如何四舍五入到 C # 中最接近的整数

如何将值舍入到最接近的整数？

例如:

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

“数学，天花板”对我没帮助，有什么想法吗？

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Use Math.Round:

double roundedValue = Math.Round(value, 0)
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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.

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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.
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最佳答案

See the official documentation for more. For example:

Basically you give the Math.Round method three parameters.

  1. The value you want to round.
  2. The number of decimals you want to keep after the value.
  3. 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).

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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

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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
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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

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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);
}
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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...

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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
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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.

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

It will give result:

var roundedVal = 3
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decimal RoundTotal = Total - (int)Total;
if ((double)RoundTotal <= .50)
Total = (int)Total;
else
Total = (int)Total + 1;
lblTotal.Text = Total.ToString();
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Using Math.Round(number) rounds to the nearest whole number.

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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


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