Develop Reference

C# String.format a number with dynamic number of significant digits [duplicate]

I have some fields returned by a collection as
2.4200
2.0044
2.0000
I want results like
2.42
2.0044
2
I tried with String.Format, but it returns 2.0000 and setting it to N0 rounds the other values as well.

I ran into the same problem but in a case where I do not have control of the output to string, which was taken care of by a library. After looking into details in the implementation of the Decimal type (see http://msdn.microsoft.com/en-us/library/system.decimal.getbits.aspx),
I came up with a neat trick (here as an extension method):
public static decimal Normalize(this decimal value)
{
return value/1.000000000000000000000000000000000m;
}
The exponent part of the decimal is reduced to just what is needed. Calling ToString() on the output decimal will write the number without any trailing 0. E.g.
1.200m.Normalize().ToString();

Is it not as simple as this, if the input IS a string? You can use one of these:
string.Format("{0:G29}", decimal.Parse("2.0044"))
decimal.Parse("2.0044").ToString("G29")
2.0m.ToString("G29")
This should work for all input.
Update Check out the Standard Numeric Formats I've had to explicitly set the precision specifier to 29 as the docs clearly state:
However, if the number is a Decimal and the precision specifier is omitted, fixed-point notation is always used and trailing zeros are preserved
Update Konrad pointed out in the comments:
Watch out for values like 0.000001. G29 format will present them in the shortest possible way so it will switch to the exponential notation. string.Format("{0:G29}", decimal.Parse("0.00000001",System.Globalization.CultureInfo.GetCultureInfo("en-US"))) will give "1E-08" as the result.

In my opinion its safer to use Custom Numeric Format Strings.
decimal d = 0.00000000000010000000000m;
string custom = d.ToString("0.#########################");
// gives: 0,0000000000001
string general = d.ToString("G29");
// gives: 1E-13

I use this code to avoid "G29" scientific notation:
public static string DecimalToString(this decimal dec)
{
string strdec = dec.ToString(CultureInfo.InvariantCulture);
return strdec.Contains(".") ? strdec.TrimEnd('0').TrimEnd('.') : strdec;
}
EDIT: using system CultureInfo.NumberFormat.NumberDecimalSeparator :
public static string DecimalToString(this decimal dec)
{
string sep = CultureInfo.CurrentCulture.NumberFormat.NumberDecimalSeparator;
string strdec = dec.ToString(CultureInfo.CurrentCulture);
return strdec.Contains(sep) ? strdec.TrimEnd('0').TrimEnd(sep.ToCharArray()) : strdec;
}

Use the hash (#) symbol to only display trailing 0's when necessary. See the tests below.
decimal num1 = 13.1534545765;
decimal num2 = 49.100145;
decimal num3 = 30.000235;
num1.ToString("0.##"); //13.15%
num2.ToString("0.##"); //49.1%
num3.ToString("0.##"); //30%

I found an elegant solution from http://dobrzanski.net/2009/05/14/c-decimaltostring-and-how-to-get-rid-of-trailing-zeros/
Basically
decimal v=2.4200M;
v.ToString("#.######"); // Will return 2.42. The number of # is how many decimal digits you support.

A very low level approach, but I belive this would be the most performant way by only using fast integer calculations (and no slow string parsing and culture sensitive methods):
public static decimal Normalize(this decimal d)
{
int[] bits = decimal.GetBits(d);
int sign = bits[3] & (1 << 31);
int exp = (bits[3] >> 16) & 0x1f;
uint a = (uint)bits[2]; // Top bits
uint b = (uint)bits[1]; // Middle bits
uint c = (uint)bits[0]; // Bottom bits
while (exp > 0 && ((a % 5) * 6 + (b % 5) * 6 + c) % 10 == 0)
{
uint r;
a = DivideBy10((uint)0, a, out r);
b = DivideBy10(r, b, out r);
c = DivideBy10(r, c, out r);
exp--;
}
bits[0] = (int)c;
bits[1] = (int)b;
bits[2] = (int)a;
bits[3] = (exp << 16) | sign;
return new decimal(bits);
}
private static uint DivideBy10(uint highBits, uint lowBits, out uint remainder)
{
ulong total = highBits;
total <<= 32;
total = total | (ulong)lowBits;
remainder = (uint)(total % 10L);
return (uint)(total / 10L);
}

