如何在java中正确获取little-endian整数

问题描述：

我需要得到64位小端整数作为字节数组，高32位归零，低32位包含一些整数，假设它是51。

I need to get 64-bit little-endian integer as byte-array with upper 32 bits zeroed and lower 32 bits containing some integer number, let say it's 51.

现在我这样做了：

 byte[] header = ByteBuffer
            .allocate(8)
            .order(ByteOrder.LITTLE_ENDIAN)
            .putInt(51)
            .array();

但我不确定这是不是正确的方法。我做得对吗？

But I'm not sure is it the right way. Am I doing it right?

答

如何尝试以下方法：

private static byte[] encodeHeader(long size) {
    if (size < 0 || size >= (1L << Integer.SIZE)) {
        throw new IllegalArgumentException("size negative or larger than 32 bits: " + size);
    }

    byte[] header = ByteBuffer
            .allocate(Long.BYTES)
            .order(ByteOrder.LITTLE_ENDIAN)
            .putInt((int) size)
            .array();
    return header;
}

就我个人而言，我认为它更清晰，你可以使用完整的32位。

Personally this I think it's even more clear and you can use the full 32 bits.

我忽略了这里的标志，你可以单独传递这些标志。我已经改变了答案，缓冲区的位置放在大小的末尾。

I'm disregarding the flags here, you could pass those separately. I've changed the answer in such a way that the position of the buffer is placed at the end of the size.

如何在java中正确获取little-endian整数

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