我使用下面的Long扩展方法将其转换为人类可读的大小字符串。这个方法是在Stack Overflow上发布的相同问题的Java解决方案的c#实现。

/// <summary>
/// Convert a byte count into a human readable size string.
/// </summary>
/// <param name="bytes">The byte count.</param>
/// <param name="si">Whether or not to use SI units.</param>
/// <returns>A human readable size string.</returns>
public static string ToHumanReadableByteCount(
    this long bytes
    , bool si
)
{
    var unit = si
        ? 1000
        : 1024;

    if (bytes < unit)
    {
        return $"{bytes} B";
    }

    var exp = (int) (Math.Log(bytes) / Math.Log(unit));

    return $"{bytes / Math.Pow(unit, exp):F2} " +
           $"{(si ? "kMGTPE" : "KMGTPE")[exp - 1] + (si ? string.Empty : "i")}B";
}

这可能不是最有效或最优化的方法，但如果您不熟悉对数数学，它更容易阅读，并且对于大多数情况来说应该足够快。

string[] sizes = { "B", "KB", "MB", "GB", "TB" };
double len = new FileInfo(filename).Length;
int order = 0;
while (len >= 1024 && order < sizes.Length - 1) {
    order++;
    len = len/1024;
}

// Adjust the format string to your preferences. For example "{0:0.#}{1}" would
// show a single decimal place, and no space.
string result = String.Format("{0:0.##} {1}", len, sizes[order]);

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 );

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

1-liner(加上前缀常量)

const String prefixes = " KMGTPEY";
/// <summary> Returns the human-readable file size for an arbitrary, 64-bit file size. </summary>
public static String HumanSize(UInt64 bytes)
    => Enumerable
    .Range(0, prefixes.Length)
    .Where(i => bytes < 1024U<<(i*10))
    .Select(i => $"{(bytes>>(10*i-10))/1024:0.###} {prefixes[i]}B")
    .First();

或者，如果你想减少LINQ对象的分配，使用相同的for循环变量:

/// <summary>
/// Returns the human-readable file size for an arbitrary, 64-bit file size.
/// </summary>
public static String HumanSize(UInt64 bytes)
{
    const String prefixes = " KMGTPEY";
    for (var i = 0; i < prefixes.Length; i++)
        if (bytes < 1024U<<(i*10))
            return $"{(bytes>>(10*i-10))/1024:0.###} {prefixes[i]}B";

    throw new ArgumentOutOfRangeException(nameof(bytes));
}