This is simple.
decimal decNumber = Convert.ToDecimal(value);
return decNumber.ToString("0.####");
Tested.
Cheers :)

Depends on what your number represents and how you want to manage the values: is it a currency, do you need rounding or truncation, do you need this rounding only for display?
If for display consider formatting the numbers are x.ToString("")
http://msdn.microsoft.com/en-us/library/dwhawy9k.aspx and
http://msdn.microsoft.com/en-us/library/0c899ak8.aspx
If it is just rounding, use Math.Round overload that requires a MidPointRounding overload
http://msdn.microsoft.com/en-us/library/ms131274.aspx)
If you get your value from a database consider casting instead of conversion:
double value = (decimal)myRecord["columnName"];

This will work:
decimal source = 2.4200m;
string output = ((double)source).ToString();
Or if your initial value is string:
string source = "2.4200";
string output = double.Parse(source).ToString();
Pay attention to this comment.

Trying to do more friendly solution of DecimalToString (https://stackoverflow.com/a/34486763/3852139):
private static decimal Trim(this decimal value)
{
var s = value.ToString(CultureInfo.InvariantCulture);
return s.Contains(CultureInfo.InvariantCulture.NumberFormat.NumberDecimalSeparator)
? Decimal.Parse(s.TrimEnd('0'), CultureInfo.InvariantCulture)
: value;
}
private static decimal? Trim(this decimal? value)
{
return value.HasValue ? (decimal?) value.Value.Trim() : null;
}
private static void Main(string[] args)
{
Console.WriteLine("=>{0}", 1.0000m.Trim());
Console.WriteLine("=>{0}", 1.000000023000m.Trim());
Console.WriteLine("=>{0}", ((decimal?) 1.000000023000m).Trim());
Console.WriteLine("=>{0}", ((decimal?) null).Trim());
}
Output:
=>1
=>1.000000023
=>1.000000023
=>

how about this:
public static string TrimEnd(this decimal d)
{
string str = d.ToString();
if (str.IndexOf(".") > 0)
{
str = System.Text.RegularExpressions.Regex.Replace(str.Trim(), "0+?$", " ");
str = System.Text.RegularExpressions.Regex.Replace(str.Trim(), "[.]$", " ");
}
return str;
}

You can just set as:
decimal decNumber = 23.45600000m;
Console.WriteLine(decNumber.ToString("0.##"));

The following code could be used to not use the string type:
int decimalResult = 789.500
while (decimalResult>0 && decimalResult % 10 == 0)
{
decimalResult = decimalResult / 10;
}
return decimalResult;
Returns 789.5

In case you want to keep decimal number, try following example:
number = Math.Floor(number * 100000000) / 100000000;

Here is an Extention method I wrote, it also removes dot or comma if it`s the last character (after the zeros were removed):
public static string RemoveZeroTail(this decimal num)
{
var result = num.ToString().TrimEnd(new char[] { '0' });
if (result[result.Length - 1].ToString() == "." || result[result.Length - 1].ToString() == ",")
{
return result.Substring(0, result.Length - 1);
}
else
{
return result;
}
}

To remove trailing zero's from a string variable dateTicks, Use
return new String(dateTicks.Take(dateTicks.LastIndexOf(dateTicks.Last(v => v != '0')) + 1).ToArray());

Additional Answer:
In a WPF Application using XAML you could use
{Binding yourDecimal, StringFormat='#,0.00#######################'}
The above answer will preserve the zero in some situations so you could still return 2.00 for example
{Binding yourDecimal, StringFormat='#,0.#########################'}
If you want to remove ALL trailing zeros, adjust accordingly.

