double-precision floating-point number.

edited Jun 20 '20 at 09:12

Community

1
1

answered Aug 07 '17 at 12:59

TomServo

7,248
5
30
47

No, I'm not looking for Floor, I'm not for the nearest integer, rather nearest float point number with specified precision. – fluter Aug 07 '17 at 13:02

Round decimal number to smaller number

3 Answers3