在Java中将long [64]转换为byte [512]？ [英] Converting long[64] to byte[512] in Java?

问题描述

我正在将进程移植到Java。在C＃和C ++中已经有了工作版本。

I'm porting a process over to Java. There's already working versions in C# and C++.

我在C＃中有一个部分，我做Marshal.Copy（...）将64个ulongs转换为512个字节C ++中的一行我使用memmove（...）来做同样的事情。 Java中有哪些可以实现相同的结果？我需要相同的二进制信息，只需要字节而不是长字。

I have a section in C# that I do Marshal.Copy(...) to convert 64 ulongs to 512 bytes and that line in C++ I use memmove(...) to do the same thing. What is available in Java to achieve the same result? I need the same binary information in the same order just as bytes instead of longs.

编辑：

我移植到Java的原因是利用Java自然具有的可移植性。我不想使用本机代码。

The reason I'm porting to Java is to take advantage of the portability that Java naturally has. I would not like to use native code.

另一件事。由于Java不包含无符号值，因此我需要稍微改变一下我要求的内容。我想从64个long（C＃和C ++中的ulongs）中的每一个获得8个无符号字节值，以便稍后我可以在数组中的索引处使用这些值。这需要发生数千次，所以最快的方式是最好的方式。

Another thing. Since Java doesn't contain unsigned values, then I need to change what I'm requesting by just a little. I would like to attain the 8 unsigned byte values from each of the 64 longs (ulongs in C# and C++) so that I can use those values at indices in arrays later. This needs to happen thousands of times so the fastest way would be the best way.

推荐答案

ByteBuffer 效果很好为此：只需输入64 long 值并使用获取 byte [] out array（）方法。 ByteOrder 类可以有效地处理字节序问题。例如，结合 wierob 评论中建议的方法：

ByteBuffer works well for this: just put in 64 long values and get a byte[] out using the array() method. The ByteOrder class can handle endian issues effectively. For example, incorporating the approach suggested in a comment by wierob:

private static byte[] xform(long[] la, ByteOrder order) {
    ByteBuffer bb = ByteBuffer.allocate(la.length * 8);
    bb.order(order);
    bb.asLongBuffer().put(la);
    return bb.array();
}

附录：产生的字节[] 组件是有符号的8位值，但Java数组需要非负整数索引值。将字节转换为 int 将导致符号扩展，但屏蔽较高位的位将给出无符号值 byte b ：

Addendum: The resulting byte[] components are signed, 8-bit values, but Java arrays require nonnegative integer index values. Casting a byte to an int will result in sign extension, but masking the higher order bits will give the unsigned value of byte b:

int i = (int) b & 0xFF;

这 answer 详细说明适用的运营商优先级规则。此相关答案演示了 double 值的类似方法。

This answer elaborates on the applicable operator precedence rules. This related answer demonstrates a similar approach for double values.

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