异步等待读和字节块包处理问题。
Asynchronous pending reads and byte chunk package processing problem.
现在我修正第9步产生的8字节块中的每一个字节的奇偶性。
Now I fix the parity of each of the bytes in the 8-byte block produced in Step 9.
请注意，所显示的大小是以512字节块为单位，而不是千字节。
Note that the size is displayed in 512 byte blocks, not kilobytes.
第9步的结果是最后8字节块的加密结果(放弃所以以前的密文块)。
The result of the encryption of the last 8-byte block (discarding all previous cipher blocks) is the outcome of Step 9.
此外，一个字节块流可以包含一个或几个包和还部分封装，如图1所示。
Furthermore, a byte chunk stream can contain one or several packages and also partial packages, as shown in figure 1.
意思是很多异步读无顺序完成(像在3.6.2讨论的)，通过异步读操作无序的返回字节块流。
This means that several asynchronous reads complete out of order (as discussed before in section 3.6.2), and byte chunk streams returned by the pending reads will not be processed in order.
统计8字节块中每字节中的1的个数，如果1的个数为偶数，那么就设置最低位为1使它成为奇数。
Count the number of ones in each byte of the 8-byte block; if the number of ones is even, make it odd by setting the least significant bit to one.
现在再次检查第10步得到的经过奇偶修正的8字节块是不是弱密钥(就像在第8步中所做的那样)。
Now I again check to see if the 8-byte parity-fixed block of Step 10 is a weak key (just like I did in Step 8).
缺省情况下，几乎所有传统的UNIX操作系统都以 512字节块为单位报告大小，而不是千字节。
Nearly all traditional UNIX operating systems report the size in 512 byte blocks rather than kilobytes by default.
如果是的话，就要用0xf0(11110000)与奇偶修正过的8字节块进行exclusiveOR。
If it is, you will exclusive OR the last byte of the 8-byte parity-fixed block with 0xf0 (11110000).
我对需要将明文数据字节放在8字节块的加密使用块密码算法，所以我用一些字符填充最后一个不完整的块。
I use a block cipher algorithm for encryption that requires the clear data bytes to be in blocks of 8 bytes, so I pad the last incomplete block with some characters.
这个数字表明，局部包(绿色)和完整的软件包(黄色)可以到达不同的异步字节块流(标1,2,3)。
The figure shows how partial packages (green) and complete packages (yellow) can arrive asynchronously in different byte chunk streams (marked 1, 2, 3).
只是一些随机的四字节的内存块，在这里是可以使用的，但是我可以在那里放置东西。
It's just some random chunk of four bytes that happens to be available at this point in time, but I can put something there.
下面说明如何做到：假定您想从文件开头算起第1000字节处开始读取一块789字节的数据。
Here is how it is done: suppose that you want to read a chunk of 789 bytes starting from byte 1000 counted from the beginning of that file.
JBD存储完整的被修改的文件系统块本身，而不是记录必定会被更改的字节范围。
Rather than recording spans of bytes that must be changed, JBD stores the complete modified filesystem blocks themselves.
图1显示一个数据块(例如一个512字节的扇区)，末尾有8字节的附加部分。
Figure 1 shows a data block (for example, a 512 byte sector) with the 8-byte footer attached to the end.
它按照块(8字节长度)对数据加密，而且还支持长度为40位、64位和128位的密钥。
It encrypts the data by blocks (8-bytes long), and supports key lengths of 40 bits, 64 bits, and 128 bits.
每个在GC堆中分配的对象有一个8字节的头(同步块+方法表指针)。
Each object allocated on the GC heap has an 8-byte overhead (sync block + method table pointer).
对于小于或等于512字节的内存块，AIXmalloc为malloc的实际后端功能提供更高效的前端。
When memory size is less than or equal to 512 bytes, AIX malloc provides a more efficient frontend to the actual backend functions of malloc.
endPoint记录最后一个内存块中第一个空闲字节的偏移位置。
The endPoint records the first free byte's offset location into the last memory block.
基于PKCS #1v1.5标准， 最开始的字节是“00”，它“保证了加密块被转换成整数之后，大小不会超过模数”。
Per the PKCS #1 v1.5 standard, the first byte is "00" and it "ensures that the encryption block, [when] converted to an integer, is less than the modulus."
结果，func函数所分配的20个字节的块就丢失了，并导致了内存泄漏。
As a result, the 20 byte block allocated by the func function is lost and results in a memory leak.
在此例中，139,456,234KB的传入文件大小没有均匀地拆分为20000000字节的块。
In this example, the incoming file size of 139,456,234 KB does not split evenly into 20000000 byte chunks.
凯文·穆瑞尔(KevinMurrell)是英国国家计算博物馆的董事，最近启用了一块456兆字节的硬盘，这块硬盘自1980年代初其断电之后就从未使用过。
Kevin Murrell, a trustee of the UK's national museum of computing, recently switched on a 456 Megabyte hard drive that had been powered down since the early 1980s.
若要将 /usr文件系统的大小增加 1000000 个 512字节的块，可输入以下命令
To increase the /usr file system size by 1000000 512-byte blocks, type
“chunk - size ”以千字节指定RAID - 0使用的块大小。
"Chunk-size" specifies the granularity of the chunks used for RAID-0 in kilobytes.
内存开销大约为(M +B)个字节(每个内存块开头的指针忽略不计)。
The overhead averages about (m + b) bytes (ignoring the Pointers at the beginning of each memory block).
传统的文件系统由匹配后端存储(512字节)的静态大小的块组成。
Traditional file systems are made up of statically sized blocks that match the back-end storage (512 bytes).
使用分区的固件可以进一步对块进行独特的分段—例如，一个块中有512字节的分段，但不包括元数据。
Firmware that USES the partitions can further apply unique segmenting to the blocks-for example, 512-byte segments within a block, not including metadata.
使用分区的固件可以进一步对块进行独特的分段—例如，一个块中有512字节的分段，但不包括元数据。
Firmware that USES the partitions can further apply unique segmenting to the blocks-for example, 512-byte segments within a block, not including metadata.
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