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);
    }
}

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);
    }
}

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 );

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;
        }

我猜你要找的是“1.4 MB”而不是“1468006字节”?

我不认为在。net中有内置的方法来做到这一点。你只需要找出哪个单元是合适的，然后格式化它。

编辑:这里有一些示例代码来做到这一点:

http://www.codeproject.com/KB/cpp/formatsize.aspx