How to Round to the nearest whole number in C#

问题

How can I round values to nearest integer?

For example:

1.1 => 1
1.5 => 2
1.9 => 2

"Math.Ceiling()" is not helping me. Any ideas?

回答1:

See the official documentation for more. For example:

Basically you give the Math.Round method three parameters.

The value you want to round.
The number of decimals you want to keep after the value.
An optional parameter you can invoke to use AwayFromZero rounding. (ignored unless rounding is ambiguous, e.g. 1.5)

Sample code:

var roundedA = Math.Round(1.1, 0); // Output: 1
var roundedB = Math.Round(1.5, 0, MidpointRounding.AwayFromZero); // Output: 2
var roundedC = Math.Round(1.9, 0); // Output: 2
var roundedD = Math.Round(2.5, 0); // Output: 2
var roundedE = Math.Round(2.5, 0, MidpointRounding.AwayFromZero); // Output: 3
var roundedF = Math.Round(3.49, 0, MidpointRounding.AwayFromZero); // Output: 3

Live Demo

You need MidpointRounding.AwayFromZero if you want a .5 value to be rounded up. Unfortunately this isn't the default behavior for Math.Round(). If using MidpointRounding.ToEven (the default) the value is rounded to the nearest even number (1.5 is rounded to 2, but 2.5 is also rounded to 2).

回答2:

Math.Ceiling

always rounds up (towards the ceiling)

Math.Floor

always rounds down (towards to floor)

what you are after is simply

Math.Round

which rounds as per this post

回答3:

You need Math.Round, not Math.Ceiling. Ceiling always "rounds" up, while Round rounds up or down depending on the value after the decimal point.

回答4:

there's this manual, and kinda cute way too:

double d1 = 1.1;
double d2 = 1.5;
double d3 = 1.9;

int i1 = (int)(d1 + 0.5);
int i2 = (int)(d2 + 0.5);
int i3 = (int)(d3 + 0.5);

simply add 0.5 to any number, and cast it to int (or floor it) and it will be mathematically correctly rounded :D

回答5:

You can use Math.Round as others have suggested (recommended), or you could add 0.5 and cast to an int (which will drop the decimal part).

double value = 1.1;
int roundedValue = (int)(value + 0.5); // equals 1

double value2 = 1.5;
int roundedValue2 = (int)(value2 + 0.5); // equals 2

回答6:

Just a reminder. Beware for double.

Math.Round(0.3 / 0.2 ) result in 1, because in double 0.3 / 0.2 = 1.49999999
Math.Round( 1.5 ) = 2

回答7:

You have the Math.Round function that does exactly what you want.

Math.Round(1.1) results with 1
Math.Round(1.8) will result with 2.... and so one.

回答8:

this will round up to the nearest 5 or not change if it already is divisible by 5

public static double R(double x)
{
    // markup to nearest 5
    return (((int)(x / 5)) * 5) + ((x % 5) > 0 ? 5 : 0);
}

回答9:

Use Math.Round:

double roundedValue = Math.Round(value, 0)

回答10:

I was looking for this, but my example was to take a number, such as 4.2769 and drop it in a span as just 4.3. Not exactly the same, but if this helps:

Model.Statistics.AverageReview   <= it's just a double from the model

Then:

@Model.Statistics.AverageReview.ToString("n1")   <=gives me 4.3
@Model.Statistics.AverageReview.ToString("n2")   <=gives me 4.28

etc...

回答11:

var roundedVal = Math.Round(2.5, 0);

It will give result:

var roundedVal = 3

回答12:

If your working with integers rather than floating point numbers, here is the way.

#define ROUNDED_FRACTION(numr,denr) ((numr/denr)+(((numr%denr)<(denr/2))?0:1))

Here both "numr" and "denr" are unsigned integers.

回答13:

Using Math.Round(number) rounds to the nearest whole number.

回答14:

Write your own round method. Something like,

function round(x) rx = Math.ceil(x) if (rx - x <= .000001) return int(rx) else return int(x) end

回答15:

decimal RoundTotal = Total - (int)Total;
if ((double)RoundTotal <= .50)
   Total = (int)Total;
else
   Total = (int)Total + 1;
lblTotal.Text = Total.ToString();

来源：https://stackoverflow.com/questions/8844674/how-to-round-to-the-nearest-whole-number-in-c-sharp

标签

rounding