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;
        }

如果你试图匹配Windows资源管理器的详细信息视图中显示的大小，这是你想要的代码:

[DllImport("shlwapi.dll", CharSet = CharSet.Unicode)]
private static extern long StrFormatKBSize(
    long qdw,
    [MarshalAs(UnmanagedType.LPTStr)] StringBuilder pszBuf,
    int cchBuf);

public static string BytesToString(long byteCount)
{
    var sb = new StringBuilder(32);
    StrFormatKBSize(byteCount, sb, sb.Capacity);
    return sb.ToString();
}

这不仅会与资源管理器完全匹配，而且还会为您提供翻译后的字符串，并匹配Windows版本的差异(例如在Win10中，K = 1000 vs.之前的版本K = 1024)。

又多了一种方法，不管怎样。我喜欢上面提到的@humbads优化解决方案，所以复制了原理，但我实现了一点不同。

我认为它是否应该是一个扩展方法是有争议的(因为不是所有的long都必须是字节大小)，但我喜欢它们，当我下次需要它时，我可以在某个地方找到它!

关于单位，我想我从来没有说过“Kibibyte”或“Mebibyte”，虽然我对这种强制而非进化的标准持怀疑态度，但我认为从长远来看，这将避免混淆。

public static class LongExtensions
{
    private static readonly long[] numberOfBytesInUnit;
    private static readonly Func<long, string>[] bytesToUnitConverters;

    static LongExtensions()
    {
        numberOfBytesInUnit = new long[6]    
        {
            1L << 10,    // Bytes in a Kibibyte
            1L << 20,    // Bytes in a Mebibyte
            1L << 30,    // Bytes in a Gibibyte
            1L << 40,    // Bytes in a Tebibyte
            1L << 50,    // Bytes in a Pebibyte
            1L << 60     // Bytes in a Exbibyte
        };

        // Shift the long (integer) down to 1024 times its number of units, convert to a double (real number), 
        // then divide to get the final number of units (units will be in the range 1 to 1023.999)
        Func<long, int, string> FormatAsProportionOfUnit = (bytes, shift) => (((double)(bytes >> shift)) / 1024).ToString("0.###");

        bytesToUnitConverters = new Func<long,string>[7]
        {
            bytes => bytes.ToString() + " B",
            bytes => FormatAsProportionOfUnit(bytes, 0) + " KiB",
            bytes => FormatAsProportionOfUnit(bytes, 10) + " MiB",
            bytes => FormatAsProportionOfUnit(bytes, 20) + " GiB",
            bytes => FormatAsProportionOfUnit(bytes, 30) + " TiB",
            bytes => FormatAsProportionOfUnit(bytes, 40) + " PiB",
            bytes => FormatAsProportionOfUnit(bytes, 50) + " EiB",
        };
    }

    public static string ToReadableByteSizeString(this long bytes)
    {
        if (bytes < 0)
            return "-" + Math.Abs(bytes).ToReadableByteSizeString();

        int counter = 0;
        while (counter < numberOfBytesInUnit.Length)
        {
            if (bytes < numberOfBytesInUnit[counter])
                return bytesToUnitConverters[counter](bytes);
            counter++;
        }
        return bytesToUnitConverters[counter](bytes);
    }
}

下面是@ deepe1的BigInteger版本的答案，它绕过了long的大小限制(因此支持yottabyte和理论上的任何后面的限制):

public static string ToBytesString(this BigInteger byteCount, string format = "N3")
{
    string[] suf = { "B", "KiB", "MiB", "GiB", "TiB", "PiB", "EiB", "YiB" };
    if (byteCount.IsZero)
    {
        return $"{0.0.ToString(format)} {suf[0]}";
    }

    var abs = BigInteger.Abs(byteCount);
    var place = Convert.ToInt32(Math.Floor(BigInteger.Log(abs, 1024)));
    var pow = Math.Pow(1024, place);

    // since we need to do this with integer math, get the quotient and remainder
    var quotient = BigInteger.DivRem(abs, new BigInteger(pow), out var remainder);
    // convert the remainder to a ratio and add both back together as doubles
    var num = byteCount.Sign * (Math.Floor((double)quotient) + ((double)remainder / pow));

    return $"{num.ToString(format)} {suf[place]}";
}

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

int size = new FileInfo( filePath ).Length / 1024;
string humanKBSize = string.Format( "{0} KB", size );
string humanMBSize = string.Format( "{0} MB", size / 1024 );
string humanGBSize = string.Format( "{0} GB", size / 1024 / 1024 );