Compare decimals to 1 decimal point? - c#

I'm new to C# (Come from Java/C++ at uni so it's not really that new I guess) but for a project I'm needing to compare decimals.
e.g.
a = 1234.123
b = 1234.142
Decimal.Compare() will of course say they're not the same as a is smaller than b. What I want to do is compare it to that first decimal place (1 and 1) so it would return true.
The only way I've been able to think of is to convert it to use Decimal.GetBits() but I was hoping there is a simpler way I just haven't thought of yet.

You can round a decimal to one fractional digit and then compare them.
if (Decimal.Round(d1,1) == Decimal.Round(d2,1))
Console.WriteLine("Close enough.");
And, if rounding (with default midpoint handling) is not what you want, Decimal types can also be used with all the other options, like those I covered in this earlier answer.

You can use Math.Truncate(Decimal) (MSDN)
Calculates the integral part of a specified decimal number.
Coding example.
Decimal a = 1234.123m;
Decimal b = 1234.142m;
Decimal A = Math.Truncate(a * 10);
Console.WriteLine(A);// <= Will out 12341
Decimal B = Math.Truncate(b * 10);
Console.WriteLine(B);// <= Will out 12341
Console.WriteLine(Decimal.Compare(A, B)); // Will out 0 ; A and B are equal. Which means a,b are equal to first decimal place
Note : This was tested and posted .
Also simple one line comparison :
Decimal a = 1234.123m;
Decimal b = 1234.142m;
if(Decimal.Compare(Math.Truncate(a*10),Math.Truncate(b*10))==0){
Console.WriteLine("Equal upto first decimal place"); // <= Will out this for a,b
}

Related

Rounding to 2 decimal places c#

Currently have a working rounding model within my c# code and is perfectly rounding numbers that have more than 2 decimal places down to 2 decimal places which is great. However, when i have lets say double value = 100.6, and i put that into double dollar_value = Math.Round(value, 2), it still returns as 100.6.
I was wondering if there was a way to transform a 1 decimal place value to 2 decimal places?
Numbers are not stored with extra zeroes (As it is a waste of memory to do so, being the numbers are the same with or without). In order to represent a number this way you will either need to display or store it as a string.
string str = value.ToString("#.00", CultureInfo.InvariantCulture);
Now str will always have 2 decimal places.
I don't know the C# method, but in C++ I'd use one of these two methods:
double value = 23.666666 ; // example
value = 0.01 * floor ( value * 100.0 ) ; // There's a "floor" function in C# too
^ See https://msdn.microsoft.com/en-us/library/e0b5f0xb(v=vs.110).aspx
Or
double value = 23.666666 ; // example
value = 0.01 * (double) ( (int)(value*100.0) ) ;
Or
double value = 23.666666 ; // example
value = 0.01 * double ( int ( value*100.0 ) ) ; // same as previous, but more C++ like
The other answers are probably better if you're looking to "print a dollar amount with two decimal places." However, if you want to transform the number to use internally, this is a way to do it.
If you want the string representation to have two decimal points use:
yourNumber.ToString ("0.00");
The number itself is always stored as a ~29 digit number regardless of its string representation.
Your value just needs to be formatted when it's display - for example value.ToString("N2") will convert it to a string with two decimal places. Check out the Standard Numeric Format Strings on MSDN to see a broader list of formatting strings.
Additionally, I'd only convert to a string when you're ready display the value to a user and would keep it as a numeric type (e.g. double) if you're passing it around between methods or planning to do any further calculations on it. Otherwise you'll be unnecessarily converting the value to and from a string multiple times.

How to check input double value contain how many digits?

