Get number of digits before decimal point

Question

I have a variable of decimal type and I want to check the number of digits before decimal point in it. What should I do? For example, 467.45 should return 3.

Answer 1

Solution without converting to string (which can be dangerous in case of exotic cultures):

static int GetNumberOfDigits(decimal d)
{
    decimal abs = Math.Abs(d);

    return abs < 1 ? 0 : (int)(Math.Log10(decimal.ToDouble(abs)) + 1);
}

Note, that this solution is valid for all decimal values

UPDATE

In fact this solution does not work with some big values, for example: 999999999999998, 999999999999999, 9999999999999939...

Obviously, the mathematical operations with double are not accurate enough for this task.

While searching wrong values I tend to use string-based alternatives proposed in this topic. As for me, that is the evidence that they are more reliable and easy-to-use (but be aware of cultures). Loop-based solutions can be faster though.

Thanks to commentators, shame on me, lesson to you.

Answer 2

Instead of converting to string, you can also divide the number by 10 until it equals 0. Interesting is, that the mathematical operations on decimals are much slower than converting the decimal to a string and returning the length (see benchmarks below).
This solution does not use the Math-methods that take a double as input; so all operations are done on decimals and no casting is involved.

using System;

public class Test
{
    public static void Main()
    {
        decimal dec = -12345678912345678912345678912.456m;
        int digits = GetDigits(dec);
        Console.WriteLine(digits.ToString());
    }

    static int GetDigits(decimal dec)
    {
        decimal d = decimal.Floor(dec < 0 ? decimal.Negate(dec) : dec);
        // As stated in the comments of the question, 
        // 0.xyz should return 0, therefore a special case
        if (d == 0m)
            return 0;
        int cnt = 1;
        while ((d = decimal.Floor(d / 10m)) != 0m)
            cnt++;
        return cnt;
    }
}

Output is 29. To run this sample, visit this link.

---

Side note: some benchmarks show surprising results (10k runs):

• while ((d = decimal.Floor(d / 10m)) != 0m): 25ms
• while ((d = d / 10m) > 1m): 32ms
• ToString with Math-double-operations: 3ms
• ToString with decimal-operations: 3ms
• BigInt (see answer of @Heinzi): 2ms

Also using random numbers instead of always the same value (to avoid possible caching of the decimal to string conversion) showed that the string-based methods are much faster.

Answer 3

I would try this:

Math.Truncate(467.45).ToString().Length

If you want to be sure not having some weird results for different cultures and with negative decimals, you better use this:

var myDecimal = 467.45m;
Math.Truncate(Math.Abs(myDecimal)).ToString(CultureInfo.InvariantCulture).Length

Answer 4

I would prefer the following instead of casting to int to ensure that you can also handle big numbers (e.g. decimal.MaxValue):

Math.Truncate ( Math.Abs ( decValue ) ).ToString( "####" ).Length

Answer 5

decimal d = 467.45M;
int i = (int)d;
Console.WriteLine(i.ToString(CultureInfo.InvariantCulture).Length); //3

As a method;

public static int GetDigitsLength(decimal d)
{
  int i = int(d);
  return i.ToString(CultureInfo.InvariantCulture).Length;
}

Note: Of course you should check first your decimals full part is bigger than Int32.MaxValue or not. Because if it is, you get an OverflowException.

Is such a case, using long instead of int can a better approach. However even a long (System.Int64) is not big enough to hold every possible decimal value.

As Rawling mentioned, your full part can hold the thousands separator and my code will be broken in such a case. Because in this way, it totally ignores my number contains NumberFormatInfo.NumberGroupSeparator or not.

That's why getting numbers only is a better approach. Like;

i.ToString().Where(c => Char.IsDigit(c)).ToArray()

Answer 6

Here's a recursive example (mostly for fun).

void Main()
{
    digitCount(0.123M); //0
    digitCount(493854289.543354345M); //10
    digitCount(4937854345454545435549.543354345M); //22
    digitCount(-4937854345454545435549.543354345M); //22
    digitCount(1.0M); //1
    //approximately the biggest number you can pass to the function that works.
    digitCount(Decimal.MaxValue + 0.4999999M); //29
}

int digitCount(decimal num, int count = 0)
{
    //divided down to last digit, return how many times that happened
    if(Math.Abs(num) < 1)
        return count;
    return digitCount(num/10, ++count); //increment the count and divide by 10 to 'remove' a digit
}

Answer 7

Math.Floor(Math.Log10((double)n) + 1); is the way to go.

