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why Math.Round(66.665d, 2) is 66.67 in c# [duplicate]

The MidpointRound.AwayFromZero overload is supposed to round the 5 up in a rounding situation.
EX: Math.Round(1.5, 1, MidpointRounding.AwayFromZero);
This rounds 1.5 to 2.
However, when I try Math.Round(1.025, 2, MidpointRounding.AwayFromZero); it rounds it down to 1.02. It is supposed to round up to 1.03 though.
Does anyone have any idea why this is happening? Thanks!

The value 1.025 cannot be represented exactly as the type double. Floating point numbers have that problem. See Why are floating point numbers inaccurate?.
When you write 1.025 as a double number, it will internally be a little bit smaller than 1.025. That's why the rounding delivers an unexpected value.
When dealing with money or other stuff where you want the value be exactly the same as displayed, use the type decimal instead of double:
decimal roundedValue = Math.Round(1.025m, 2, MidpointRounding.AwayFromZero);
Note that the m after the number turns the number into a literal of the decimal type.

Rounding-up a decimal to 2 decimal places

I have a problem to round a number to two decimal places.
I have the number 3106.4647771976413339683766317M.
The correct round to two decimal places is 3106.47, but using Math.Round(value, 2, MidpointRounding.AwayFromZero) the number is 3106.46.
The problem is the method look to third decimal place to round, but if it look to fourth decimal place will generate the correct number.
Someone has something like that?

Mathematically, the correct round to two decimal places is 3106.46.
What you probably want is a ceiling:
Math.Ceiling(3106.4647771976413339683766317M * 100) / 100
produces 3106.47. There is no version of Math.Ceiling accepting a number of decimal places, that's why there are multiplication and division.
In addition, note that there is a caveat in this expression:
Math.Round(value, 2, MidpointRounding.AwayFromZero)
Math.Round does not have a variant with three arguments, where the first one is a decimal. It works, because the value is implicitly converted to double. However, this is unwanted.

If you really want to round that way (more of a common mistake than actual rounding in the traditional sense) you can round up from the furthest right digit and move left one place at a time.
double numberToRound = 3106.4647771976413339683766317;
int placesToRoundTo = 2;
int smallestPlaceIndex = 24; // you would need to determine this value
for (int i = smallestPlaceIndex; i >= placesToRoundTo)
{
numberToRound = Math.Round(numberToRound, i, MidpointRounding.AwayFromZero);
}
There isn't a simple built-in method to do this because it isn't normal rounding.

Casting int to double and rounding off to last whole number

I'm stuck with a little issue here, say you have the following code:
int whole = 0;
double decimal = 88.00
whole = decimal / 1.5; //now the answer is 58.66
So here's the issue, explicitly casting a double to an int is easy enough. But if I do this now 'whole' is going to be set to 59. This - is not so good, I want it to be set to the last whole number (being 58).
How do you do this in C#?

To round doubles to integers, you have 4 basic math functions:
Math.Round() - Rounds to the nearest whole number (or user specified number of deciml places), and lets you choose to round middle points up or down.
Math.Floor() - Rounds to the first whole number toward negative infinity.
Math.Ceiling() - Rounds to the first whole number toward positive infinity.
Math.Truncate() - Rounds to the first whole number toward zero.
I think you want either Floor or Truncate. Both round down for positive numbers, but Truncate rounds -3.6 to -3, while Floor rounds it to -4.
Casting to int does the same as truncating, so you can use that if you prefer.

Math.Floor:
whole = (int)Math.Floor(decimal / 1.5);

If you cast double to int, the answer will NOT be 59 -- it will be 58. When casting double to int, the value will be rounded towards zero. So, this is sufficient:
int whole = 0;
double x = 88.00;
whole = (int)(x / 1.5); // whole will be 58

Use Math.Floor if you want to round to the last whole number, and Math.Ceiling if you want to round to the next.

If you need rounding, it's pretty common to use:
whole = (int)(decimal / 1.5 + 0.5);
Without the 0.5, you're truncating, not rounding.
If you have a rounding function in your math libraries, that's good too. Some of these will do the odd/even thing for rounding 0.5 to avoid a little bit of data skew.

This will convert the double value into int:
whole = (int)(decimal / 1.5);
Also, you can use Math.Floor(doubleValue).