I need a double value to contain 2 digits after ".", such as 2.15, 20.15. If the input value is 3.125, then it should print an error message.
My code is:
private static bool isTwoDigits(double num)
{
return (num - Math.Floor(num)).ToString().Length <= 4;
}
If you input 2.15, then it will be 2.15 -2 = 0.15 <= 4 - which works. But when I change num to 20.15 it doesn't, because (num - Math.Floor(num)) here will return 0.14999999999.
Any other good ideas?
This is the nature of binary floating points number. Just like 1/3 can't be exactly written out as a finite decimal number, 0.1 can't be exactly represented by a finite binary expansion.
So depending on what you are trying to achieve exactly, you could:
If you are validating some string input (e.g. a textbox), you can process the information at the string level, e.g. with a RegEx.
You can store your numbers in the decimal datatype, which can store decimal values exactly.
You can do your computation on a double but you have to give yourself a tolerance. If you expect only 2 digits of precision, you can do something like Math.Abs(x - Math.Round(x, 2)) < 0.00000001). The definition of this tolerance margin depends on your use case.
If you're really worried about the number of decimal places, on a base-10 number, use decimal instead of double.
the decimal is for calculating financial calculations, and the reason it's called decimal in the first place is so that it can better handle base-10 calculations such as dollars and cents.
And you can also check if the number is 2 digits a bit more simply.
return num % 0.01m == 0.0m;
SO as has already been said, you can use regexp to ensure the entire format is correct.
But if you know there will only be 1 decimal because its already a number you can also just use String.IndexOf
eg
double foo = .... ;
string fooString = foo.ToString();
if (fooString.Length - fooString.IndexOf(".") != 3) => error.
(Its 3 because Length is max index + 1 )

