字的大小和寻址 [英] Size of a Word and addressing

问题描述

我正在刷新自己的记忆信息，但对单词的大小感到困惑.根据我的理解，一个Word并不是一个通用定义的大小，而是特定系统定义的大小(以字节数来衡量).

I was refreshing myself on memory information and I am confused on the size of a Word. From my understanding, a Word is not a universally defined size, but is a size defined by the specific system (in terms of number of bytes).

根据维基百科:

因此，具有 32 位内存地址的处理器可以直接访问 4 GiB 的字节可寻址内存.

Hence, a processor with 32-bit memory addresses can directly access 4 GiB of byte-addressable memory.

这是否意味着 32 位处理器可以寻址 4,294,967,295 个字?32 位 Windows 仅限于 4GB 内存，但阅读单词的含义让我感到疑惑.窗口中的每个字是否等于 1 个字节?一个字的大小是否可能只是更大的字节数，而 32 位处理器能够寻址 8GB、10GB、12GB 甚至更多的内存?

Does that mean that a 32-bit processor can address 4,294,967,295 Words? 32-bit windows is limited to 4GB of RAM, but reading over the meaning of a word had me wondering. Does each Word in windows equate to 1 byte? Could the size of a word just be a larger number of bytes and a 32-bit processor be able to address 8GB, 10GB, 12GB or even more memory?

推荐答案

这是否意味着 32 位处理器可以寻址 4,294,967,295 个字?

Does that mean that a 32-bit processor can address 4,294,967,295 Words?

这取决于 CPU 以及您如何看待它.

It depends on the CPU and on how you look at it.

有些 CPU 无法处理比单词更小的任何内容.16 位德州仪器数字信号处理器就是一个很好的例子.它们的 C/C++ char (AKA byte)、short 和 int 类型都是一样的大小，16位.这是可以用唯一地址(指针)寻址的最小内存单元，同时也是机器字.由于这些处理器上的数据地址/指针是 16 位的，因此它们最多可以寻址数据存储器中的 2¹⁶ 个 16 位字.

There are CPUs that can't address anything smaller than a word. 16-bit Texas Instruments Digital Signal Processors are a good example. Their C/C++ char (AKA byte), short and int types are all of the same size, 16 bits. And that is the smallest unit of memory that can be addressed with a unique address (pointer) and that's the machine word at the same time. Since data addresses/pointers are 16-bit on these processors, they can address at most 2¹⁶ 16-bit words in the data memory.

现在，如果您回到 32 位操作模式下的 x86 CPU，情况就会有所不同.内存的最小可寻址单元是 8 位字节，最大的是 32 位字(机器字).地址和指针也是 32 位的(如果我们忽略分段和页面转换).这让我们有 2³² 个唯一的内存地址.而且，很简单，使用它们您可以访问多达 2³² 8 位字节的内存.但是，您可以使用 2³² 个唯一地址来寻址多少个 32 位字?答案取决于您是想要非重叠词还是重叠词.你看，x86 CPU 可以在任何地址访问 32 位内存单元，而不仅仅是在 4 字节的倍数的地址.

Now, if you go back to x86 CPUs in 32-bit modes of operation, things are a little different. The smallest addressable unit of memory is an 8-bit byte and the largest is a 32-bit word (machine word). Addresses and pointers are 32-bit as well (if we ignore segmentation and page translation). This lets us have 2³² unique memory addresses. And, trivially, with them you can access up to 2³² 8-bit bytes of memory. But how many 32-bit words can you address with 2³² unique addresses? The answer depends on whether you want non-overlapping or overlapping words. You see, the x86 CPU can access 32-bit units of memory at any address, not just at addresses that are multiple of 4 bytes.

在 x86 上您不仅限于此:

You aren't just limited to this on x86:

  0 1 2 3 4 5 6 7  <- address
  \word/  \word/

这些都是 x86 上 32 位字访问的有效地址:

These all are all valid addresses for 32-bit word accesses on x86:

  0 1 2 3 4 5 6 7  <- address
  \word/  | | | |
    \word/  / | |
      \word/  / |
        \word/  /
          \word/
            ...

那么，在 x86 上用 2³² 个唯一地址可以寻址多少个 32 位字?如果你说的是所有唯一可寻址和重叠的，那就是其中的 2³².OTOH，如果您谈论的是所有唯一可寻址且不重叠的地址，那么就是其中的 2³⁰.

So, how many 32-bit words can you address with 2³² unique addresses on x86? If you're talking about all uniquely addressable and overlapping ones, that's 2³² of them. If, OTOH, you're talking about all uniquely addressable and non-overlapping ones, that's 2³⁰ of them.

OTOH，如果您的 32 位 CPU 使用非 32 位地址，则总数会有所不同.

OTOH, if your 32-bit CPU uses non-32-bit addresses, the total count will be different.

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