However, this new code now assumes that the starting address is 16-byte aligned, and also that it has enough padding on the end that the next data element in memory is also 16-byte aligned.
然而，这段新代码假定起始地址是按照16字节对齐的，并且末尾也有足够的填充位，因此内存中下一个数据元素也是16字节对齐的。
Remember that DMA transfers of 16 bytes or more must be in 16-byte multiples and aligned to 16-byte boundaries on both the SPE and the PPE.
记住，传输16字节或更多字节的dma在spe和PPE上必须是16字节的整数倍，并按照16字节边界进行对齐。
By keeping the data aligned and padded to 16-byte boundaries, vector operations can be performed effortlessly.
通过保持数据对齐并针对16字节边界进行填充，向量操作就可以毫不费力地执行了。
That means that words will be 4-byte aligned, halfwords will be 2-byte aligned, and bytes don't have to be aligned at all.
这意味着字将是4字节对齐的，半字将是2字节对齐的，而字节则不需要对齐。
If it is given an address to load from that is not at a 16-byte boundary, it simply zeroes out the last four bits of the address before loading it so that it will be an aligned load.
如果要加载自的地址不是一个 16字节的边界，它就会将该地址的最后四位截断然后再加载，以便使其能够加载。
Thus, if a partition begins on a 4096-byte (8-sector) boundary, it's properly aligned.
因此，如果一个分区起始于一个4096字节(8个扇区)边界，则表示它得到合理对齐。
This is critical, as the SPU can only load exactly 16 bytes at a time, aligned to exactly a 16-byte boundary.
这十分关键，因为SPU一次只能加载16字节，并对齐到16字节的边界。
Remember, the LQD instruction only loads from 16-byte boundaries. It will therefore ignore the four least significant bits during the load, and just load an aligned quadword from memory.
记住，lqd指令只能从16字节边界加载，所以它会在加载期间忽略最低有效的四位，而只会从内存加载已对齐的四字。
In this example, the structure s is aligned to an 8-byte boundary, and so is the s.var2 variable.
在这个示例中，结构s 以 8 个字节为单位进行对齐，s.var2变量同样也是如此。
The ideal situation is that variables are aligned at 4 byte if they are 4 byte big, 8 if 8 byte etc.
理想的情况是变量大小是4 个字节则按4 个字节对齐，是8 个字节则按8 个字节对齐。
Modern architectures (Pentium and newer) would prefer "long double" to be aligned to an 8 - or 16-byte boundary.
现代的体系架构(Pentium及更新的体系架构)更喜欢将“longdouble”按照8字节或16字节进行对齐。
Packing values of 2 and higher will cause each field to be aligned on a byte offset relative to the beginning of the structure.
封装值2和更高的值将导致字节上要对齐的每个字段相对于结构的开头进行偏置。
Why can't you access a 4 byte long variable in a single memory access on an unaligned address(i. e. not divisible by 4), as it's the case with aligned addresses?
为什么你不能访问一个4字节长的变量在未对齐地址（即不被4整除）一个内存访问，因为它与对齐地址的情况？
PDCP PDU is a bit string that is byte aligned (i. e. multiple of 8 bits) in length.
一个PDCPPDU是由字节(8比特组成)队列表示长度的位串。
The memory allocator must give you a 16-byte aligned block of space, because, in general, memory allocation must provide suitable alignment so that the memory can be used for any purpose.
内存分配器必须给你一个16字节对齐块的空间，因为，在一般情况下，内存分配必须提供适当的对齐，内存可用于任何目的。
The pointer you receive will be aligned with at least an8- byte boundary.
你接受到的指针将至少和一个八位的边界相伴随。
The pointer you receive will be aligned with at least an8- byte boundary.
你接受到的指针将至少和一个八位的边界相伴随。
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