The following code will be able to remove the trailing 0's. I know it's the hard way but it works.
private static string RemoveTrailingZeros(string input)
{
for (int i = input.Length - 1; i > 0; i-- )
{
if (!input.Contains(".")) break;
if (input[i].Equals('0'))
{
input= input.Remove(i);
}
else break;
}
return input;
}

string.Format("{0:G29}", decimal.Parse("2.00"))
string.Format("{0:G29}", decimal.Parse(Your_Variable))

try this code:
string value = "100";
value = value.Contains(".") ? value.TrimStart('0').TrimEnd('0').TrimEnd('.') : value.TrimStart('0');

Very simple answer is to use TrimEnd(). Here is the result,
double value = 1.00;
string output = value.ToString().TrimEnd('0');
Output is 1
If my value is 1.01 then my output will be 1.01

try like this
string s = "2.4200";
s = s.TrimStart("0").TrimEnd("0", ".");
and then convert that to float

Converting decimal number to hexadecimal number using recursive method

I am trying to convert decimal to hexadecimal. I have found many codes online. I used
int decValue = int.Parse(hexValue, System.Globalization.NumberStyles.HexNumber);
but my instructor told me I can't use any of those, just use recursive method. I am new to programming and little confused about recursive method.
I did find other methods to convert it, I am using below method, and I used switch statement to change numbers to letters. Program works fine. But not sure if it is recursive method? Can someone let me know if it is recursive method, if not help me understand how recursive method work.
static void HexadecimalConversion(int decimals)
{
if (decimals == 0)
return;
else
{
int hexadecimals = decimals % 16;
decimals = decimals / 16;
HexadecimalConversion(decimals);

With most recursive problems, you have 1 or 2 special cases and a general case. For this problem there are 3 cases:
Special Case #1. The value to be converted is 0.
The General Case. The value to be converted is greater than 0.
The Terminating Case. When the value to be converted is finally decremented to 0.
You need to distinguish between the two 'zero' conditions, lest you always append a trailing zero to the result, so...you need a 2-layered approach, something like this:
static string Int2Hex( int value )
{
if ( value < 0 ) throw new ArgumentOutOfRangeException("value") ;
if ( value == 0 ) return "0" ;
string result = ToHex( (uint) value ).ToString() ;
return result ;
}
static StringBuilder ToHex ( uint value )
{
StringBuilder buffer ;
if ( value <= 0 )
{
buffer = new StringBuilder() ;
}
else
{
buffer = ToHex( value / 16 ).Append( "0123456789ABCDEF"[ (int)(value % 16 ) ] ) ;
}
return buffer ;
}

Yet another implementation:
public string ConvertToHexa(int number)
{
if (number == 0)
return String.Empty;
var head = ConvertToHexa(number / 16);
var remainder = number % 16;
var tail = (char)(remainder + (remainder >= 10 ? 'A' - 10 : '0'));
return head + tail;
}
Console.WriteLine(ConvertToHexa(202)) gives "CA" (which is correct).

Another implementation
public void ConvertToHexa(int number)
{
if (number == 0)
return;
ConvertToHexa(number / 16);
var remainder = number % 16;
Console.Write(remainder >= 10 ? ((char)(remainder - 10 + 'A')).ToString() : remainder.ToString());
}

C# how to check same sign of 2 decimal values using bit?

I have 2 decimal values: a and b. How do I use bit operator to check if two value is same sign?

You can use Math.Sign(). When you use Math.Sign(x), if x is negative it returns -1 else if its positive, the function returns 1 or when its 0 it returns 0. So :
if(Math.Sign(a) == Math.Sign(b))
{
// Code when sign matched.
}
else
{
// Code when sign not matched.
}

Do you mean if both are positive or both are negative?
bool bothSameSign = (d1 >= 0 && d2 >= 0) || (d1 < 0 && d2 < 0);

I don't think you really need to use the bit operator for this, but if for some reason you must (e.g. this is a school question):
Firstly you can use Decimal.GetBits() get all the bits in the two Decimals to compare, as an array of 4 ints.
Then you can inspect the sign bit which is at bit 31 in the int at offset 3 in the array of ints.
Decimal d1 = 1;
Decimal d2 = -1;
var bits1 = Decimal.GetBits(d1);
var bits2 = Decimal.GetBits(d2);
const int signMask = 1 << 31;
const int signWord = 3;
bool sameSign = ((bits1[signWord] & signMask) == (bits2[signWord] & signMask));

You could make,
static int Sign(this decimal value)
{
return Decimal.GetBits(value)[3] & 0x8000;
}
and do
a.Sign == b.Sign;

Bitwise shift is required for the sign-checking you want to accomplish:
if ( ( number >> sizeof(byte) * sizeof(numberType) -1 ) & 1)
{ /* < 0 */ }
else
{ /* >= 0 */ }
// you can of course use magic numbers
// example for int: if ( ( number >> 31 ) & 1) { /* < 0 */ }
Problem is, you can't bitshift a decimal. You would have to do something like this:
var shiftableNumber = Int.Parse(Math.Truncate(yourDecimal));
I can't verify it, but I suspect it would defeat the purpose of optimizing through bitwise operators. You might aswell use the builtin Math.Sign() directly.

