如何在Java中将字节转换为long？ [英] How do I convert a byte to a long in Java?

问题描述

我正在从硬件设备读取8个字节的数据。我需要将它们转换为数值。我想我想将它们转换为长度，因为它应该适合8个字节。我对Java和低级数据类型操作不是很熟悉。我似乎有两个问题（除了几乎没有相关硬件的文档），字节期望是无符号的，所以我不能进行直接整数转换。我不确定它们是什么字节序。

I am reading 8 bytes of data in from a hardware device. I need to convert them into a numeric value. I think I want to convert them to a long as that should fit 8 bytes. I am not very familiar with Java and low level data type operations. I seem to have two problems (apart from the fact there is almost no documentation for the hardware in question), The bytes are expecting to be unsigned, so I can't do a straight integer conversion. I am not sure what endianness they are.

任何建议都将受到赞赏。

Any advice would be appreciated.

结束了这个（取自一些我可能应该在一周前读过的一些源代码）：

Ended up with this (taken from some source code I probably should have read a week ago):

public static final long toLong (byte[] byteArray, int offset, int len)
{
   long val = 0;
   len = Math.min(len, 8);
   for (int i = (len - 1); i >= 0; i--)
   {
      val <<= 8;
      val |= (byteArray [offset + i] & 0x00FF);
   }
   return val;
}

推荐答案

对于字节序，请测试你知道一些数字，然后你将使用字节移位将它们移动到长整数。

For the endianness, test with some numbers you know, and then you will be using a byte shifting to move them into the long.

你可能会发现这是一个起点。
http://www.janeg.ca/scjp/oper/shift。 html

You may find this to be a starting point. http://www.janeg.ca/scjp/oper/shift.html

难点在于取决于结尾会改变你的方式，但你会改变24,16,8然后加上最后一个，基本上，如果做32位，但你会更长，所以只需做额外的转移。

The difficulty is that depending on the endianess will change how you do it, but you will shift by 24, 16, 8 then add the last one, basically, if doing 32 bits, but you are going longer, so just do extra shifting.

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