我喜欢使用以下方法(它支持高达tb，这在大多数情况下已经足够了，但它可以很容易地扩展):

private string GetSizeString(long length)
{
    long B = 0, KB = 1024, MB = KB * 1024, GB = MB * 1024, TB = GB * 1024;
    double size = length;
    string suffix = nameof(B);

    if (length >= TB) {
        size = Math.Round((double)length / TB, 2);
        suffix = nameof(TB);
    }
    else if (length >= GB) {
        size = Math.Round((double)length / GB, 2);
        suffix = nameof(GB);
    }
    else if (length >= MB) {
        size = Math.Round((double)length / MB, 2);
        suffix = nameof(MB);
    }
    else if (length >= KB) {
        size = Math.Round((double)length / KB, 2);
        suffix = nameof(KB);
    }

    return $"{size} {suffix}";
}

请记住，这是为c# 6.0(2015)编写的，因此对于较早的版本可能需要进行一些编辑。

那么递归呢:

private static string ReturnSize(double size, string sizeLabel)
{
  if (size > 1024)
  {
    if (sizeLabel.Length == 0)
      return ReturnSize(size / 1024, "KB");
    else if (sizeLabel == "KB")
      return ReturnSize(size / 1024, "MB");
    else if (sizeLabel == "MB")
      return ReturnSize(size / 1024, "GB");
    else if (sizeLabel == "GB")
      return ReturnSize(size / 1024, "TB");
    else
      return ReturnSize(size / 1024, "PB");
  }
  else
  {
    if (sizeLabel.Length > 0)
      return string.Concat(size.ToString("0.00"), sizeLabel);
    else
      return string.Concat(size.ToString("0.00"), "Bytes");
  }
}

然后你称之为:

return ReturnSize(size, string.Empty);

为了获得人类可读的字符串，就像用户在Windows环境中习惯的那样，你应该使用StrFormatByteSize():

using System.Runtime.InteropServices;

...

private long mFileSize;

[DllImport("Shlwapi.dll", CharSet = CharSet.Auto)]
public static extern int StrFormatByteSize(
    long fileSize,
    [MarshalAs(UnmanagedType.LPTStr)] StringBuilder buffer,
    int bufferSize);
    
public string HumanReadableFileSize
{
    get
    {
        var sb = new StringBuilder(20);
        StrFormatByteSize(mFileSize, sb, 20);
        return sb.ToString();
    }
}

我在这里找到了这个: http://csharphelper.com/blog/2014/07/format-file-sizes-in-kb-mb-gb-and-so-forth-in-c/

另一种皮肤的方法，没有任何类型的循环和负大小支持(对文件大小增量有意义):

public static class Format
{
    static string[] sizeSuffixes = {
        "B", "KB", "MB", "GB", "TB", "PB", "EB", "ZB", "YB" };

    public static string ByteSize(long size)
    {
        Debug.Assert(sizeSuffixes.Length > 0);

        const string formatTemplate = "{0}{1:0.#} {2}";

        if (size == 0)
        {
            return string.Format(formatTemplate, null, 0, sizeSuffixes[0]);
        }

        var absSize = Math.Abs((double)size);
        var fpPower = Math.Log(absSize, 1000);
        var intPower = (int)fpPower;
        var iUnit = intPower >= sizeSuffixes.Length
            ? sizeSuffixes.Length - 1
            : intPower;
        var normSize = absSize / Math.Pow(1000, iUnit);

        return string.Format(
            formatTemplate,
            size < 0 ? "-" : null, normSize, sizeSuffixes[iUnit]);
    }
}

下面是测试套件:

[TestFixture] public class ByteSize
{
    [TestCase(0, Result="0 B")]
    [TestCase(1, Result = "1 B")]
    [TestCase(1000, Result = "1 KB")]
    [TestCase(1500000, Result = "1.5 MB")]
    [TestCase(-1000, Result = "-1 KB")]
    [TestCase(int.MaxValue, Result = "2.1 GB")]
    [TestCase(int.MinValue, Result = "-2.1 GB")]
    [TestCase(long.MaxValue, Result = "9.2 EB")]
    [TestCase(long.MinValue, Result = "-9.2 EB")]
    public string Format_byte_size(long size)
    {
        return Format.ByteSize(size);
    }
}

这是我编的，效果很好。

public string[] DetermineDigitalSize(string filename)
        {
            string[] result = new string[2];
            string[] sizes = { "B", "KB", "MB", "GB", "GB" };
            double len = new FileInfo(filename).Length;
             double adjustedSize = len;
            double testSize = 0;
            int order = 0;
            while (order< sizes.Length-1)
            {
                testSize = adjustedSize / 1024;
                if (testSize >= 1) { adjustedSize = testSize; order++; }
                else { break; }
            }
            result[0] = $"{adjustedSize:f2}";
            result[1] = sizes[order];
            return result;
        }

下面是一个Log10的方法:

using System;

class Program {
   static string NumberFormat(double n) {
      var n2 = (int)Math.Log10(n) / 3;
      var n3 = n / Math.Pow(1e3, n2);
      return String.Format("{0:f3}", n3) + new[]{"", " k", " M", " G"}[n2];
   }

   static void Main() {
      var s = NumberFormat(9012345678);
      Console.WriteLine(s == "9.012 G");
   }
}

https://learn.microsoft.com/dotnet/api/system.math.log10