Comparing double values in C#

I've a double variable called x.
In the code, x gets assigned a value of 0.1 and I check it in an 'if' statement comparing x and 0.1
if (x==0.1)
{
----
}
Unfortunately it does not enter the if statement
Should I use Double or double?
What's the reason behind this? Can you suggest a solution for this?
It's a standard problem due to how the computer stores floating point values. Search here for "floating point problem" and you'll find tons of information.
In short – a float/double can't store 0.1 precisely. It will always be a little off.
You can try using the decimal type which stores numbers in decimal notation. Thus 0.1 will be representable precisely.
You wanted to know the reason:
Float/double are stored as binary fractions, not decimal fractions. To illustrate:
12.34 in decimal notation (what we use) means
1 * 101 + 2 * 100 + 3 * 10-1 + 4 * 10-2
The computer stores floating point numbers in the same way, except it uses base 2: 10.01 means
1 * 21 + 0 * 20 + 0 * 2-1 + 1 * 2-2
Now, you probably know that there are some numbers that cannot be represented fully with our decimal notation. For example, 1/3 in decimal notation is 0.3333333…. The same thing happens in binary notation, except that the numbers that cannot be represented precisely are different. Among them is the number 1/10. In binary notation that is 0.000110011001100….
Since the binary notation cannot store it precisely, it is stored in a rounded-off way. Hence your problem.
double and Double are the same (double is an alias for Double) and can be used interchangeably.
The problem with comparing a double with another value is that doubles are approximate values, not exact values. So when you set x to 0.1 it may in reality be stored as 0.100000001 or something like that.
Instead of checking for equality, you should check that the difference is less than a defined minimum difference (tolerance). Something like:
if (Math.Abs(x - 0.1) < 0.0000001)
{
...
}
You need a combination of Math.Abs on X-Y and a value to compare with.
You can use following Extension method approach
public static class DoubleExtensions
{
const double _3 = 0.001;
const double _4 = 0.0001;
const double _5 = 0.00001;
const double _6 = 0.000001;
const double _7 = 0.0000001;
public static bool Equals3DigitPrecision(this double left, double right)
{
return Math.Abs(left - right) < _3;
}
public static bool Equals4DigitPrecision(this double left, double right)
{
return Math.Abs(left - right) < _4;
}
...
Since you rarely call methods on double except ToString I believe its pretty safe extension.
Then you can compare x and y like
if(x.Equals4DigitPrecision(y))
Comparing floating point number can't always be done precisely because of rounding. To compare
(x == .1)
the computer really compares
(x - .1) vs 0
Result of sybtraction can not always be represeted precisely because of how floating point number are represented on the machine. Therefore you get some nonzero value and the condition evaluates to false.
To overcome this compare
Math.Abs(x- .1) vs some very small threshold ( like 1E-9)
From the documentation:
Precision in Comparisons
The Equals method should be used with caution, because two apparently equivalent values can be unequal due to the differing precision of the two values. The following example reports that the Double value .3333 and the Double returned by dividing 1 by 3 are unequal.
...
Rather than comparing for equality, one recommended technique involves defining an acceptable margin of difference between two values (such as .01% of one of the values). If the absolute value of the difference between the two values is less than or equal to that margin, the difference is likely to be due to differences in precision and, therefore, the values are likely to be equal. The following example uses this technique to compare .33333 and 1/3, the two Double values that the previous code example found to be unequal.
So if you really need a double, you should use the techique described on the documentation.
If you can, change it to a decimal. It' will be slower, but you won't have this type of problem.
Use decimal. It doesn't have this "problem".
Exact comparison of floating point values is know to not always work due to the rounding and internal representation issue.
Try imprecise comparison:
if (x >= 0.099 && x <= 0.101)
{
}
The other alternative is to use the decimal data type.
double (lowercase) is just an alias for System.Double, so they are identical.
For the reason, see Binary floating point and .NET.
In short: a double is not an exact type and a minute difference between "x" and "0.1" will throw it off.
Double (called float in some languages) is fraut with problems due to rounding issues, it's good only if you need approximate values.
The Decimal data type does what you want.
