问题
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?
回答1:
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:
_Address_ -Value_
00ff2254 45
00ff2255 00
00ff2256 09
00ff2257 ff
00ff2258 02
00ff2259 00
00ff225a 00
00ff225b 00
(assuming you're on a "little-endian" machine, which most modern desktop PCs are.)
回答2:
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?
回答3:
To complete the answer of RH, you may be surprised to have so many numbers for a given address.
You should consider
Address Byte (8 bits)
00ff2254 45
00ff2255 00
00ff2256 09
00ff2257 ff
00ff2258 02
...
(On a cpu architecture used by XP)
回答4:
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.
来源：https://stackoverflow.com/questions/3819998/bits-in-a-memory-address