问题描述
所以昨天在工作中，我不得不编写一个应用程序来计算 AFP 文件中的页数.所以我翻阅了我的 MO:DCA 规范 PDF 并找到了结构化字段 BPG (Begin Page) 及其 3 字节标识符.该应用程序需要在 AIX 机器上运行，因此我决定用 Java 编写它.
So at work yesterday, I had to write an application to count the pages in an AFP file. So I dusted off my MO:DCA spec PDF and found the structured field BPG (Begin Page) and its 3-byte identifier. The app needs to run on an AIX box, so I decided to write it in Java.
为了获得最大效率，我决定读取每个结构化字段的前 6 个字节，然后跳过该字段中的剩余字节.这会让我:
For maximum efficiency, I decided that I would read the first 6 bytes of each structured field and then skip the remaining bytes in the field. This would get me:
0: Start of field byte
1-2: 2-byte length of field
3-5: 3-byte sequence identifying the type of field
所以我检查字段类型，如果它是 BPG，我会增加一个页面计数器，如果不是，我不会.然后我跳过字段中的剩余字节而不是通读它们.在这里，在跳过(实际上是在字段长度中)是我发现 Java 使用有符号字节的地方.
So I check the field type and increment a page counter if it's BPG, and I don't if it's not. Then I skip the remaining bytes in the field rather than read through them. And here, in the skipping (and really in the field length) is where I discovered that Java uses signed bytes.
我做了一些谷歌搜索，发现了很多有用的信息.当然，最有用的指令是按位执行 & 到 0xff 以获得 unsigned int 值.这对于我获得可用于计算要跳过的字节数的长度是必要的.
I did some googling and found quite a bit of useful information. Most useful, of course, was the instruction to do a bitwise & to 0xff to get the unsigned int value. This was necessary for me to get a length that could be used in the calculation for the number of bytes to skip.
我现在知道在 128 时，我们从 -128 开始倒数.我想知道的是这里的按位运算是如何工作的——更具体地说，我是如何得出负数的二进制表示的.
I now know that at 128, we start counting backwards from -128. What I want to know is how the bitwise operation works here--more specifically, how I arrive at the binary representation for a negative number.
如果我正确理解按位 & ，则您的结果等于一个仅设置两个数字的公共位的数字.所以假设 byte b = -128，我们会有:
If I understand the bitwise & properly, your result is equal to a number where only the common bits of your two numbers are set. So assuming byte b = -128, we would have:
b & 0xff // 128
1000 0000-128
1111 1111 255
---------
1000 0000 128
那么我如何以 -128 获得 1000 0000?我如何获得像 -72 或 -64 这样不太明显的东西的二进制表示?
So how would I arrive at 1000 0000 for -128? How would I get the binary representation of something less obvious like -72 or -64?
推荐答案
为了获得负数的二进制表示，你需要计算二进制补码:
In order to obtain the binary representation of a negative number you calculate two's complement:
- 获取正数的二进制表示
- 反转所有位
- 添加一个
我们以-72为例:
0100 1000 72
1011 0111 All bits inverted
1011 1000 Add one
所以-72的二进制(8位)表示是10111000.
So the binary (8-bit) representation of -72 is 10111000.
实际发生在您身上的是:您的文件有一个值为 10111000 的字节.当解释为无符号字节(这可能是您想要的)时，这是 88.
What is actually happening to you is the following: You file has a byte with value 10111000. When interpreted as an unsigned byte (which is probably what you want), this is 88.
在Java中，当这个字节被用作一个int时(例如因为read()返回一个int，或者因为隐式提升)，它会被解释为一个有符号字节，并且符号- 扩展至 11111111 11111111 11111111 10111000.这是一个值为 -72 的整数.
In Java, when this byte is used as an int (for example because read() returns an int, or because of implicit promotion), it will be interpreted as a signed byte, and sign-extended to 11111111 11111111 11111111 10111000. This is an integer with value -72.
通过与 0xff 进行与运算，您只保留最低 8 位，因此您的整数现在是 00000000 00000000 00000000 10111000，即 88.
By ANDing with 0xff you retain only the lowest 8 bits, so your integer is now 00000000 00000000 00000000 10111000, which is 88.
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