For reference decimal and Decimal are the same in .NET C#, as are the double and Double types, they both refer to the same type (decimal and double are very different though, as you've seen).
Beware that the Decimal data type has some costs associated with it, so use it with caution if you're looking at loops etc.
Official MS help, especially interested "Precision in Comparisons" part in context of the question.
https://learn.microsoft.com/en-us/dotnet/api/system.double.equals
// Initialize two doubles with apparently identical values
double double1 = .333333;
double double2 = (double) 1/3;
// Define the tolerance for variation in their values
double difference = Math.Abs(double1 * .00001);
// Compare the values
// The output to the console indicates that the two values are equal
if (Math.Abs(double1 - double2) <= difference)
Console.WriteLine("double1 and double2 are equal.");
else
Console.WriteLine("double1 and double2 are unequal.");
1) Should i use Double or double???
Double and double is the same thing. double is just a C# keyword working as alias for the class System.Double
The most common thing is to use the aliases! The same for string (System.String), int(System.Int32)
Also see Built-In Types Table (C# Reference)
Taking a tip from the Java code base, try using .CompareTo and test for the zero comparison. This assumes the .CompareTo function takes in to account floating point equality in an accurate manner. For instance,
System.Math.PI.CompareTo(System.Math.PI) == 0
This predicate should return true.
// number of digits to be compared
public int n = 12
// n+1 because b/a tends to 1 with n leading digits
public double MyEpsilon { get; } = Math.Pow(10, -(n+1));
public bool IsEqual(double a, double b)
{
// Avoiding division by zero
if (Math.Abs(a)<= double.Epsilon || Math.Abs(b) <= double.Epsilon)
return Math.Abs(a - b) <= double.Epsilon;
// Comparison
return Math.Abs(1.0 - a / b) <= MyEpsilon;
}
Explanation
The main comparison function done using division a/b which should go toward 1. But why division? it simply puts one number as reference defines the second one. For example
a = 0.00000012345
b = 0.00000012346
a/b = 0.999919002
b/a = 1.000081004
(a/b)-1 = 8.099789405475458e-5‬
1-(b/a) = 8.100445524503848e-5‬
or
a=12345*10^8
b=12346*10^8
a/b = 0.999919002
b/a = 1.000081004
(a/b)-1 = 8.099789405475458e-5‬
1-(b/a) = 8.100445524503848e-5‬
by division we get rid of trailing or leading zeros (or relatively small numbers) that pollute our judgement of number precision. In the example, the comparison is of order 10^-5, and we have 4 number accuracy, because of that in the beginning code I wrote comparison with 10^(n+1) where n is number accuracy.
Adding onto Valentin Kuzub's answer above:
we could use a single method that supports providing nth precision number:
public static bool EqualsNthDigitPrecision(this double value, double compareTo, int precisionPoint) =>
Math.Abs(value - compareTo) < Math.Pow(10, -Math.Abs(precisionPoint));
Note: This method is built for simplicity without added bulk and not with performance in mind.
As a general rule:
Double representation is good enough in most cases but can miserably fail in some situations. Use decimal values if you need complete precision (as in financial applications).
Most problems with doubles doesn't come from direct comparison, it use to be a result of the accumulation of several math operations which exponentially disturb the value due to rounding and fractional errors (especially with multiplications and divisions).
Check your logic, if the code is:
x = 0.1
if (x == 0.1)
it should not fail, it's to simple to fail, if X value is calculated by more complex means or operations it's quite possible the ToString method used by the debugger is using an smart rounding, maybe you can do the same (if that's too risky go back to using decimal):
if (x.ToString() == "0.1")
Floating point number representations are notoriously inaccurate because of the way floats are stored internally. E.g. x may actually be 0.0999999999 or 0.100000001 and your condition will fail. If you want to determine if floats are equal you need to specify whether they're equal to within a certain tolerance.
I.e.:
if(Math.Abs(x - 0.1) < tol) {
// Do something
}
My extensions method for double comparison:
public static bool IsEqual(this double value1, double value2, int precision = 2)
{
var dif = Math.Abs(Math.Round(value1, precision) - Math.Round(value2, precision));
while (precision > 0)
{
dif *= 10;
precision--;
}
return dif < 1;
}
To compare floating point, double or float types, use the specific method of CSharp:
if (double1.CompareTo(double2) > 0)
{
// double1 is greater than double2
}
if (double1.CompareTo(double2) < 0)
{
// double1 is less than double2
}
if (double1.CompareTo(double2) == 0)
{
// double1 equals double2
}
https://learn.microsoft.com/en-us/dotnet/api/system.double.compareto?view=netcore-3.1

