.net 截断十进制数不四舍五入
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Truncate Decimal number not Round Off
提问by ManojN
Possible Duplicate:
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
我想截断小数如下
i.e.
IE
- 2.22939393 -> 2.229
- 2.22977777 -> 2.229
- 2.22939393 -> 2.229
- 2.22977777 -> 2.229
采纳答案by Bill the Lizard
double d = 2.22977777;
d = ( (double) ( (int) (d * 1000.0) ) ) / 1000.0 ;
Of course, this won't work if you're trying to truncate rounding error, but it should work fine with the values you give in your examples. See the first two answers to this questionfor details on why it won't work sometimes.
当然，如果您试图截断舍入误差，这将不起作用，但它应该适用于您在示例中给出的值。有关为什么有时不起作用的详细信息，请参阅此问题的前两个答案。
回答by CMS
You can use Math.Round:
您可以使用Math.Round：
decimal rounded = Math.Round(2.22939393, 3); //Returns 2.229
Or you can use ToString with the N3 numeric format.
或者您可以将 ToString 与 N3数字格式一起使用。
string roundedNumber = number.ToString("N3");
EDIT:Since you don't want rounding, you can easily use Math.Truncate:
编辑：由于您不想四舍五入，您可以轻松使用Math.Truncate：
Math.Truncate(2.22977777 * 1000) / 1000; //Returns 2.229
回答by Carl H?rberg
A function to truncate an arbitrary number of decimals:
截断任意小数位数的函数：
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;
}
回答by Glenn Slayden
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.
这是一个不受整数溢出影响的扩展方法（就像上面的一些答案一样）。它还缓存了一些 10 的幂以提高效率。
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;
}
回答by Jim
This is similar to TcKs suggestion above, but using math.truncate rather than int conversions
这类似于上面的 TcKs 建议，但使用 math.truncate 而不是 int 转换
VB: but you'll get the idea
VB：但你会明白的
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
回答by Pavlo Neiman
Try this
尝试这个
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);
回答by magicrebirth
Maybe another quick solution could be:
也许另一个快速解决方案可能是：
>>> float("%.1f" % 1.00001)
1.0
>>> float("%.3f" % 1.23001)
1.23
>>> float("%.5f" % 1.23001)
1.23001
回答by mezoid
What format are you wanting the output?
你想要什么格式的输出？
If you're happy with a string then consider the following C# code:
如果您对字符串感到满意，请考虑以下 C# 代码：
double num = 3.12345;
num.ToString("G3");
The result will be "3.12".
结果将是“3.12”。
This link might be of use if you're using .NET. http://msdn.microsoft.com/en-us/library/dwhawy9k.aspx
如果您使用的是 .NET，则此链接可能有用。 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.
我希望这会有所帮助....但是除非您确定您使用的语言和您想要的输出格式，否则很难提出适当的解决方案。
回答by TcKs
Try this:
尝试这个：
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.
它可以写得更短，但这更具描述性。
EDIT:Short way:
编辑：简短的方法：
decimal original = GetSomeDecimal(); // 22222.22939393
int decimalPlaces = GetDecimalPlaces(); // 3
decimal result = ((int)original) + (((int)(original * Math.Pow(10, decimalPlaces)) / (Math.Pow(10, decimalPlaces));
回答by user274105
Forget Everything just check out this
忘记一切，看看这个
double num = 2.22939393;
num = Convert.ToDouble(num.ToString("#0.000"));