Converting to int is BAD because decimal may be bigger than int:

Decimal.MaxValue = 79,228,162,514,264,337,593,543,950,335;
Int32.MaxValue = 2,147,483,647; //that is, hexadecimal 0x7FFFFFFF;

Math.Floor(n).ToString().Count(); is bad because it may include thousands seperators.

Answer 8

If you have a bias towards smaller numbers, you can use something more simple like this.

It is split up into two methods, so the first method is smaller and can be inlined.

Performance is about the same as the solution with the Log10, but without the rounding errors. The Method using Log10, is still the fastest (a bit) specially for numbers > 1 million.

    public static int CountNrOfDigitsIfs(decimal d) {

        var absD = Math.Abs(d);
        // 1
        if (absD < 10M) return 1;
        // 2
        if (absD < 100M) return 2;
        // 3
        if (absD < 1000M) return 3;
        // 4
        if (absD < 10000M) return 4;

        return CountNrOfDigitsIfsLarge(d);
    }

    private static int CountNrOfDigitsIfsLarge(decimal d) {

        // 5
        if (d < 100000M) return 5;
        // 6
        if (d < 1000000M) return 6;
        // 7
        if (d < 10000000M) return 7;
        // 8
        if (d < 100000000M) return 8;
        // 9
        if (d < 1000000000M) return 9;
        // 10
        if (d < 10000000000M) return 10;
        // 11
        if (d < 100000000000M) return 11;
        // 12
        if (d < 1000000000000M) return 12;
        // 13
        if (d < 10000000000000M) return 13;
        // 14
        if (d < 100000000000000M) return 14;
        // 15
        if (d < 1000000000000000M) return 15;
        // 16
        if (d < 10000000000000000M) return 16;
        // 17
        if (d < 100000000000000000M) return 17;
        // 18
        if (d < 1000000000000000000M) return 18;
        // 19
        if (d < 10000000000000000000M) return 19;
        // 20
        if (d < 100000000000000000000M) return 20;
        // 21
        if (d < 1000000000000000000000M) return 21;
        // 22
        if (d < 10000000000000000000000M) return 22;
        // 23
        if (d < 100000000000000000000000M) return 23;
        // 24
        if (d < 1000000000000000000000000M) return 24;
        // 25
        if (d < 10000000000000000000000000M) return 25;
        // 26
        if (d < 100000000000000000000000000M) return 26;
        // 27
        if (d < 1000000000000000000000000000M) return 27;
        // 28
        if (d < 10000000000000000000000000000M) return 28;

        return 29; // Max nr of digits in decimal
    }

This code is generated using the following T4 template:

<#
   const int SIGNIFICANT_DECIMALS = 29;
   const int SPLIT = 5;
#>

namespace Study.NrOfDigits {
    static partial class DigitCounter {

        public static int CountNrOfDigitsIfs(decimal d) {

            var absD = Math.Abs(d);
<#          
            for (int i = 1; i < SPLIT; i++) { // Only 29 significant digits
               var zeroes = new String('0', i);
#>
            // <#= i #>
            if (absD < 1<#= zeroes #>M) return <#= i #>;
<# 
            }
#>

            return CountNrOfDigitsIfsLarge(d);
        }

        private static int CountNrOfDigitsIfsLarge(decimal d) {

<#          
            for (int i = SPLIT; i < SIGNIFICANT_DECIMALS; i++) { // Only 29 significant digits
               var zeroes = new String('0', i);
#>
            // <#= i #>
            if (d < 1<#= zeroes #>M) return <#= i #>;
<# 
            }
#>

            return <#= SIGNIFICANT_DECIMALS #>; // Max nr of digits in decimal
        }

    }
}

Answer 9

So, I've run into this before, and solved it with this code:

SqlDecimal d = new SqlDecimal(467.45M);
int digits = d.Precision - d.Scale;

SqlDecimal is part of the System.Data.SqlTypes namespace. "Precision" is the total number of significant digits, while "Scale" is the number of digits after the decimal point.

Now, I know one objection to going this route is that SqlDecimal is part of the SQL Server-specific code. It's a valid point, but I would also point out that it's a part of the .NET framework itself, and has been since at least version 1.1, so it seems like it would be still be applicable no matter what the code around it is doing.