How do you round a number to two decimal places in C#?

I want to do this using the Math.Round function
Here's some examples:
decimal a = 1.994444M;
Math.Round(a, 2); //returns 1.99
decimal b = 1.995555M;
Math.Round(b, 2); //returns 2.00
You might also want to look at bankers rounding / round-to-even with the following overload:
Math.Round(a, 2, MidpointRounding.ToEven);
There's more information on it here.
Try this:
twoDec = Math.Round(val, 2)
If you'd like a string
> (1.7289).ToString("#.##")
"1.73"
Or a decimal
> Math.Round((Decimal)x, 2)
1.73m
But remember! Rounding is not distributive, ie. round(x*y) != round(x) * round(y). So don't do any rounding until the very end of a calculation, else you'll lose accuracy.
Personally I never round anything. Keep it as resolute as possible, since rounding is a bit of a red herring in CS anyway. But you do want to format data for your users, and to that end, I find that string.Format("{0:0.00}", number) is a good approach.
Wikipedia has a nice page on rounding in general.
All .NET (managed) languages can use any of the common language run time's (the CLR) rounding mechanisms. For example, the Math.Round() (as mentioned above) method allows the developer to specify the type of rounding (Round-to-even or Away-from-zero). The Convert.ToInt32() method and its variations use round-to-even. The Ceiling() and Floor() methods are related.
You can round with custom numeric formatting as well.
Note that Decimal.Round() uses a different method than Math.Round();
Here is a useful post on the banker's rounding algorithm.
See one of Raymond's humorous posts here about rounding...
// convert upto two decimal places
String.Format("{0:0.00}", 140.6767554); // "140.67"
String.Format("{0:0.00}", 140.1); // "140.10"
String.Format("{0:0.00}", 140); // "140.00"
Double d = 140.6767554;
Double dc = Math.Round((Double)d, 2); // 140.67
decimal d = 140.6767554M;
decimal dc = Math.Round(d, 2); // 140.67
=========
// just two decimal places
String.Format("{0:0.##}", 123.4567); // "123.46"
String.Format("{0:0.##}", 123.4); // "123.4"
String.Format("{0:0.##}", 123.0); // "123"
can also combine "0" with "#".
String.Format("{0:0.0#}", 123.4567) // "123.46"
String.Format("{0:0.0#}", 123.4) // "123.4"
String.Format("{0:0.0#}", 123.0) // "123.0"
If you want to round a number, you can obtain different results depending on: how you use the Math.Round() function (if for a round-up or round-down), you're working with doubles and/or floats numbers, and you apply the midpoint rounding. Especially, when using with operations inside of it or the variable to round comes from an operation. Let's say, you want to multiply these two numbers: 0.75 * 0.95 = 0.7125. Right? Not in C#
Let's see what happens if you want to round to the 3rd decimal:
double result = 0.75d * 0.95d; // result = 0.71249999999999991
double result = 0.75f * 0.95f; // result = 0.71249997615814209
result = Math.Round(result, 3, MidpointRounding.ToEven); // result = 0.712. Ok
result = Math.Round(result, 3, MidpointRounding.AwayFromZero); // result = 0.712. Should be 0.713
As you see, the first Round() is correct if you want to round down the midpoint. But the second Round() it's wrong if you want to round up.
This applies to negative numbers:
double result = -0.75 * 0.95; //result = -0.71249999999999991
result = Math.Round(result, 3, MidpointRounding.ToEven); // result = -0.712. Ok
result = Math.Round(result, 3, MidpointRounding.AwayFromZero); // result = -0.712. Should be -0.713
So, IMHO, you should create your own wrap function for Math.Round() that fit your requirements. I created a function in which, the parameter 'roundUp=true' means to round to next greater number. That is: 0.7125 rounds to 0.713 and -0.7125 rounds to -0.712 (because -0.712 > -0.713). This is the function I created and works for any number of decimals:
double Redondea(double value, int precision, bool roundUp = true)
{
if ((decimal)value == 0.0m)
return 0.0;
double corrector = 1 / Math.Pow(10, precision + 2);
if ((decimal)value < 0.0m)
{
if (roundUp)
return Math.Round(value, precision, MidpointRounding.ToEven);
else
return Math.Round(value - corrector, precision, MidpointRounding.AwayFromZero);
}
else
{
if (roundUp)
return Math.Round(value + corrector, precision, MidpointRounding.AwayFromZero);
else
return Math.Round(value, precision, MidpointRounding.ToEven);
}
}
The variable 'corrector' is for fixing the inaccuracy of operating with floating or double numbers.
This is for rounding to 2 decimal places in C#:
label8.Text = valor_cuota .ToString("N2") ;
In VB.NET:
Imports System.Math
round(label8.text,2)
I know its an old question but please note for the following differences between Math round and String format round:
decimal d1 = (decimal)1.125;
Math.Round(d1, 2).Dump(); // returns 1.12
d1.ToString("#.##").Dump(); // returns "1.13"
decimal d2 = (decimal)1.1251;
Math.Round(d2, 2).Dump(); // returns 1.13
d2.ToString("#.##").Dump(); // returns "1.13"
Had a weird situation where I had a decimal variable, when serializing 55.50 it always sets default value mathematically as 55.5. But whereas, our client system is seriously expecting 55.50 for some reason and they definitely expected decimal. Thats when I had write the below helper, which always converts any decimal value padded to 2 digits with zeros instead of sending a string.
public static class DecimalExtensions
{
public static decimal WithTwoDecimalPoints(this decimal val)
{
return decimal.Parse(val.ToString("0.00"));
}
}
Usage should be
var sampleDecimalValueV1 = 2.5m;
Console.WriteLine(sampleDecimalValueV1.WithTwoDecimalPoints());
decimal sampleDecimalValueV1 = 2;
Console.WriteLine(sampleDecimalValueV1.WithTwoDecimalPoints());
Output:
2.50
2.00
One thing you may want to check is the Rounding Mechanism of Math.Round:
http://msdn.microsoft.com/en-us/library/system.midpointrounding.aspx
Other than that, I recommend the Math.Round(inputNumer, numberOfPlaces) approach over the *100/100 one because it's cleaner.
You should be able to specify the number of digits you want to round to using Math.Round(YourNumber, 2)
You can read more here.
Math.Floor(123456.646 * 100) / 100
Would return 123456.64
string a = "10.65678";
decimal d = Math.Round(Convert.ToDouble(a.ToString()),2)
public double RoundDown(double number, int decimalPlaces)
{
return Math.Floor(number * Math.Pow(10, decimalPlaces)) / Math.Pow(10, decimalPlaces);
}

How do I display a decimal value to 2 decimal places?

