C# 十进制(字符串类型)在最后一个字符处四舍五入

Question

这似乎是重复的，但我找不到合适的答案(问题足够接近但是..)我有一个代表十进制数的字符串，它总是有很多小数位，至少 20，有时最多 2000 (代表特定的验证计算，即像'是数字 135 到 147 素数 X 等 - 只是为了给你一些上下文)

例如:

123.829743892473218762329384241002373824970132871283923423961723816273823447623528347662123874999

我正在尝试四舍五入到倒数第二位。我建立了一个(有点)有效的小方法。但是(如上例所示)..如果最后一位数字> 4且倒数第二位数字为9，这意味着我还必须提高前一位数字，并删除最后一位0，如果-那么前一位数字也是9，这意味着我还必须提高前一位数字，依此类推。

例如

123.99999 // must become 124
123.823467283762378  // must become 123.82346728376238
123.823467283762398  // must become 123.8234672837624 (notice the last 0 has gone)
123.09999999 // must become 123.1 (notice no trailing zeros needed..)
122.00000009 // must become 122.0000001
124.81379281 // must become simply 124.8137928
129.07872345 // must become 129.0787235
129.07872344 // must become 129.0787234

等等等等。换句话说，它只是通过仅删除最后一位数字来舍入数字(这是一个字符串!!)，但继续向左舍入直到不需要为止。仅在最后一个 小数 位需要舍入(如果数字是整数则不需要)并且规则是如果最后一位的数字 >4 则数字被截断并且前一个(左边)数字被保留增加 1，忽略尾随零(如果有)，规则继续到整数部分的最后一位(即 123.99999 变为 124，但整数 124 保持原样等)。

有人可以帮我为此构建一个字符串扩展吗？

using System;
    public class Example
    {
        public string round(string LargeDecimal)
        {
            Console.WriteLine("Number as string is: " + LargeDecimal);
            
            int lastDigit = (int)char.GetNumericValue(LargeDecimal[LargeDecimal.Length -1]);   // get last character
            Console.WriteLine("lastDigit = " + lastDigit.ToString());
            
            string number = LargeDecimal.Remove(LargeDecimal.Length - 1);  // delete last character
            Console.WriteLine("Now number is " + number);
            
            if (lastDigit > 4)
            {
                Console.WriteLine("Last digit {0} was >4", lastDigit.ToString());
                int newLastDigit = (int)char.GetNumericValue(number[number.Length -1]);
                Console.WriteLine("Next to left last digit is {0} which will be raised by 1 and become {1}", newLastDigit.ToString(), (newLastDigit +1).ToString());
                newLastDigit += 1;                             //increase by one
                number = number.Remove(number.Length - 1);  // delete ex-penultimate (and now last) character
                number = number + newLastDigit.ToString(); 
                return number;       // and add a digit increased by 1
            }
            else
            {
                return LargeDecimal.Remove(LargeDecimal.Length - 2);
            }
        }
        public Example()
        {}
    }

public class Program
{
    public static void Main(string[] args)
    {
        string myNumber = "124.2398478278268985738276523548769";

        Example myExample = new Example();
    
        string result = myExample.round(myNumber);

        Console.WriteLine("Now I have " + result);
    }
}

Answer 1

嗯，这很棘手，因为有 2000 位小数，您不能使用 decimal。所以也许有比这更简单的东西，但它可能对你有用(.NET fiddle demo):

public static string RoundLongNumber(this string input)
{
    if (!ValidLongNumber(input, out bool isInteger) || isInteger) return input;
    int index = input.IndexOf('.');
    string part1 = input.Remove(index);
    string part2 = input.Substring(index + 1);
    StringBuilder sb = new StringBuilder(part2);

    while(true)
    {
        if (LastInt() <= 4)
        {
            return BuildNumber();
        }

        sb.Length = sb.Length - 1; // remove last
        int lastInt = LastInt() + 1;
        while (lastInt == 10)
        {
            sb.Length = sb.Length - 1;
            if (sb.Length == 0)
            {
                // just integer remaining
                int num = int.Parse(part1);
                return (++num).ToString();
            }
            lastInt = LastInt() + 1;
        }

        sb[sb.Length - 1] = (char)(lastInt + '0');
        if (lastInt != 9) return BuildNumber();
    }

    int LastInt() => sb[sb.Length - 1] - '0';
    string BuildNumber() => part1 + "." + sb.ToString();
}

private static bool ValidLongNumber(string number, out bool isInteger)
{
    isInteger = true;
    if (string.IsNullOrWhiteSpace(number)) return false;
    int pointCount = 0;
    foreach(char c in number)
    {
        bool isDigit = char.IsDigit(c);
        bool isPoint = c == '.';
        if (!isDigit) isInteger = false;
        if (isPoint) pointCount++;
        if(pointCount > 1 || (!isPoint && !isDigit)) return false;
    }
    return true;
}

这是您的示例:

public static void Main()
{
    var strings = new List<string> {
    "123.99999", // must become 124
    "129.99999", // must become 130
    "123.823467283762378",  // must become 123.82346728376238
    "123.823467283762398",  // must become 123.8234672837624 (notice the last 0 has gone)
    "123.09999999", // must become 123.1 (notice no trailing zeros needed..)
    "122.00000009", // must become 122.0000001
    "124.81379281", // must become simply 124.8137928 >>> Why? Should remain same
    "129.07872345", // must become 129.0787235
    "129.07872344", // must become 129.0787234        >>> Why? Should remain same
    };

    IEnumerable<string> results = strings.Select(s => s.RoundLongNumber());
    foreach(var res in results)
    {
        Console.WriteLine(res);
    }
}

请注意，这两个结果是不同的，但要么您没有解释该规则，要么您的期望是错误的:

124.81379281 // must become simply 124.8137928
129.07872344 // must become 129.0787234

为什么？据我了解，两者应该保持不变。