c# - 解析小数并过滤右边多余的0？

Question

我从一个 XML 文件中收到以下格式的小数:

1.132000
6.000000

目前我正在像这样使用 Decimal.Parse:

decimal myDecimal = Decimal.Parse(node.Element("myElementName").Value, System.Globalization.CultureInfo.InvariantCulture);

• 如何将 myDecimal 打印成字符串 看起来像下面？

  1.132
  6
  

Answer 1

我不认为有任何标准的数字格式字符串总是会忽略尾随的无意义的零，恐怕。

您可以尝试编写自己的十进制规范化方法，但这可能非常棘手。使用 .NET 4 中的 BigInteger 类，这将是合理可行的，但如果没有它(或类似的东西)，它确实会非常困难。

编辑:好的，我想这就是你想要的:

using System;
using System.Numerics;

public static class DecimalExtensions
{
    // Avoiding implicit conversions just for clarity
    private static readonly BigInteger Ten = new BigInteger(10);
    private static readonly BigInteger UInt32Mask = new BigInteger(0xffffffffU);

    public static decimal Normalize(this decimal input)
    {
        unchecked
        {
            int[] bits = decimal.GetBits(input);
            BigInteger mantissa = 
                new BigInteger((uint) bits[0]) +
                (new BigInteger((uint) bits[1]) << 32) +
                (new BigInteger((uint) bits[2]) << 64);

            int sign = bits[3] & int.MinValue;            
            int exponent = (bits[3] & 0xff0000) >> 16;

            // The loop condition here is ugly, because we want
            // to do both the DivRem part and the exponent check :(
            while (exponent > 0)
            {
                BigInteger remainder;
                BigInteger divided = BigInteger.DivRem(mantissa, Ten, out remainder);
                if (remainder != BigInteger.Zero)
                {
                    break;
                }
                exponent--;
                mantissa = divided;
            }
            // Okay, now put it all back together again...
            bits[3] = (exponent << 16) | sign;
            // For each 32 bits, convert the bottom 32 bits into a uint (which won't
            // overflow) and then cast to int (which will respect the bits, which
            // is what we want)
            bits[0] = (int) (uint) (mantissa & UInt32Mask);
            mantissa >>= 32;
            bits[1] = (int) (uint) (mantissa & UInt32Mask);
            mantissa >>= 32;
            bits[2] = (int) (uint) (mantissa & UInt32Mask);

            return new decimal(bits);
        }
    }

    class Program
    {
        static void Main(string[] args)
        {
            Check(6.000m);
            Check(6000m);
            Check(6m);
            Check(60.00m);
            Check(12345.00100m);
            Check(-100.00m);
        }

        static void Check(decimal d)
        {
            Console.WriteLine("Before: {0}  -  after: {1}", d, d.Normalize());
        }
    }
}