When displaying the value of a decimal currently with .ToString(), it's accurate to like 15 decimal places, and since I'm using it to represent dollars and cents, I only want the output to be 2 decimal places.
Do I use a variation of .ToString() for this?
decimalVar.ToString("#.##"); // returns ".5" when decimalVar == 0.5m
or
decimalVar.ToString("0.##"); // returns "0.5" when decimalVar == 0.5m
or
decimalVar.ToString("0.00"); // returns "0.50" when decimalVar == 0.5m
I know this is an old question, but I was surprised to see that no one seemed to post an answer that;
Didn't use bankers rounding
Keeps the value as a decimal.
This is what I would use:
decimal.Round(yourValue, 2, MidpointRounding.AwayFromZero);
http://msdn.microsoft.com/en-us/library/9s0xa85y.aspx
decimalVar.ToString("F");
This will:
Round off to 2 decimal places eg. 23.456 → 23.46
Ensure that there
are always 2 decimal places eg. 23 → 23.00; 12.5 → 12.50
Ideal for displaying currency.
Check out the documentation on ToString("F") (thanks to Jon Schneider).
If you just need this for display use string.Format
String.Format("{0:0.00}", 123.4567m); // "123.46"
http://www.csharp-examples.net/string-format-double/
The "m" is a decimal suffix. About the decimal suffix:
http://msdn.microsoft.com/en-us/library/364x0z75.aspx
Given decimal d=12.345; the expressions d.ToString("C") or String.Format("{0:C}", d) yield $12.35 - note that the current culture's currency settings including the symbol are used.
Note that "C" uses number of digits from current culture. You can always override default to force necessary precision with C{Precision specifier} like String.Format("{0:C2}", 5.123d).
If you want it formatted with commas as well as a decimal point (but no currency symbol), such as 3,456,789.12...
decimalVar.ToString("n2");
There's a very important characteristic of Decimal that isn't obvious:
A Decimal 'knows' how many decimal places it has based upon where it came from
The following may be unexpected :
Decimal.Parse("25").ToString() => "25"
Decimal.Parse("25.").ToString() => "25"
Decimal.Parse("25.0").ToString() => "25.0"
Decimal.Parse("25.0000").ToString() => "25.0000"
25m.ToString() => "25"
25.000m.ToString() => "25.000"
Doing the same operations with Double will result in zero decimal places ("25") for all of the above examples.
If you want a decimal to 2 decimal places there's a high likelyhood it's because it's currency in which case this is probably fine for 95% of the time:
Decimal.Parse("25.0").ToString("c") => "$25.00"
Or in XAML you would use {Binding Price, StringFormat=c}
One case I ran into where I needed a decimal AS a decimal was when sending XML to Amazon's webservice. The service was complaining because a Decimal value (originally from SQL Server) was being sent as 25.1200 and rejected, (25.12 was the expected format).
All I needed to do was Decimal.Round(...) with 2 decimal places to fix the problem regardless of the source of the value.
// generated code by XSD.exe
StandardPrice = new OverrideCurrencyAmount()
{
TypedValue = Decimal.Round(product.StandardPrice, 2),
currency = "USD"
}
TypedValue is of type Decimal so I couldn't just do ToString("N2") and needed to round it and keep it as a decimal.
Here is a little Linqpad program to show different formats:
void Main()
{
FormatDecimal(2345.94742M);
FormatDecimal(43M);
FormatDecimal(0M);
FormatDecimal(0.007M);
}
public void FormatDecimal(decimal val)
{
Console.WriteLine("ToString: {0}", val);
Console.WriteLine("c: {0:c}", val);
Console.WriteLine("0.00: {0:0.00}", val);
Console.WriteLine("0.##: {0:0.##}", val);
Console.WriteLine("===================");
}
Here are the results:
ToString: 2345.94742
c: $2,345.95
0.00: 2345.95
0.##: 2345.95
===================
ToString: 43
c: $43.00
0.00: 43.00
0.##: 43
===================
ToString: 0
c: $0.00
0.00: 0.00
0.##: 0
===================
ToString: 0.007
c: $0.01
0.00: 0.01
0.##: 0.01
===================
Math.Round Method (Decimal, Int32)
Very rarely would you want an empty string if the value is 0.
decimal test = 5.00;
test.ToString("0.00"); //"5.00"
decimal? test2 = 5.05;
test2.ToString("0.00"); //"5.05"
decimal? test3 = 0;
test3.ToString("0.00"); //"0.00"
The top rated answer is incorrect and has wasted 10 minutes of (most) people's time.
Mike M.'s answer was perfect for me on .NET, but .NET Core doesn't have a decimal.Round method at the time of writing.