Modulo from very large int C#

I'm having a problem with modulo from int which has 31 chars. It seems to bug out on
Int64 convertedNumber = Int64.Parse(mergedNumber); with Value was either too large or too small for an Int64. (Overflow Exception). How to fix it so that modulo doesn't bug out ?
class GeneratorRachunkow {
private static string numerRozliczeniowyBanku = "11111155"; // 8 chars
private static string identyfikatorNumeruRachunku = "7244"; // 4 chars
private static string stalaBanku = "562100"; // 6 chars
public static string generator(string pesel, string varKlientID) {
string peselSubstring = pesel.Substring(pesel.Length - 5); // 5 chars (from the end of the string);
string toAttach = varKlientID + peselSubstring;
string indywidualnyNumerRachunku = string.Format("{0}", toAttach.ToString().PadLeft(13, '0')); // merging pesel with klient id and adding 0 to the begining to match 13 chars
string mergedNumber = numerRozliczeniowyBanku + identyfikatorNumeruRachunku + indywidualnyNumerRachunku + stalaBanku; // merging everything -> 31 chars
Int64 convertedNumber = Int64.Parse(mergedNumber);
Int64 modulo = MathMod(convertedNumber, 97);
Int64 wynik = 98 - modulo;
string wynikString = string.Format("{0}", wynik.ToString().PadLeft(2, '0')); // must be 2 chars
indywidualnyNumerRachunku = wynikString + numerRozliczeniowyBanku + identyfikatorNumeruRachunku + indywidualnyNumerRachunku;
return indywidualnyNumerRachunku;
}
private static Int64 MathMod(Int64 a, Int64 b) {
return (Math.Abs(a * b) + a) % b;
}
}

The max value for Int64 is 9223372036854775807 (19 characters when printed). You will probably want to use BigInteger instead (which was introduced in .NET 4):
public static string generator(string pesel, string varKlientID) {
// I have cut some code here to keep it short
BigInteger convertedNumber;
if (BigInteger.TryParse(mergedNumber , out convertedNumber))
{
BigInteger modulo = convertedNumber % 97;
// The rest of the method goes here...
}
else
{
// string could not be parsed to BigInteger; handle gracefully
}
}
private static BigInteger MathMod(BigInteger a, BigInteger b)
{
return (BigInteger.Abs(a * b) + a) % b;
}

Int64.MaxValue is 9,223,372,036,854,775,807 that's 19 characters. So you just can't fit that in. I suggest looking at this question for working with big numbers.

Try this function instead of "MathMod":
static int ModString(string x, int y)
{
if (x.Length == 0)
return 0;
string x2 = x.Substring(0,x.Length - 1); // first digits
int x3 = int.Parse(x.Substring(x.Length - 1)); // last digit
return (ModString(x2, y) * 10 + x3) % y;
}
(since all of your numbers are positive, there is no point in using Math.Abs, as in your original MathMod function).
Use it this way:
modulo = ModString(mergedNumber,97);
This should works with all versions of .NET since 1.1, without the need of BigInteger.

The answer you are looking for is demonstrated here. It includes various manners to calculate the modulus for huge numbers. I used similar methods as described here for international bank account numbers.
A direct link to someone who has a copy pastable method is here.

Remove trailing zeros

I have some fields returned by a collection as
2.4200
2.0044
2.0000
I want results like
2.42
2.0044
2
I tried with String.Format, but it returns 2.0000 and setting it to N0 rounds the other values as well.