I looked under the hood with a decompiler (JetBrains' dotPeek in this instance), to see if maybe the code for calculating precision and scale could be easily extracted and just used, without pulling in SqlDecimal. The code to calculate scale is very simple, but the method to calculate precision is non-trivial, so if it were me, I'd just go through SqlDecimal.

Answer 10

var sep = Convert.ToChar(CultureInfo.CurrentCulture.NumberFormat.NumberDecimalSeparator);
var count = d.ToString().TakeWhile(c => c != sep).Count();

Answer 11

This will do if you really don't want to use the Log method (which IMO is the best way). It's about the clearest way I can think of of doing this using ToString():

Math.Abs(val).ToString("f0", CultureInfo.InvariantCulture).Length

Or alternatively, if you don't want to count 0.123M as having one digit:

Math.Abs(val).ToString("#", CultureInfo.InvariantCulture).Length

Answer 12

You could use ToString function with a custom format.

Decimal value = 467.45m;
int count = Math.Abs(value).ToString("#", System.Globalization.CultureInfo.InvariantCulture).Length;

The # specify you only want the characters before the .

The System.Globalization.CultureInfo.InvariantCulture ensure you won't get any formating from the Region Option.

Answer 13

This answer is pretty much lifted from Calculate System.Decimal Precision and Scale but with a minor change to fit the question asked.

class Program
{
    static void Main()
    {
        decimal dec = 467.45m;
        Console.WriteLine(dec.GetNumberOfDigitsBeforeDecimalPlace());
    }
}

public static class DecimalEx
{
    public static int GetNumberOfDigitsBeforeDecimalPlace(this decimal dec)
    {
        var x = new System.Data.SqlTypes.SqlDecimal(dec);
        return x.Precision - x.Scale;
    }
}

Also if you want to do it without using the SqlDecimal class check out Jon Skeet's answer for the same question.

Answer 14

TLDR all the other answers. I wrote this in PHP, and the math would be the same. (If I knew C# I'd have written in that language.)

$input=21689584.999;

    $input=abs($input);
$exp=0;
do{
  $test=pow(10,$exp);

  if($test > $input){
    $digits=$exp;
  }
  if($test == $input){
    $digits=$exp+1;
  }
  $exp++;
}while(!$digits);
if($input < 1){$digits=0;}
echo $digits;

I don't doubt there's a better way, but I wanted to throw in my $.02

EDIT:

I php-ized the code I mentioned in my comments, but did away with the int conversion.

function digitCount($input){
  $digits=0;
  $input=abs($input);
    while ($input >= 1) {
      $digits++;
      $input=$input/10;
      //echo $input."<br>";
    }
  return $digits;   
}
$big=(float)(PHP_INT_MAX * 1.1);
echo digitCount($big);

Answer 15

The mathematical way of doing this (and probably the fastest) is to get a logarytm of base 10 of a absolute value of this number and round it up.

Math.Floor(Math.Log10(Math.Abs(val)) + 1);

Answer 16

Use modulo, i'm not a C# programmer, but I'm pretty sure this solution work:

double i = 1;
int numberOfDecimals = 0;


while (varDouble % i != varDouble)
{
numberOfDecimals++;
i*=10;
}

return numberOfDecimals;

Answer 17

This would be the Java solution

public class test {
    public static void main(String args[]) {
        float f = 1.123f;
        int a = (int) f;
        int digits = 0;
        while (a > 0) {
            digits++;
            a=a/10;
        }
        System.out.println("No Of digits before decimal="+digits);
    }
}

Answer 18

If you treat zeros or lack of zeroes as 1 number, this is OK. If you want zero to return zero or lack of zero to return zero, then there are a few edge cases to work out which shouldn't be too hard to add. Also, should Absolute value to handle negative numbers. Added that test case as well.

        const decimal d = 123.45m; 
        const decimal d1 = 0.123m;
        const decimal d2 = .567m;
        const decimal d3 = .333m;
        const decimal d4 = -123.45m;

