内存地址中的位
在使用 immunity 调试器在 Windows XP 32 位上进行调试时，我在堆栈上看到以下内容：
_Address_ -Value_
00ff2254 ff090045
00ff2258 00000002
我的理解是每个地址位置都包含 8 位。
这是正确的吗？
While debugging on Windows XP 32-bit using the immunity debugger, I see the following on the stack:
_Address_ -Value_
00ff2254 ff090045
00ff2258 00000002
My understanding is that every address location contains 8 bits.
Is this correct?
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评论（4）
如果我正确理解你的问题，答案是肯定的，每个单独的内存位置都包含 8 位。
调试器一次显示 4 个字节（32 位），以使显示更加紧凑（并且因为许多数据类型占用 32 位，所以查看 32 位值通常很有用）。这就是左列中的地址相距 4 个位置的原因。
如果调试器一次显示一个字节（8 位），则显示将如下所示：（
假设您位于“little-endian”机器，大多数现代台式电脑都是这样。）
If I'm understanding your question correctly, the answer is yes, every individual memory location contains 8 bits.
The debugger is showing you 4 bytes (32 bits) at a time, to make the display more compact (and because many data types take up 32 bits, so it's often useful to see 32-bit values). That's why the addresses in the left column are 4 locations apart.
If the debugger showed one byte (8 bits) at a time, the display would look like this:
(assuming you're on a "little-endian" machine, which most modern desktop PCs are.)
我认为你的问题的主要问题是你要求一件事，但我发现在阴影中潜伏着另一个不同的问题。
首先，也是最重要的，计算机内存中的可寻址实体被组织为字节，每个字节有 8 位，所以是的，每个地址可以说是指 8 位或一个字节。
但是，您可以轻松地将更多字节组合在一起以形成更大、更复杂的数据结构。
如果您的问题确实是“为什么我在堆栈转储中的某个地址处看到 8 位值的内容”，那么原因是它转储 32 位（4 字节）值。
换句话说，您可以获取地址、地址+1、地址+2 和地址+3，从每个地址中获取字节，然后组合成一个 32 位值。
这真的是你的问题吗？
I think the main problem with your question is that you ask for one thing, but I detect a different question lurking in the shadows.
First, and foremost, addressable entities in the memory of a computer is organized as bytes, which are 8 bits each, so yes, each address can be said to refer to 8 bits, or a byte.
However, you can easily group more bytes together to form bigger and more complex data structures.
If your question is really "Why am I seeing an 8-digit value as the contents at an address in my stack dump", then the reason for that is that it dumps 32-bit (4 bytes) values.
In other words, you can take the address, the address+1, the address+2, and the address+3, grab the bytes from each of those, and combine to a 32-bit value.
Is that really your question?
为了完成 RH 的答案，您可能会惊讶于给定地址有这么多数字。
你应该考虑
（在XP使用的cpu架构上）
To complete the answer of RH, you may be surprised to have so many numbers for a given address.
You should consider
(On a cpu architecture used by XP)
内存位置指的是内存的一个位置，每个连续的内存位置指的是内存中的下一个字节。所以，你只能在单字节边界上寻址内存，每个人都应该知道一个字节是 8 位宽。
A memory location refers to a location of memory, and each consecutive memory location refers to the next byte in memory. So, you can only address memory on a one byte boundary, and everyone should know that a byte is 8 bits wide.