I ran into the same problem but in a case where I do not have control of the output to string, which was taken care of by a library. After looking into details in the implementation of the Decimal type (see http://msdn.microsoft.com/en-us/library/system.decimal.getbits.aspx),
I came up with a neat trick (here as an extension method):
public static decimal Normalize(this decimal value)
{
return value/1.000000000000000000000000000000000m;
}
The exponent part of the decimal is reduced to just what is needed. Calling ToString() on the output decimal will write the number without any trailing 0. E.g.
1.200m.Normalize().ToString();

Is it not as simple as this, if the input IS a string? You can use one of these:
string.Format("{0:G29}", decimal.Parse("2.0044"))
decimal.Parse("2.0044").ToString("G29")
2.0m.ToString("G29")
This should work for all input.
Update Check out the Standard Numeric Formats I've had to explicitly set the precision specifier to 29 as the docs clearly state:
However, if the number is a Decimal and the precision specifier is omitted, fixed-point notation is always used and trailing zeros are preserved
Update Konrad pointed out in the comments:
Watch out for values like 0.000001. G29 format will present them in the shortest possible way so it will switch to the exponential notation. string.Format("{0:G29}", decimal.Parse("0.00000001",System.Globalization.CultureInfo.GetCultureInfo("en-US"))) will give "1E-08" as the result.

In my opinion its safer to use Custom Numeric Format Strings.
decimal d = 0.00000000000010000000000m;
string custom = d.ToString("0.#########################");
// gives: 0,0000000000001
string general = d.ToString("G29");
// gives: 1E-13

I use this code to avoid "G29" scientific notation:
public static string DecimalToString(this decimal dec)
{
string strdec = dec.ToString(CultureInfo.InvariantCulture);
return strdec.Contains(".") ? strdec.TrimEnd('0').TrimEnd('.') : strdec;
}
EDIT: using system CultureInfo.NumberFormat.NumberDecimalSeparator :
public static string DecimalToString(this decimal dec)
{
string sep = CultureInfo.CurrentCulture.NumberFormat.NumberDecimalSeparator;
string strdec = dec.ToString(CultureInfo.CurrentCulture);
return strdec.Contains(sep) ? strdec.TrimEnd('0').TrimEnd(sep.ToCharArray()) : strdec;
}

Use the hash (#) symbol to only display trailing 0's when necessary. See the tests below.
decimal num1 = 13.1534545765;
decimal num2 = 49.100145;
decimal num3 = 30.000235;
num1.ToString("0.##"); //13.15%
num2.ToString("0.##"); //49.1%
num3.ToString("0.##"); //30%

I found an elegant solution from http://dobrzanski.net/2009/05/14/c-decimaltostring-and-how-to-get-rid-of-trailing-zeros/
Basically
decimal v=2.4200M;
v.ToString("#.######"); // Will return 2.42. The number of # is how many decimal digits you support.

A very low level approach, but I belive this would be the most performant way by only using fast integer calculations (and no slow string parsing and culture sensitive methods):
public static decimal Normalize(this decimal d)
{
int[] bits = decimal.GetBits(d);
int sign = bits[3] & (1 << 31);
int exp = (bits[3] >> 16) & 0x1f;
uint a = (uint)bits[2]; // Top bits
uint b = (uint)bits[1]; // Middle bits
uint c = (uint)bits[0]; // Bottom bits
while (exp > 0 && ((a % 5) * 6 + (b % 5) * 6 + c) % 10 == 0)
{
uint r;
a = DivideBy10((uint)0, a, out r);
b = DivideBy10(r, b, out r);
c = DivideBy10(r, c, out r);
exp--;
}
bits[0] = (int)c;
bits[1] = (int)b;
bits[2] = (int)a;
bits[3] = (exp << 16) | sign;
return new decimal(bits);
}
private static uint DivideBy10(uint highBits, uint lowBits, out uint remainder)
{
ulong total = highBits;
total <<= 32;
total = total | (ulong)lowBits;
remainder = (uint)(total % 10L);
return (uint)(total / 10L);
}

This is simple.
decimal decNumber = Convert.ToDecimal(value);
return decNumber.ToString("0.####");
Tested.
Cheers :)

Depends on what your number represents and how you want to manage the values: is it a currency, do you need rounding or truncation, do you need this rounding only for display?
If for display consider formatting the numbers are x.ToString("")
http://msdn.microsoft.com/en-us/library/dwhawy9k.aspx and
http://msdn.microsoft.com/en-us/library/0c899ak8.aspx
If it is just rounding, use Math.Round overload that requires a MidPointRounding overload
http://msdn.microsoft.com/en-us/library/ms131274.aspx)
If you get your value from a database consider casting instead of conversion:
double value = (decimal)myRecord["columnName"];

This will work:
decimal source = 2.4200m;
string output = ((double)source).ToString();
Or if your initial value is string:
string source = "2.4200";
string output = double.Parse(source).ToString();
Pay attention to this comment.

