比如@NET3的解决方案。使用shift而不是除法来测试字节的范围，因为除法占用更多的CPU成本。

private static readonly string[] UNITS = new string[] { "B", "KB", "MB", "GB", "TB", "PB", "EB" };

public static string FormatSize(ulong bytes)
{
    int c = 0;
    for (c = 0; c < UNITS.Length; c++)
    {
        ulong m = (ulong)1 << ((c + 1) * 10);
        if (bytes < m)
            break;
    }

    double n = bytes / (double)((ulong)1 << (c * 10));
    return string.Format("{0:0.##} {1}", n, UNITS[c]);
}

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);
    }
}

使用日志解决问题....

static String BytesToString(long byteCount)
{
    string[] suf = { "B", "KB", "MB", "GB", "TB", "PB", "EB" }; //Longs run out around EB
    if (byteCount == 0)
        return "0" + suf[0];
    long bytes = Math.Abs(byteCount);
    int place = Convert.ToInt32(Math.Floor(Math.Log(bytes, 1024)));
    double num = Math.Round(bytes / Math.Pow(1024, place), 1);
    return (Math.Sign(byteCount) * num).ToString() + suf[place];
}

同样是在c#中，但是转换起来应该很简单。为了便于阅读，我还四舍五入到小数点后1位。

基本上确定以1024为基数的小数位数，然后除以1024^小数位。

以及一些使用和输出的示例:

Console.WriteLine(BytesToString(9223372036854775807));  //Results in 8EB
Console.WriteLine(BytesToString(0));                    //Results in 0B
Console.WriteLine(BytesToString(1024));                 //Results in 1KB
Console.WriteLine(BytesToString(2000000));              //Results in 1.9MB
Console.WriteLine(BytesToString(-9023372036854775807)); //Results in -7.8EB

编辑: 有人指出我漏了一门数学课。地板，所以我把它合并了。(转换。ToInt32使用舍入，而不是截断，这就是为什么需要使用Floor。)谢谢你的帮助。

Edit2: 有一些关于负大小和0字节大小的注释，所以我更新以处理这些情况。

这个问题很老了，但是一个非常快速的c#函数可以是:

public static string PrettyPrintBytes(long numBytes)
{
    if (numBytes < 1024)
        return $"{numBytes} B";
            
    if (numBytes < 1048576)
        return $"{numBytes / 1024d:0.##} KB";

    if (numBytes < 1073741824)
        return $"{numBytes / 1048576d:0.##} MB";

    if (numBytes < 1099511627776)
        return $"{numBytes / 1073741824d:0.##} GB";

    if (numBytes < 1125899906842624)
        return $"{numBytes / 1099511627776d:0.##} TB";
            
    if (numBytes < 1152921504606846976)
        return $"{numBytes / 1125899906842624d:0.##} PB";

    return $"{numBytes / 1152921504606846976d:0.##} EB";
}

每次调用只有一次强制转换和一次除法，最多只能进行6次比较。在进行基准测试时，我发现字符串插值比使用string . format()快得多。

string[] suffixes = { "B", "KB", "MB", "GB", "TB", "PB", "EB", "ZB", "YB" };
int s = 0;
long size = fileInfo.Length;

while (size >= 1024)
{
    s++;
    size /= 1024;
}

string humanReadable = String.Format("{0} {1}", size, suffixes[s]);

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";
}