删除小数点，转换为int [英] removing decimal, to an int

问题描述

嗨

我有尝试删除小数的这种方法

Hi

I got this method trying to remove decimal

public static int[] encode(String orig)

       {

           String int_part = " ";

           String dec_part = " ";

           Regex digitsOnly = new Regex(@"[^(\d|\.)]");

           //orig = Regex.Replace(orig,"[^\\.0-8]", ""); //remove non-numerical characters



           int dot_index = orig.Trim().IndexOf("0");



           if (dot_index > 0)

           { //do we have a decimal place?



               int_part = orig.Substring(0, dot_index); //get integer portion of string

               dec_part = orig.Substring(dot_index); //get part after decimal



               int_part = Regex.Replace(int_part, "^0+", ""); //remove leading 0's

               dec_part = Regex.Replace(dec_part, "0+$", ""); //remove trailing 0's



               int[] result = new int[2];

               int.TryParse("0", out result[0]);

               int.TryParse("1", out result[1]);



               return result;

           }

           return new int[] { 0, 1 };

       }

推荐答案

"， ); // 删除尾随的0's int []结果= 新 int [ 2 ]; int .TryParse(" ， 输出结果[ 0 ])； int .TryParse(" ， 输出结果[ 1 ])； 返回结果； } 返回 新 int [] { 0 ， 1 }; }

", ""); //remove trailing 0's int[] result = new int[2]; int.TryParse("0", out result[0]); int.TryParse("1", out result[1]); return result; } return new int[] { 0, 1 }; }

有一种比字符串处理更简单的方法:

There is a simpler way to do this than string manipulation:

1. 删除数字的非数字部分，保留小数点.
2. 使用 ^ ](或 Decimal.TryParse [ ^ ])将您的字符串转换为十进制
3. 使用 Decimal.Truncate [ ^ ]到获取整数部分，其结果是小数，因此如果需要，您需要将其强制转换为int.您还可以使用 Math.Floor [

1. Remove the non-numeric part of the number, preserving the decimal point.
2. Use Decimal.Parse[^] (or Decimal.TryParse[^]) to convert your string into a decimal
3. Use Decimal.Truncate[^] to get the integer part, the result of this is a decimal, so you will need cast to int if needed. You could also use Math.Floor[^] if you want
4. Subtract the integer part from the original value & perform a ToString(). Stripping the leading "0." from this will give you the fractional part as a string

注意将小数部分存储为int可能会导致问题:
7.01-> int = 7，分数字符串="01"如果将01解析为一个int，则将得到1，因此您的数组将为7,1
7.1-> int = 7，分数字符串="1"如果将1解析为一个int，则将得到1，所以您的数组也将是7,1


实际上，我曾经是个白痴:一旦有了十进制"字符串，就可以在其上执行字符串 .split(''.''.''并解析生成的两个元素中的两个元素-element字符串数组:doh:

Note storing the fractional part as an int might cause problems:
7.01 --> int = 7, fratctional string = "01" if you parse 01 to an int, you''ll get 1 so your array would be 7,1
7.1 --> int = 7, fratctional string = "1" if you parse 1 to an int, you''ll get 1 so your array would be 7,1 also


Actualy I''ve been an idiot: once you have the "decimal" string, you can do a string .split(''.'') on it and parse the two elements in the resulting two-element array of string :doh:

不完全知道您遇到了什么问题，我想您想将Int转换为十进制值.因此，为什么不使用诸如convert或parse之类的inbuld方法，这可能对您有用.

谢谢，
Ambesha

Not exactly know that what problem you are getting , I think you want to convert your decimal value in Int. so why not you use inbuld method like convert or parse, that might be useful for you.

Thanks,
Ambesha

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