How much memory can a n-bit system address?

[RapidSnail]

I've been doing some wiki reading about computer topics (logic gates, architechture, programming, etc.) and something's confusing me about addressable memory. On the wiki page about 32-bit systems, it says that such systems have the ability to address 4 GB of physical memory:

"The range of integer values that can be stored in 32 bits is 0 through 4,294,967,295 or -2,147,483,648 through 2,147,483,647 using two's complement encoding. Hence, a processor with 32-bit memory addresses can directly access 4 GB of byte-addressable memory."

This follows from 2^32 = 4 294 967 296 bits of data. However, the article says 4 GB, or gigibytes, of memory; a byte traditionally defined as an 8-bit word. So it would seem as if the maximum amount of byte-addressable memory would be 536 870 912 bytes (2^32 ÷ 8). Which doesn't make much sense, since 32-bit systems can address 4 GB of RAM.

?

 

[Aluvus]

Each address is used to address one byte, rather than one bit. So with 32 addressing bits, you can address 2^32 memory locations, each of which is one byte (8 bits) in size. So 4 GiB.

 

[RapidSnail]

Originally posted by: Aluvus
Each address is used to address one byte, rather than one bit. So with 32 addressing bits, you can address 2^32 memory locations, each of which is one byte (8 bits) in size. So 4 GiB.

1) So a memory address in a 32-bit system is a 32-bit integer mapped to one byte of memory located in system RAM? (The total amount of bytes addressable is therefore 2^32 bytes.)

2) Why use byte-addressing instead of larger sizes of word-addressing? Is the issue how small the data chunks can be rather than how large?

3) Could you increase the amount of addressable memory in 32-bit systems by creating memory sticks with, say, 4-byte word sizes instead of single byte?

4) Is the amount of data addressable in one operation limited by the bus width? (i.e. Only 32 bits or 4 bytes or data can be transferred in one operation on a 32-bit system.)

 

[Markbnj]

They actually do not use byte addressing. A 32-bit system addresses dwords, and a 64-bit system addresses quadwords. A byte is the smallest piece of memory that can be assigned a unit address. On a 32-bit system addresses like 0, 4, 8, 12 are dword alligned and can be accessed at the lowest cost; there is some cost for addressing, say, byte 5 or 9. I used to know a lot more about this than I do now, but I think that's the gist of it.