Thus, to access a byte, a multiplier of 1 is used, for a word, 2, and for a dword, 4.
因此，要想访问一个字节，就使用乘数1；对于字，乘数是2；对于双字，乘数是4。
This enables access to the underlying byte array without having to create additional copies after the data is written to the network.
这实现了对底层字节数组的访问，而无需在数据写到网络后创建额外的副本。
The byte ordering (Big-Endian or Little-Endian) for a storage access is specified by the operating system.
存储器存取的字节顺序(Big -Endian或Little - Endian)是由操作系统指定的。
Indicates the last access time for this storage stream or byte array.
指示此存储、流或字节数组的上次访问时间。
Specifies the last access time for this storage, stream, or byte array.
指定此存储、流或字节数组的上次访问时间。
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？
In addition, the scratchpad provides access to the 1-byte upper and lower alarm trigger registers.
此外，暂存器可以访问的1个字节的上限和下限报警触发寄存器。
In addition, the scratchpad provides access to the 1-byte upper and lower alarm trigger registers.
此外，暂存器可以访问的1个字节的上限和下限报警触发寄存器。
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