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

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

Math.Round

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

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

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

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

Another option:

string strVal = "32.11"; // will return 33// string strVal = "32.00" // returns 32// string strVal = "32.98" // returns 33string[] 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 negd = ((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 * 3/4int 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 doubledd = 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 purposedd =(double)c * ((double)a / (double)b);printf("%.2f\t",dd);c=f;// d = c * 3/4 with integer roundingd = ((c * a) + ((c>0? (b>>1):-(b>>1)))) / b;printf("%d\t",d);puts("");}return 0;}/* test outputin 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