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。

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);
    }
}

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));
}

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]}";
}

请求函数的测试和显著优化版本发布在这里:

c#人类可读文件大小-优化的函数

源代码:

// Returns the human-readable file size for an arbitrary, 64-bit file size 
// The default format is "0.### XB", e.g. "4.2 KB" or "1.434 GB"
public string GetBytesReadable(long i)
{
    // Get absolute value
    long absolute_i = (i < 0 ? -i : i);
    // Determine the suffix and readable value
    string suffix;
    double readable;
    if (absolute_i >= 0x1000000000000000) // Exabyte
    {
        suffix = "EB";
        readable = (i >> 50);
    }
    else if (absolute_i >= 0x4000000000000) // Petabyte
    {
        suffix = "PB";
        readable = (i >> 40);
    }
    else if (absolute_i >= 0x10000000000) // Terabyte
    {
        suffix = "TB";
        readable = (i >> 30);
    }
    else if (absolute_i >= 0x40000000) // Gigabyte
    {
        suffix = "GB";
        readable = (i >> 20);
    }
    else if (absolute_i >= 0x100000) // Megabyte
    {
        suffix = "MB";
        readable = (i >> 10);
    }
    else if (absolute_i >= 0x400) // Kilobyte
    {
        suffix = "KB";
        readable = i;
    }
    else
    {
        return i.ToString("0 B"); // Byte
    }
    // Divide by 1024 to get fractional value
    readable = (readable / 1024);
    // Return formatted number with suffix
    return readable.ToString("0.### ") + suffix;
}

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。