Limiting double to 3 decimal places

This i what I am trying to achieve:
If a double has more than 3 decimal places, I want to truncate any decimal places beyond the third. (do not round.)
Eg.: 12.878999 -> 12.878
If a double has less than 3 decimals, leave unchanged
Eg.: 125 -> 125
89.24 -> 89.24
I came across this command:
double example = 12.34567;
double output = Math.Round(example, 3);
But I do not want to round. According to the command posted above,
12.34567 -> 12.346
I want to truncate the value so that it becomes: 12.345

Doubles don't have decimal places - they're not based on decimal digits to start with. You could get "the closest double to the current value when truncated to three decimal digits", but it still wouldn't be exactly the same. You'd be better off using decimal.
Having said that, if it's only the way that rounding happens that's a problem, you can use Math.Truncate(value * 1000) / 1000; which may do what you want. (You don't want rounding at all, by the sounds of it.) It's still potentially "dodgy" though, as the result still won't really just have three decimal places. If you did the same thing with a decimal value, however, it would work:
decimal m = 12.878999m;
m = Math.Truncate(m * 1000m) / 1000m;
Console.WriteLine(m); // 12.878
EDIT: As LBushkin pointed out, you should be clear between truncating for display purposes (which can usually be done in a format specifier) and truncating for further calculations (in which case the above should work).

I can't think of a reason to explicitly lose precision outside of display purposes. In that case, simply use string formatting.
double example = 12.34567;
Console.Out.WriteLine(example.ToString("#.000"));

double example = 3.1416789645;
double output = Convert.ToDouble(example.ToString("N3"));

Multiply by 1000 then use Truncate then divide by 1000.

If your purpose in truncating the digits is for display reasons, then you just just use an appropriate formatting when you convert the double to a string.
Methods like String.Format() and Console.WriteLine() (and others) allow you to limit the number of digits of precision a value is formatted with.
Attempting to "truncate" floating point numbers is ill advised - floating point numbers don't have a precise decimal representation in many cases. Applying an approach like scaling the number up, truncating it, and then scaling it down could easily change the value to something quite different from what you'd expected for the "truncated" value.
If you need precise decimal representations of a number you should be using decimal rather than double or float.

You can use:
double example = 12.34567;
double output = ( (double) ( (int) (example * 1000.0) ) ) / 1000.0 ;

Good answers above- if you're looking for something reusable here is the code. Note that you might want to check the decimal places value, and this may overflow.
public static decimal TruncateToDecimalPlace(this decimal numberToTruncate, int decimalPlaces)
{
decimal power = (decimal)(Math.Pow(10.0, (double)decimalPlaces));
return Math.Truncate((power * numberToTruncate)) / power;
}

In C lang:
double truncKeepDecimalPlaces(double value, int numDecimals)
{
int x = pow(10, numDecimals);
return (double)trunc(value * x) / x;
}

Round double in two decimal places in C#?

I want to round up double value in two decimal places in c# how can i do that?
double inputValue = 48.485;
after round up
inputValue = 48.49;
Related: c# - How do I round a decimal value to 2 decimal places (for output on a page)

This works:
inputValue = Math.Round(inputValue, 2);

Math.Round(inputValue, 2, MidpointRounding.AwayFromZero)

Another easy way is to use ToString with a parameter.
Example:
float d = 54.9700F;
string s = d.ToString("N2");
Console.WriteLine(s);
Result:
54.97

Use Math.Round
value = Math.Round(48.485, 2);

You should use
inputvalue=Math.Round(inputValue, 2, MidpointRounding.AwayFromZero)
Math.Round
Math.Round rounds a double-precision floating-point value to a
specified number of fractional digits.
MidpointRounding
Specifies how mathematical rounding methods should process a number
that is midway between two numbers.
Basically the function above will take your inputvalue and round it to 2 (or whichever number you specify) decimal places. With MidpointRounding.AwayFromZero when a number is halfway between two others, it is rounded toward the nearest number that is away from zero. There is also another option you can use that rounds towards the nearest even number.

you can try one from below.there are many way for this.
1.
value=Math.Round(123.4567, 2, MidpointRounding.AwayFromZero) //"123.46"
2.
inputvalue=Math.Round(123.4567, 2) //"123.46"
3.
String.Format("{0:0.00}", 123.4567); // "123.46"
4.
string.Format("{0:F2}", 123.456789); //123.46
string.Format("{0:F3}", 123.456789); //123.457
string.Format("{0:F4}", 123.456789); //123.4568

Use an interpolated string, this generates a rounded up string:
var strlen = 6;
$"{48.485:F2}"
Output
"48.49"

I think all these answers are missing the question. The problem was to "Round UP", not just "Round". It is my understanding that Round Up means that ANY fractional value about a whole digit rounds up to the next WHOLE digit. ie: 48.0000000 = 48 but 25.00001 = 26. Is this not the definition of rounding up? (or have my past 60 years in accounting been misplaced?