大端字节序的位掩码
这不是一个问题，而是一个健全性检查！
如果您需要将 4 个字节作为 Big endian 中的位掩码读入 Java，这些字节为：
0x00、0x01、0xB6、0x02。
将其转换为 int 将为： 112130
二进制为： 00000000000000011010011000000010
一系列字节的字节序不会影响位位置，不是吗？
谢谢托尼
This isn't a question as much as it's a sanity check!
If you needed to read 4 bytes into Java as a bitmask in Big endian and those bytes were:
0x00, 0x01, 0xB6, 0x02.
Making that into an int would be: 112130
The binary would be: 00000000000000011010011000000010
The endian of a series of bytes wouldn't affect the bit position, would it?
Thanks
Tony
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字节顺序反映了字节的顺序，但不反映这些字节内位的顺序。
假设我想表示（两字节）字 0x9001。
如果我只是以二进制形式输入，则为 1001000000000001。
如果我将字节（从低地址到高地址）转储到大端机器上，我会看到
10010000 00000001。如果我在小端机器上转储字节（从低地址到高地址），我会看到
00000001 10010000。Endian-ness reflects the ordering of bytes, but not the ordering of the bits within those bytes.
Let's say I want to represent the (two-byte) word 0x9001.
If I just type this out in binary, that would be 1001000000000001.
If I dump the bytes (from lower address to higher) on a big-endian machine, I would see
10010000 00000001.If I dump the bytes (from lower address to higher) on a little-endian machine, I would see
00000001 10010000.一般来说，如果您正在读取的内容为您提供了整个字节，那么您无需担心组成这些字节的位的顺序：正如您正确假设的那样，重要的是字节的顺序。
您可能需要担心各个位的“字节顺序”的时间是您实际读取/写入位流而不是整个字节的时候（例如，如果您正在编写在位级别运行的压缩算法，那么您' d 必须决定以什么顺序写入位）。
In general, if the thing you're reading from is giving you whole bytes, then you don't need to worry about the order of bits making up those bytes: it is just the order of the bytes that matters, as you correctly suppose.
The time you might have to worry about the "endianness" of individual bits is where you're actually reading/writing a stream of bits rather than whole bytes (e.g. if you were writing a compression algorithm that operated at the bit level, you'd have to make a decision about what order to write the bits in).
您唯一需要注意的是如何准确地“将 4 个字节读入 Java”——这就是字节序很重要的地方，您可能会搞砸它（
DataInputStream假定为大字节序）。一旦你读到的值变成了 int 112130，你就设置好了。The only thing you have to pay attention is how exactly you "read 4 bytes into Java" - that's where endianness matters and you can mess it up (
DataInputStreamassumes big endian). Once the value you've read has become the int 112130, you're set.