In .NET Core, I had to use:
decimal roundedValue = Math.Round(rawNumber, 2, MidpointRounding.AwayFromZero);
A hacky method, including conversion to string, is:
public string FormatTo2Dp(decimal myNumber)
{
// Use schoolboy rounding, not bankers.
myNumber = Math.Round(myNumber, 2, MidpointRounding.AwayFromZero);
return string.Format("{0:0.00}", myNumber);
}
None of these did exactly what I needed, to force 2 d.p. and round up as 0.005 -> 0.01
Forcing 2 d.p. requires increasing the precision by 2 d.p. to ensure we have at least 2 d.p.
then rounding to ensure we do not have more than 2 d.p.
Math.Round(exactResult * 1.00m, 2, MidpointRounding.AwayFromZero)
6.665m.ToString() -> "6.67"
6.6m.ToString() -> "6.60"
The top-rated answer describes a method for formatting the string representation of the decimal value, and it works.
However, if you actually want to change the precision saved to the actual value, you need to write something like the following:
public static class PrecisionHelper
{
public static decimal TwoDecimalPlaces(this decimal value)
{
// These first lines eliminate all digits past two places.
var timesHundred = (int) (value * 100);
var removeZeroes = timesHundred / 100m;
// In this implementation, I don't want to alter the underlying
// value. As such, if it needs greater precision to stay unaltered,
// I return it.
if (removeZeroes != value)
return value;
// Addition and subtraction can reliably change precision.
// For two decimal values A and B, (A + B) will have at least as
// many digits past the decimal point as A or B.
return removeZeroes + 0.01m - 0.01m;
}
}
An example unit test:
[Test]
public void PrecisionExampleUnitTest()
{
decimal a = 500m;
decimal b = 99.99m;
decimal c = 123.4m;
decimal d = 10101.1000000m;
decimal e = 908.7650m
Assert.That(a.TwoDecimalPlaces().ToString(CultureInfo.InvariantCulture),
Is.EqualTo("500.00"));
Assert.That(b.TwoDecimalPlaces().ToString(CultureInfo.InvariantCulture),
Is.EqualTo("99.99"));
Assert.That(c.TwoDecimalPlaces().ToString(CultureInfo.InvariantCulture),
Is.EqualTo("123.40"));
Assert.That(d.TwoDecimalPlaces().ToString(CultureInfo.InvariantCulture),
Is.EqualTo("10101.10"));
// In this particular implementation, values that can't be expressed in
// two decimal places are unaltered, so this remains as-is.
Assert.That(e.TwoDecimalPlaces().ToString(CultureInfo.InvariantCulture),
Is.EqualTo("908.7650"));
}
You can use system.globalization to format a number in any required format.
For example:
system.globalization.cultureinfo ci = new system.globalization.cultureinfo("en-ca");
If you have a decimal d = 1.2300000 and you need to trim it to 2 decimal places then it can be printed like this d.Tostring("F2",ci); where F2 is string formating to 2 decimal places and ci is the locale or cultureinfo.
for more info check this link
http://msdn.microsoft.com/en-us/library/dwhawy9k.aspx
https://msdn.microsoft.com/en-us/library/dwhawy9k%28v=vs.110%29.aspx
This link explains in detail how you can handle your problem and what you can do if you want to learn more. For simplicity purposes, what you want to do is
double whateverYouWantToChange = whateverYouWantToChange.ToString("F2");
if you want this for a currency, you can make it easier by typing "C2" instead of "F2"
The most applicable solution is
decimalVar.ToString("#.##");
Double Amount = 0;
string amount;
amount=string.Format("{0:F2}", Decimal.Parse(Amount.ToString()));
If you need to keep only 2 decimal places (i.e. cut off all the rest of decimal digits):
decimal val = 3.14789m;
decimal result = Math.Floor(val * 100) / 100; // result = 3.14
If you need to keep only 3 decimal places:
decimal val = 3.14789m;
decimal result = Math.Floor(val * 1000) / 1000; // result = 3.147
var arr = new List<int>() { -4, 3, -9, 0, 4, 1 };
decimal result1 = arr.Where(p => p > 0).Count();
var responseResult1 = result1 / arr.Count();
decimal result2 = arr.Where(p => p < 0).Count();
var responseResult2 = result2 / arr.Count();
decimal result3 = arr.Where(p => p == 0).Count();
var responseResult3 = result3 / arr.Count();
Console.WriteLine(String.Format("{0:#,0.000}", responseResult1));
Console.WriteLine(String.Format("{0:#,0.0000}", responseResult2));
Console.WriteLine(String.Format("{0:#,0.00000}", responseResult3));
you can put as many 0 as you want.

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