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Keep one number after decimal point, not rounding

I have a double variable in c#:
DoubleVar = 3.488;
I want to keep only one digit after decimal point, not rounding. For example:
3.4
Thanks so much in advance.

Since you specified "no rounding", and did not specify that this was being displayed as text. Then the following will Truncate to 1 decimal place
value = Math.Truncate(value * 10) / 10
Note: with floats you are likely going to get artefacts anyway, as there are numbers which can't be represented in base 2.

public static void Main()
{
double val = 1.33;
Console.WriteLine(String.Format("{0:0.#}", val));
}

how to leave decimal places to 1 without rounding

I need to convert my value 2.8634 to 2.8. I tried the following ,
var no = Math.Round(2.8634,2,MidpointRounding.AwayFromZero)
I'm getting 2.87.
Suggest me some ideas how to convert.
Thanks

This might do the trick for you
decimal dsd = 2.8634m;
var no = Math.Truncate(dsd * 10) / 10;
Math.Truncate calculates the integral part of a specified decimal number. The number is rounded to the nearest integer towards zero.
You can also have a look on the difference between Math.Floor, Math.Ceiling, Math.Truncate, Math.Round with an amazing explanation.

Use this one.Hope this will work for you.
var no = Math.Round(2.8634,1,MidpointRounding.AwayFromZero)

It's a tad more cryptic (but more efficient) than calling a Math method, but you can simply multiply the value by 10, cast to an integer (which effectively truncates the decimal portion), and then divide by 10.0 (or 10d/10f, all just to ensure we don't get integer division) to get back the value you are after. I.e.:
float val = 2.8634;
val = ((int)(val * 10)) / 10.0;

Imitate Math.Round for GnuMP library in C#

I downloaded for myself GnuMP library: https://gnumpnet.codeplex.com/ which behind the curtains uses gmp.dll which is wrapper for https://gmplib.org.
It has type Real, which is used for high precision calculations. In other project I have double and decimal type, which I want to replace with Real.
I need to replace Math.Round() with customized round for Real ( type in gnump.net). Did anybody tried to implement Round but for Real of Gnump.

Today I found an mathematically correct answer:
public static Real Round(Real r, int precision)
{
Real scaled = Real.Pow(10, precision + 1);
Real multiplied = r*scaled;
Real truncated = Trunc(multiplied);
Real lastNumber = truncated - Trunc(truncated/10)*10;
if (lastNumber >= 5)
{
truncated += 10;
}
truncated = Trunc(truncated/10);
return truncated * 10 /(scaled);
}
When I say mathematically correct I mean that following code:
Real r = 2.5;
r = Real.Round(r, 0);
will give 3. According to http://msdn.microsoft.com/en-us/library/wyk4d9cy.aspx Math.Round will give "round to even" (so called banker's rounding), but I for my task need mathematical round.

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);
}