        NumberFormatInfo currentProvider = NumberFormatInfo.InvariantInfo;
        var newProvider = (NumberFormatInfo) currentProvider.Clone();
        newProvider.NumberDecimalDigits = 0;
        string number = d.ToString("N", newProvider);  //returns 123 =  .Length = 3
        string number1 = d1.ToString("N", newProvider); //returns 0 = .Length = 1
        string number2 = d2.ToString("N", newProvider); //returns 1 =  .Length = 1
        string number3 = d3.ToString("N", newProvider); //returns 0 =  .Length = 1
        string number4 = Math.Abs(d4).ToString("N", newProvider); //returns 123 =  .Length = 3

Here's a somewhat final solution, if you find a test case that doesn't work, let me know. It should return 3,0,0,0,3 for the test cases provided.

        public static int NumbersInFrontOfDecimal(decimal input)
        {
            NumberFormatInfo currentProvider = NumberFormatInfo.InvariantInfo;
            var newProvider = (NumberFormatInfo)currentProvider.Clone();
            newProvider.NumberDecimalDigits = 0;

            var absInput = Math.Abs(input);
            var numbers =  absInput.ToString("N", newProvider);

            //Handle Zero and < 1
            if (numbers.Length == 1 && input < 1.0m)
            {
                return 0;
            }

            return numbers.Length;
        }

Answer 19

Here is my optimized version of the code inspired by Gray's answer:

    static int GetNumOfDigits(decimal dTest)
    {
        int nAnswer = 0;

        dTest = Math.Abs(dTest);

        //For loop version
        for (nAnswer = 0; nAnswer < 29 && dTest > 1; ++nAnswer)
        {
            dTest *= 0.1M;
        }

        //While loop version
        //while (dTest > 1)
        //{
        //    nAnswer++;
        //    dTest *= 0.1M;
        //}

        return (nAnswer);
    }

If you don't want the Math.Abs to be called inside this function then be sure to use it outside the function on the parameter before calling GetNumOfDigits.

I decided to remove the other codes to reduce clutter in my answer, even though they helped me get to this point...

If there is any improvements needed, then let me know and I will update it :).

Answer 20

In order to get an accurate and culturally agnostic answer I do the following:

1. Use System.Numerics.BigInteger, whose constructor accepts a decimal and doesn't seem to produce any rounding errors.
2. Use BigInteger.Abs() to remove any sign.
3. Use BigInteger.ToString() with the "#" format to suppress any separators that might occur.

Code

decimal num = 123213123.123123M;
int length = BigInteger.Abs((BigInteger)num).ToString("#").Length;

Answer 21

You could do this by rounding the number, then getting the length of the new number. You could do it like this:

var number = 476.43;
var newNumber = Math.round(number);

//get the length of the rounded number, and subtract 1 if the
//number is negative (remove the negative sign from the count)
int digits = newNumber.ToString().Length - (number < 0 ? 1 : 0);

Answer 22

The other solutions will lose digits if the number is too large.

public int Digits(Decimal i)
{
    NumberFormatInfo format = CultureInfo.CurrentCulture.NumberFormat;
    var str = Math.Abs(i).ToString().Replace(format.NumberGroupSeparator, "");
    var index = str.IndexOf(format.NumberDecimalSeparator);
    var digits = index == -1 ? str.Length : index;
}

Answer 23

Algorithm:

• Convert |decimal| to String.
• If "." exist in the decimal, cut before it, otherwise consider the whole number.
• Return string length.

Example:

3.14 --> 3.14 --> "3.14" --> "3.14".Substring(0,1) --> "3".Length --> 1

-1024 --> 1024 --> "1024" --> IndexOf(".") = -1 --> "1024" --> 4

Code:

static int getNumOfDigits (decimal num)
{
    string d = Math.Abs(num).ToString();

    if (d.IndexOf(".") > -1)
    {
        d = d.Substring(0, d.IndexOf("."));
    }

    return d.Length;
}

Answer 24

I haven't tested this but I would keep it straightforward and do:

decimal value = 467.45;
string str = Convert.toString(value); // convert your decimal type to a string
string[] splitStr = str.split('.'); // split it into an array (use comma separator assuming you know your cultural context)
Console.WriteLine(splitStr[0].Count); // get the first element. You can also get the number of figures after the point by indexing the next value in the array.

This does not handle negative numbers. If you care about those then considering taking the absolute value. Furthermore, if you want 0 before the decimal place to not be counted then you can use a simple if statement to check it.

Answer 25

simple :

string value = "467.45";
int count =  value.split('.')[0] == "0" ? 0 : value.split('.')[0].ToString().Replace("-","").Count();