How to round up value C# to the nearest integer?

I want to round up double to int.

Eg,

double a=0.4, b=0.5;

I want to change them both to integer.

so that

int aa=0, bb=1;

aa is from a and bb is from b.

Any formula to do that?

Use Math.Ceiling to round up

Math.Ceiling(0.5); // 1

Use Math.Round to just round

Math.Round(0.5, MidpointRounding.AwayFromZero); // 1

And Math.Floor to round down

Math.Floor(0.5); // 0

Check out Math.Round. You can then cast the result to an int.

The .NET framework uses banker's rounding in Math.Round by default. You should use this overload:

Math.Round(0.5d, MidpointRounding.AwayFromZero)  //1
Math.Round(0.4d, MidpointRounding.AwayFromZero)  //0

Math.Round

> Rounds a double-precision floating-point value to the nearest integral value.

Use a function in place of MidpointRounding.AwayFromZero:

myRound(1.11125,4)

Answer:- 1.1114

public static Double myRound(Double Value, int places = 1000)
{
    Double myvalue = (Double)Value;
    if (places == 1000)
    {
        if (myvalue - (int)myvalue == 0.5)
        {
            myvalue = myvalue + 0.1;
            return (Double)Math.Round(myvalue);
        }
        return (Double)Math.Round(myvalue);
        places = myvalue.ToString().Substring(myvalue.ToString().IndexOf(".") + 1).Length - 1;
    } if ((myvalue * Math.Pow(10, places)) - (int)(myvalue * Math.Pow(10, places)) > 0.49)
    {
        myvalue = (myvalue * Math.Pow(10, places + 1)) + 1;
        myvalue = (myvalue / Math.Pow(10, places + 1));
    }
    return (Double)Math.Round(myvalue, places);
}

Math.Round(0.5) returns zero due to floating point rounding errors, so you'll need to add a rounding error amount to the original value to ensure it doesn't round down, eg.

Console.WriteLine(Math.Round(0.5, 0).ToString()); // outputs 0 (!!)
Console.WriteLine(Math.Round(1.5, 0).ToString()); // outputs 2
Console.WriteLine(Math.Round(0.5 + 0.00000001, 0).ToString()); // outputs 1
Console.WriteLine(Math.Round(1.5 + 0.00000001, 0).ToString()); // outputs 2
Console.ReadKey();

Just some adjusting @BrunoLM's answer with more samples :

Math.Round(0.4); // =0
Math.Round(0.5); // =0
Math.Round(0.6); // =1

Math.Round(0.4, MidpointRounding.AwayFromZero); // = 0
Math.Round(0.5, MidpointRounding.AwayFromZero); // = 1
Math.Round(0.6, MidpointRounding.AwayFromZero); // = 1

Math.Round(0.4, MidpointRounding.ToEven); // = 0
Math.Round(0.5, MidpointRounding.ToEven); // = 0
Math.Round(0.6, MidpointRounding.ToEven); // = 1

Another option:

string strVal = "32.11"; // will return 33
// string strVal = "32.00" // returns 32
// string strVal = "32.98" // returns 33

string[] valStr = strVal.Split('.');

int32 leftSide = Convert.ToInt32(valStr[0]);
int32 rightSide = Convert.ToInt32(valStr[1]);

if (rightSide > 0)
    leftSide = leftSide + 1;


return (leftSide);

It is also possible to round negative integers

// performing d = c * 3/4 where d can be pos or neg
d = ((c * a) + ((c>0? (b>>1):-(b>>1)))) / b;
// explanation:
// 1.) multiply:          c * a  
// 2.) if c is negative:  (c>0? subtract half of the dividend 
//                              (b>>1) is bit shift right = (b/2)
//     if c is positive:  else  add half of the dividend 
// 3.) do the division
// on a C51/52 (8bit embedded) or similar like ATmega the below code may execute in approx 12cpu cycles (not tested)

Extended from a tip somewhere else in here. Sorry, missed from where.

/* Example test: integer rounding example including negative*/
#include <stdio.h>
#include <string.h>

int main () {
   //rounding negative int
   // doing something like d = c * 3/4
   int a=3;
   int b=4;
   int c=-5;
   int d;
   int s=c;
   int e=c+10;


   for(int f=s; f<=e; f++) {
      printf("%d\t",f);

      double cd=f, ad=a, bd=b , dd;

      // d = c * 3/4  with double
      dd = cd * ad / bd;
    
      printf("%.2f\t",dd);
      printf("%.1f\t",dd);        
      printf("%.0f\t",dd);

      // try again with typecast have used that a lot in Borland C++ 35 years ago....... maybe evolution has overtaken it ;) ***
      // doing div before mul on purpose
      dd =(double)c * ((double)a / (double)b);
      printf("%.2f\t",dd);

      c=f;
      // d = c * 3/4  with integer rounding
      d = ((c * a) + ((c>0? (b>>1):-(b>>1)))) / b;
      printf("%d\t",d);
      puts("");
  }
 return 0;
}

/* test output
in  2f     1f   0f cast int   
-5	-3.75	-3.8	-4	-3.75	-4	
-4	-3.00	-3.0	-3	-3.75	-3	
-3	-2.25	-2.2	-2	-3.00	-2	
-2	-1.50	-1.5	-2	-2.25	-2	
-1	-0.75	-0.8	-1	-1.50	-1	
 0	 0.00	 0.0	 0	-0.75	 0	
 1	 0.75	 0.8	 1	 0.00	 1	
 2	 1.50	 1.5	 2	 0.75	 2	
 3	 2.25	 2.2	 2	 1.50	 2	
 4	 3.00	 3.0	 3	 2.25  3	
 5	 3.75	 3.8	 4	 3.00	

// by the way evolution: 
// Is there any decent small integer library out there for that by now?

It is simple. So follow this code.

decimal d = 10.5;
int roundNumber = (int)Math.Floor(d + 0.5);

Result is 11