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c# - How do I round a decimal value to 2 decimal places (for output on a page)
I want to truncate the decimals like below
i.e.
This question is related to
.net
floating-point
truncate
This is similar to TcKs suggestion above, but using math.truncate rather than int conversions
VB: but you'll get the idea
Private Function TruncateToDecimalPlace(byval ToTruncate as decimal, byval DecimalPlaces as integer) as double
dim power as decimal = Math.Pow(10, decimalplaces)
return math.truncate(totruncate * power) / power
end function
Try this:
decimal original = GetSomeDecimal(); // 22222.22939393
int number1 = (int)original; // contains only integer value of origina number
decimal temporary = original - number1; // contains only decimal value of original number
int decimalPlaces = GetDecimalPlaces(); // 3
temporary *= (Math.Pow(10, decimalPlaces)); // moves some decimal places to integer
temporary = (int)temporary; // removes all decimal places
temporary /= (Math.Pow(10, decimalPlaces)); // moves integer back to decimal places
decimal result = original + temporary; // add integer and decimal places together
It can be writen shorter, but this is more descriptive.
EDIT: Short way:
decimal original = GetSomeDecimal(); // 22222.22939393
int decimalPlaces = GetDecimalPlaces(); // 3
decimal result = ((int)original) + (((int)(original * Math.Pow(10, decimalPlaces)) / (Math.Pow(10, decimalPlaces));
Here's an extension method which does not suffer from integer overflow (like some of the above answers do). It also caches some powers of 10 for efficiency.
static double[] pow10 = { 1e0, 1e1, 1e2, 1e3, 1e4, 1e5, 1e6, 1e7, 1e8, 1e9, 1e10 };
public static double Truncate(this double x, int precision)
{
if (precision < 0)
throw new ArgumentException();
if (precision == 0)
return Math.Truncate(x);
double m = precision >= pow10.Length ? Math.Pow(10, precision) : pow10[precision];
return Math.Truncate(x * m) / m;
}
Try this:
decimal original = GetSomeDecimal(); // 22222.22939393
int number1 = (int)original; // contains only integer value of origina number
decimal temporary = original - number1; // contains only decimal value of original number
int decimalPlaces = GetDecimalPlaces(); // 3
temporary *= (Math.Pow(10, decimalPlaces)); // moves some decimal places to integer
temporary = (int)temporary; // removes all decimal places
temporary /= (Math.Pow(10, decimalPlaces)); // moves integer back to decimal places
decimal result = original + temporary; // add integer and decimal places together
It can be writen shorter, but this is more descriptive.
EDIT: Short way:
decimal original = GetSomeDecimal(); // 22222.22939393
int decimalPlaces = GetDecimalPlaces(); // 3
decimal result = ((int)original) + (((int)(original * Math.Pow(10, decimalPlaces)) / (Math.Pow(10, decimalPlaces));
This is similar to TcKs suggestion above, but using math.truncate rather than int conversions
VB: but you'll get the idea
Private Function TruncateToDecimalPlace(byval ToTruncate as decimal, byval DecimalPlaces as integer) as double
dim power as decimal = Math.Pow(10, decimalplaces)
return math.truncate(totruncate * power) / power
end function
You can use Math.Round:
decimal rounded = Math.Round(2.22939393, 3); //Returns 2.229
Or you can use ToString with the N3 numeric format.
string roundedNumber = number.ToString("N3");
EDIT: Since you don't want rounding, you can easily use Math.Truncate:
Math.Truncate(2.22977777 * 1000) / 1000; //Returns 2.229
What format are you wanting the output?
If you're happy with a string then consider the following C# code:
double num = 3.12345;
num.ToString("G3");
The result will be "3.12".
This link might be of use if you're using .NET. http://msdn.microsoft.com/en-us/library/dwhawy9k.aspx
I hope that helps....but unless you identify than language you are using and the format in which you want the output it is difficult to suggest an appropriate solution.
Maybe another quick solution could be:
>>> float("%.1f" % 1.00001)
1.0
>>> float("%.3f" % 1.23001)
1.23
>>> float("%.5f" % 1.23001)
1.23001
Try this
double d = 2.22912312515;
int demention = 3;
double truncate = Math.Truncate(d) + Math.Truncate((d - Math.Truncate(d)) * Math.Pow(10.0, demention)) / Math.Pow(10.0, demention);
Forget Everything just check out this
double num = 2.22939393;
num = Convert.ToDouble(num.ToString("#0.000"));
What format are you wanting the output?
If you're happy with a string then consider the following C# code:
double num = 3.12345;
num.ToString("G3");
The result will be "3.12".
This link might be of use if you're using .NET. http://msdn.microsoft.com/en-us/library/dwhawy9k.aspx
I hope that helps....but unless you identify than language you are using and the format in which you want the output it is difficult to suggest an appropriate solution.
