如何在以下代码中舍入数字 [英] How to round off the numbers in the following code

问题描述

string[] final = new string[result.Length];             

                for (int splitCount = 0; splitCount < result.Length; splitCount++)
                {
                    MatchCollection match = Regex.Matches(result[splitCount], @"\d+(\.\d+)?");

                    foreach (Match m in match)
                    {
                        final[splitCount] += (Double.Parse(m.Value, new System.Globalization.NumberFormatInfo() { NumberDecimalSeparator = "." })).ToString()+"            ";
                    }
                }

如何在上面的代码中将double值四舍五入为4位数

How to round off the double value to 4 digits in the above code

推荐答案

string[] final = new string[result.Length];             
for (int splitCount = 0; splitCount < result.Length; splitCount++)
{
  MatchCollection match = Regex.Matches(result[splitCount], @"\d+(?:\.\d*)?");
  final[splitCount] = string.Empty;      // Don't forget to initialize the entries in final.
  foreach (Match m in match)
  {
    final[splitCount] += Double.Parse(m.Value, System.Globalization.CultureInfo.InvariantCulture).ToString("F4")+"            ";
  }
}

只需使用.ToString（）中的格式信息来控制精度。

I更改了正则表达式，以便它是一个非捕获组并允许小数点而不跟随数字。例如，1234。

Just use the format info in the .ToString() to control the precision.
I changed the regex so that it was a noncapturing group and to allow for decimal point without following digits. E.g., "1234."

使用 Math.Round [ ^ ]功能。

int round（双号）

{

返回（数字> = 0）？ （int）（数字+ 0.5）:( int）（数字 - 0.5）;

}







使用Long而不是Int如果你有很长的否



另一种方式是



double a = 2.333333;

System.Math.Round（a）

使用数学函数

int round(double number)
{
return (number >= 0) ? (int)(number + 0.5) : (int)(number - 0.5);
}



Use Long Instead Of Int If You have A long No

And Another Way Is

double a = 2.333333;
System.Math.Round(a)
Use Math Funtion

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