'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?
Solution 1:[1]
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
Solution 2:[2]
Check out Math.Round. You can then cast the result to an int.
Solution 3:[3]
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
Solution 4:[4]
Rounds a double-precision floating-point value to the nearest integral value.
Solution 5:[5]
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);
}
Solution 6:[6]
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();
Solution 7:[7]
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
Solution 8:[8]
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);
Solution 9:[9]
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?
Solution 10:[10]
It is simple. So follow this code.
decimal d = 10.5;
int roundNumber = (int)Math.Floor(d + 0.5);
Result is 11
Sources
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Source: Stack Overflow