Trying to do more friendly solution of DecimalToString (https://stackoverflow.com/a/34486763/3852139):
private static decimal Trim(this decimal value)
{
var s = value.ToString(CultureInfo.InvariantCulture);
return s.Contains(CultureInfo.InvariantCulture.NumberFormat.NumberDecimalSeparator)
? Decimal.Parse(s.TrimEnd('0'), CultureInfo.InvariantCulture)
: value;
}
private static decimal? Trim(this decimal? value)
{
return value.HasValue ? (decimal?) value.Value.Trim() : null;
}
private static void Main(string[] args)
{
Console.WriteLine("=>{0}", 1.0000m.Trim());
Console.WriteLine("=>{0}", 1.000000023000m.Trim());
Console.WriteLine("=>{0}", ((decimal?) 1.000000023000m).Trim());
Console.WriteLine("=>{0}", ((decimal?) null).Trim());
}
Output:
=>1
=>1.000000023
=>1.000000023
=>

how about this:
public static string TrimEnd(this decimal d)
{
string str = d.ToString();
if (str.IndexOf(".") > 0)
{
str = System.Text.RegularExpressions.Regex.Replace(str.Trim(), "0+?$", " ");
str = System.Text.RegularExpressions.Regex.Replace(str.Trim(), "[.]$", " ");
}
return str;
}

You can just set as:
decimal decNumber = 23.45600000m;
Console.WriteLine(decNumber.ToString("0.##"));

The following code could be used to not use the string type:
int decimalResult = 789.500
while (decimalResult>0 && decimalResult % 10 == 0)
{
decimalResult = decimalResult / 10;
}
return decimalResult;
Returns 789.5

In case you want to keep decimal number, try following example:
number = Math.Floor(number * 100000000) / 100000000;

Here is an Extention method I wrote, it also removes dot or comma if it`s the last character (after the zeros were removed):
public static string RemoveZeroTail(this decimal num)
{
var result = num.ToString().TrimEnd(new char[] { '0' });
if (result[result.Length - 1].ToString() == "." || result[result.Length - 1].ToString() == ",")
{
return result.Substring(0, result.Length - 1);
}
else
{
return result;
}
}

To remove trailing zero's from a string variable dateTicks, Use
return new String(dateTicks.Take(dateTicks.LastIndexOf(dateTicks.Last(v => v != '0')) + 1).ToArray());

Additional Answer:
In a WPF Application using XAML you could use
{Binding yourDecimal, StringFormat='#,0.00#######################'}
The above answer will preserve the zero in some situations so you could still return 2.00 for example
{Binding yourDecimal, StringFormat='#,0.#########################'}
If you want to remove ALL trailing zeros, adjust accordingly.

The following code will be able to remove the trailing 0's. I know it's the hard way but it works.
private static string RemoveTrailingZeros(string input)
{
for (int i = input.Length - 1; i > 0; i-- )
{
if (!input.Contains(".")) break;
if (input[i].Equals('0'))
{
input= input.Remove(i);
}
else break;
}
return input;
}

string.Format("{0:G29}", decimal.Parse("2.00"))
string.Format("{0:G29}", decimal.Parse(Your_Variable))

try this code:
string value = "100";
value = value.Contains(".") ? value.TrimStart('0').TrimEnd('0').TrimEnd('.') : value.TrimStart('0');

Very simple answer is to use TrimEnd(). Here is the result,
double value = 1.00;
string output = value.ToString().TrimEnd('0');
Output is 1
If my value is 1.01 then my output will be 1.01

try like this
string s = "2.4200";
s = s.TrimStart("0").TrimEnd("0", ".");
and then convert that to float