下面是从aioobe转换到Kotlin的转换:

/**
 * https://stackoverflow.com/a/3758880/1006741
 */
fun Long.humanReadableByteCountBinary(): String {
    val b = when (this) {
        Long.MIN_VALUE -> Long.MAX_VALUE
        else -> abs(this)
    }
    return when {
        b < 1024L -> "$this B"
        b <= 0xfffccccccccccccL shr 40 -> "%.1f KiB".format(Locale.UK, this / 1024.0)
        b <= 0xfffccccccccccccL shr 30 -> "%.1f MiB".format(Locale.UK, this / 1048576.0)
        b <= 0xfffccccccccccccL shr 20 -> "%.1f GiB".format(Locale.UK, this / 1.073741824E9)
        b <= 0xfffccccccccccccL shr 10 -> "%.1f TiB".format(Locale.UK, this / 1.099511627776E12)
        b <= 0xfffccccccccccccL -> "%.1f PiB".format(Locale.UK, (this shr 10) / 1.099511627776E12)
        else -> "%.1f EiB".format(Locale.UK, (this shr 20) / 1.099511627776E12)
    }
}

有趣的事实:这里发布的原始代码片段是Stack Overflow上被复制最多的Java代码片段，它是有缺陷的。它被修好了，但却变得一团糟。 本文的完整故事:有史以来复制最多的堆栈溢出代码片段是有缺陷的!

来源:格式化字节大小到人类可读的格式|编程。指南

SI（1 k = 1，000）

public static String humanReadableByteCountSI(long bytes) {
    if (-1000 < bytes && bytes < 1000) {
        return bytes + " B";
    }
    CharacterIterator ci = new StringCharacterIterator("kMGTPE");
    while (bytes <= -999_950 || bytes >= 999_950) {
        bytes /= 1000;
        ci.next();
    }
    return String.format("%.1f %cB", bytes / 1000.0, ci.current());
}

二进制(1's = 1,024)

public static String humanReadableByteCountBin(long bytes) {
    long absB = bytes == Long.MIN_VALUE ? Long.MAX_VALUE : Math.abs(bytes);
    if (absB < 1024) {
        return bytes + " B";
    }
    long value = absB;
    CharacterIterator ci = new StringCharacterIterator("KMGTPE");
    for (int i = 40; i >= 0 && absB > 0xfffccccccccccccL >> i; i -= 10) {
        value >>= 10;
        ci.next();
    }
    value *= Long.signum(bytes);
    return String.format("%.1f %ciB", value / 1024.0, ci.current());
}

示例输出:

                             SI     BINARY

                  0:        0 B        0 B
                 27:       27 B       27 B
                999:      999 B      999 B
               1000:     1.0 kB     1000 B
               1023:     1.0 kB     1023 B
               1024:     1.0 kB    1.0 KiB
               1728:     1.7 kB    1.7 KiB
             110592:   110.6 kB  108.0 KiB
            7077888:     7.1 MB    6.8 MiB
          452984832:   453.0 MB  432.0 MiB
        28991029248:    29.0 GB   27.0 GiB
      1855425871872:     1.9 TB    1.7 TiB
9223372036854775807:     9.2 EB    8.0 EiB   (Long.MAX_VALUE)

我最近问了同样的问题:

格式文件大小为MB, GB等。

虽然没有开箱即用的答案，但我可以接受这个解决方案:

private static final long K = 1024;
private static final long M = K * K;
private static final long G = M * K;
private static final long T = G * K;

public static String convertToStringRepresentation(final long value){
    final long[] dividers = new long[] { T, G, M, K, 1 };
    final String[] units = new String[] { "TB", "GB", "MB", "KB", "B" };
    if(value < 1)
        throw new IllegalArgumentException("Invalid file size: " + value);
    String result = null;
    for(int i = 0; i < dividers.length; i++){
        final long divider = dividers[i];
        if(value >= divider){
            result = format(value, divider, units[i]);
            break;
        }
    }
    return result;
}

private static String format(final long value,
    final long divider,
    final String unit){
    final double result =
        divider > 1 ? (double) value / (double) divider : (double) value;
    return new DecimalFormat("#,##0.#").format(result) + " " + unit;
}

测试代码:

public static void main(final String[] args){
    final long[] l = new long[] { 1l, 4343l, 43434334l, 3563543743l };
    for(final long ll : l){
        System.out.println(convertToStringRepresentation(ll));
    }
}

输出(在我的德语地区):

1 B
4,2 KB
41,4 MB
3,3 GB

我已经打开了一个问题，要求谷歌番石榴的这个功能。也许有人愿意支持它。

filename=filedilg.getSelectedFile().getAbsolutePath();
File file=new File(filename);

String disp=FileUtils.byteCountToDisplaySize(file.length());
System.out.println("THE FILE PATH IS "+file+"THIS File SIZE IS IN MB "+disp);

private static final String[] Q = new String[]{"", "K", "M", "G", "T", "P", "E"};

public String getAsString(long bytes)
{
    for (int i = 6; i > 0; i--)
    {
        double step = Math.pow(1024, i);
        if (bytes > step) return String.format("%3.1f %s", bytes / step, Q[i]);
    }
    return Long.toString(bytes);
}

试试JSR 363。它的单元扩展模块，如Unicode CLDR(在GitHub: uom-systems中)，为您完成所有这些。

你可以使用每个实现中包含的MetricPrefix或BinaryPrefix(与上面的一些例子相比)，如果你在印度或附近的国家生活和工作，IndianPrefix(也在uom-系统的公共模块中)允许你使用和格式化“千万字节”或“Lakh字节”。