Here's an extension method which does not suffer from integer overflow (like some of the above answers do). It also caches some powers of 10 for efficiency.
static double[] pow10 = { 1e0, 1e1, 1e2, 1e3, 1e4, 1e5, 1e6, 1e7, 1e8, 1e9, 1e10 };
public static double Truncate(this double x, int precision)
{
if (precision < 0)
throw new ArgumentException();
if (precision == 0)
return Math.Truncate(x);
double m = precision >= pow10.Length ? Math.Pow(10, precision) : pow10[precision];
return Math.Truncate(x * m) / m;
}
You can use Math.Round:
decimal rounded = Math.Round(2.22939393, 3); //Returns 2.229
Or you can use ToString with the N3 numeric format.
string roundedNumber = number.ToString("N3");
EDIT: Since you don't want rounding, you can easily use Math.Truncate:
Math.Truncate(2.22977777 * 1000) / 1000; //Returns 2.229
Forget Everything just check out this
double num = 2.22939393;
num = Convert.ToDouble(num.ToString("#0.000"));
A function to truncate an arbitrary number of decimals:
public decimal Truncate(decimal number, int digits)
{
decimal stepper = (decimal)(Math.Pow(10.0, (double)digits));
int temp = (int)(stepper * number);
return (decimal)temp / stepper;
}
Maybe another quick solution could be:
>>> float("%.1f" % 1.00001)
1.0
>>> float("%.3f" % 1.23001)
1.23
>>> float("%.5f" % 1.23001)
1.23001
You can use Math.Round:
decimal rounded = Math.Round(2.22939393, 3); //Returns 2.229
Or you can use ToString with the N3 numeric format.
string roundedNumber = number.ToString("N3");
EDIT: Since you don't want rounding, you can easily use Math.Truncate:
Math.Truncate(2.22977777 * 1000) / 1000; //Returns 2.229
You can use Math.Round:
decimal rounded = Math.Round(2.22939393, 3); //Returns 2.229
Or you can use ToString with the N3 numeric format.
string roundedNumber = number.ToString("N3");
EDIT: Since you don't want rounding, you can easily use Math.Truncate:
Math.Truncate(2.22977777 * 1000) / 1000; //Returns 2.229
A function to truncate an arbitrary number of decimals:
public decimal Truncate(decimal number, int digits)
{
decimal stepper = (decimal)(Math.Pow(10.0, (double)digits));
int temp = (int)(stepper * number);
return (decimal)temp / stepper;
}
Try this
double d = 2.22912312515;
int demention = 3;
double truncate = Math.Truncate(d) + Math.Truncate((d - Math.Truncate(d)) * Math.Pow(10.0, demention)) / Math.Pow(10.0, demention);
Try this:
decimal original = GetSomeDecimal(); // 22222.22939393
int number1 = (int)original; // contains only integer value of origina number
decimal temporary = original - number1; // contains only decimal value of original number
int decimalPlaces = GetDecimalPlaces(); // 3
temporary *= (Math.Pow(10, decimalPlaces)); // moves some decimal places to integer
temporary = (int)temporary; // removes all decimal places
temporary /= (Math.Pow(10, decimalPlaces)); // moves integer back to decimal places
decimal result = original + temporary; // add integer and decimal places together
It can be writen shorter, but this is more descriptive.
EDIT: Short way:
decimal original = GetSomeDecimal(); // 22222.22939393
int decimalPlaces = GetDecimalPlaces(); // 3
decimal result = ((int)original) + (((int)(original * Math.Pow(10, decimalPlaces)) / (Math.Pow(10, decimalPlaces));
What format are you wanting the output?
If you're happy with a string then consider the following C# code:
double num = 3.12345;
num.ToString("G3");
The result will be "3.12".
This link might be of use if you're using .NET. http://msdn.microsoft.com/en-us/library/dwhawy9k.aspx
I hope that helps....but unless you identify than language you are using and the format in which you want the output it is difficult to suggest an appropriate solution.
Try this:
decimal original = GetSomeDecimal(); // 22222.22939393
int number1 = (int)original; // contains only integer value of origina number
decimal temporary = original - number1; // contains only decimal value of original number
int decimalPlaces = GetDecimalPlaces(); // 3
temporary *= (Math.Pow(10, decimalPlaces)); // moves some decimal places to integer
temporary = (int)temporary; // removes all decimal places
temporary /= (Math.Pow(10, decimalPlaces)); // moves integer back to decimal places
decimal result = original + temporary; // add integer and decimal places together
It can be writen shorter, but this is more descriptive.
EDIT: Short way:
decimal original = GetSomeDecimal(); // 22222.22939393
int decimalPlaces = GetDecimalPlaces(); // 3
decimal result = ((int)original) + (((int)(original * Math.Pow(10, decimalPlaces)) / (Math.Pow(10, decimalPlaces));
Source: Stackoverflow.com