Generating a decimal with trailing zeros

Question

I need to round a decimal to a specific number of significant figures. I have a library that can be simplified as shown.

decimal a = Math.Round(20.051M * 100M) / 100M; // 20.05
decimal b = Math.Round(20.001M * 100M) / 100M; // 20;

Clearly I lost the trailing zeros when performing the Round, I can get them back by adding .00M Is there a way to do this dynamically, if I were rounding to five significant figures I would want to use .000M, likewise if my decimal was < 10 .000M would be appropriate.

b = b + .00M; // 20.00

For reference here is the actual rounding method.

DecimalExtensions.cs

private static readonly decimal[] Pows = Enumerable.Range(-28, 57).Select(p => (decimal)Math.Pow(10, p)).ToArray();

/// <remarks>https://stackoverflow.com/a/18146056/3194005</remarks>
public static decimal RoundToSignificantFigures(this decimal value, int significantFigures)
{
    if (value == 0)
        return 0;

    int d = Log10Ceiling(Math.Abs(value));
    int power = significantFigures - d;

    decimal magnitude = (decimal)Math.Pow(10, power);

    return Math.Round(value * magnitude) / magnitude;
}

private static int Log10Ceiling(decimal value)
{
    int log10 = Array.BinarySearch(Pows, value);

    return (log10 >= 0 ? log10 : ~log10) - 28;
}

It isn't for display, it's needed to enforce significant figure rules. For example downstream `a` and `b` could get averaged. The result can only be as precise as the least precise input. In this case that's a difference between 20 and 20.02 — Derrick Moeller, Dec 15 '17 at 18:43

Sergey Kalinichenko · Accepted Answer · 2017-12-15T18:55:41.040

2

You can make a value with the required number of zeros after the decimal point by computing the value of mag^-1, and subtracting it from itself:

var res = Math.Round(value * magnitude) / magnitude;
var invMag = 1/magnitude;
return res + invMag - invMag;

In your example mag=10², so invMag=10^-2, or 0.01. When you subtract 0.01M - 0.01M you end up with 0.00M, which you can add to result to force it to have the desired number of trailing zeros without changing the value.

Demo

edited Dec 15 '17 at 18:55

answered Dec 15 '17 at 18:43

Sergey Kalinichenko

714,442
84
1,110
1,523

This appears to be working, as soon as I've consumed what it's doing I'll accept. Thanks. – Derrick Moeller Dec 15 '17 at 18:47

score 2 · Answer 2 · answered Dec 15 '17 at 19:10

This is not an answer, is too long for a comment, and too specific for a new question, so...

I asked a similar question here, and discovered an anomaly, whereby the dotnetfiddle compiler returns 20 for your example:

decimal b = Math.Round(20.001M * 100M) / 100M; 
b = b + .00M;

Whereas it (correctly) returns 20.00 for:

decimal b = Math.Round(20.001M * 100M) / 100M; 
b = Decimal.Add(b, .00M);

Both return 20.00 with the Visual Studio compilers I have tried (VS2013, VS2015).

I don't know if this is a compiler bug, but in the absence of an explanation I'd prefer the second version.

It seems to be a bug in C# 5.0 compiler. It optimize `b + .00M` as `b`. — user4003407, Dec 15 '17 at 19:35

score -1 · Answer 3 · answered Dec 15 '17 at 18:40

-1

The value of 20M is the same as 20.00M, if it's formatting for the user just do

20M.ToString("0.00");

Checkout NumberFormatInfo - https://msdn.microsoft.com/en-us/library/system.globalization.numberformatinfo(v=vs.110).aspx

answered Dec 15 '17 at 18:40

Kevin Smith

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4
52
77

1

`20M` and `20.00M` are equal, but not the same: `decimal.GetBits(20M)` return `int[4] { 20, 0, 0, 0 }`, while `decimal.GetBits(20.00M)` return `int[4] { 2000, 0, 0, 131072 }`. – user4003407 Dec 15 '17 at 18:49
1

What if you want to preserve decimals? E.g. if you want to negate a numerical string value, preserving decimals. An easy way to do this is `(-Decimal.Parse(stringValue)).ToString()` which gives the right result without the caller needing to know how many decimals are in the input. – Joe Dec 15 '17 at 20:24

Generating a decimal with trailing zeros

3 Answers3