Memory Addressing

问题:

I was reading http://duartes.org/gustavo/blog/post/motherboard-chipsets-memory-map and in specific, the following section:

In a motherboard the CPU’s gateway to the world is the front-side bus connecting it to the northbridge. Whenever the CPU needs to read or write memory it does so via this bus. It uses some pins to transmit the physical memory address it wants to write or read, while other pins send the value to be written or receive the value being read. An Intel Core 2 QX6600 has 33 pins to transmit the physical memory address (so there are 2^33 choices of memory locations) and 64 pins to send or receive data (so data is transmitted in a 64-bit data path, or 8-byte chunks). This allows the CPU to physically address 64 gigabytes of memory (2^33 locations * 8 bytes) although most chipsets only handle up to 8 gigs of RAM.

Now the math above states that since there are 33 pins for addressing, 2^33 * 8 bytes = 64 GB. All good, but now I get a bit confused. Let's say I install a 64 bit OS, I'll be able to address 64 GB total or 2^64Gb * 8 = 2^64GB (which is much more)? Also, assuming I'm using the same cpu above on a 32 bit cpu, I can address only 4 GB still (2^32 bits = 4Gb * 8 = 4GB)?

I think the physical vs "OS Allowable" is getting me confused.

Thanks!

回答1:

You're confusing a bunch of things:

• The size of a pointer limits the amount of virtual memory a user process can access. Not all of these will actually be usable by your process (it is traditional to reserve the "high" 1 or 2 GB for use by the kernel).
• Not all virtual address bits are valid. The original AMD64 implementation effectively uses 48-bit sign-extended addresses (i.e. addresses in the range [0x0000800000000000,0xFFFF7FFFFFFFFFFF] are invalid). This exists largely to limit page tables to 4 levels, which decreases the cost of a page fault; you need 6-level page tables to address the full 2^64 bits, assuming 4K pages. For comparison, i386 has 2-level page tables.
• Not all virtual addresses need to correspond to physical addresses at any given time. This is the whole point of virtual memory: you can address memory which doesn't "physically" exist, and the OS pages it in for you.
• Not all physical addresses correspond to virtual addresses. They might not be mapped, for one, but it's also possible to have more physical memory than you can address. PAE supports up to 64 GB of physical addresses, and was common on servers before AMD64. While an indivial process can't address 64 GB, it means you can run a lot of multi-gigabyte processes without swapping all the time.
• And finally: There's no point having more physical addresses than your RAM slots can handle. I have a D945GCLF2 board which supports AMD64, but only 2 GB of RAM. There's no point having extra physical address lines which can't be used anyway. (I'm handwaving over memory-mapped devices and the funky two-DIMMs-one-slot thing which I forget the name of.)

Also, note a few other things:

• For memory-mapped I/O (in the hardware sense), the CPU needs to address individual bytes. It can't just do a 64-bit access. This seems to have been glossed over.
• Modern processors include the memory controller on the CPU instead using the traditional northbridge and FSB (see HyperTransport and QuickPath).

回答2:

Yes, number of bits in physical and virtual addresses can be different. Say, here is what 64-bit Linux says about the cores here (cat /proc/cpuinfo):

...
processor       : 3
vendor_id       : AuthenticAMD
cpu family      : 15
model           : 33
model name      : Dual Core AMD Opteron(tm) Processor 280
stepping        : 2
cpu MHz         : 2392.623
cache size      : 1024 KB
...
bogomips        : 4784.41
TLB size        : 1088 4K pages
clflush size    : 64
cache_alignment : 64
address sizes   : 40 bits physical, 48 bits virtual
power management: ts fid vid ttp

回答3:

There are a few things to consider about the physical address wires:

1. Each physical address wire ("pin") references a front-side-bus-word, not a byte address. If the CPU fetches 64-bit words, then the physical address wires will be aligned to that 8-byte boundary. Therefore, address lines A0-A2 are not wired because they would always be zero. Thus, the byte address range of the physical wires is increased by the width of the front-side bus.

2. The virtual memory system can maintain a map of 64-bit virtual addresses to n-bit physical addresses. In practice, the OS maintains a "physical max address" value which the VM mappings do not exceed.

3. Some memory architectures allow memory bank paging, where off-CPU hardware increases the effective physical memory address range by re-using some physical addresses for different "banks" of memory.

回答4:

Imagine that in a 64-bit OS some of the wires to address memory don't go anywhere. The OS understands that this is pretty confusing, so it takes the standard 64-bit address and uses virtual memory mapping to make you believe that you're living in a flat 64-bit space.

回答5:

The chipset limit is a big factor -- the hardware on the motherboard has to be able to pass the addresses from the CPU to the RAM. So the 8GB limit will apply unless you have a motherboard designed to handle more.

For reference, current 64-bit CPUs have the upper x-number-of bits (somewhere between 8 and 24 bits) of the address space wired together, as 64 bits is simply too much address space for now (you'd need 8 billion 2GB modules to take up that much address space). AMDs, for example, have a 48-bit limit (IIRC) on address space in a single segment. Which is more than enough, but nowhere near the theoretical max.

回答6:

The main difference between a 64-bit and 32-bit OS is that one simply regards the primitive datatype (e.g. a word) as being wider. If the CPU can only physically address 2^33 locations, that won't change just because you're using a 64-bit OS. On the other hand, using a 32-bit OS will generally limit your addressable memory since 32-bit pointers can't represent all the possible values that your CPU could use to address memory (in your example, a 32-bit pointer is one bit short).

Long story short, your addressable memory is limited by both the pointer width (an OS restriction) and the data address bus width (a physical restriction). Some architectures have clever ways of getting around the OS pointer width by using two pointers, one to address a "bank" of memory and another to locally address within the bank. These schemes have sort of fallen out vogue lately, though.

Also, modern OSes generally use a virtual memory subsystem that translates logical addresses into their corresponding physical ones. With caching, the actual physical location of the memory could be in one (or several!) components along a memory heirarchy (e.g. processor cache, main memory, hard disk, etc.) Don't know how I completely forgot to mention VM, but it definitely would help your understanding to investigate it.

回答7:

I believe that if you have a 64 bit operating system you can (theoretically) address 2^64 * 8 bytes = 16 EB (exabytes), but you will be limited by the hardware to 2^33 * 8 bytes = 64 GB. If you have a 32 bit OS you will not be able to utilize the full hardware capacity since the OS is the limiting factor, only being able to express 2^32 different addresses. I might be off but that's my current understanding.

回答8:

I think you are getting confused by the fact the memory store 8 bytes at the same time , but an address (at the CPU level) refer to 1 byte (and not a bunch of 8). So with 32 bits you can "refer" to 2^32 bytes = 4GB. If you prefer +8 on pointer correspond to +1 on the number of the "physical" line. You can then have access to more memory using pagination (not sure if it still used in modern computer).

To do an analogy with a library, you (or the CPU) can enumerate 32^2 books, but the librarian (the chipset) deals with shelves of book. So what is for you book #10, is book #2 or of the shelf #2 but you never see the shelves number. That's the job of the librarian to go to the good shelf and bring you the good book. For me (another program on the same computer) book #10 could be a different one : book #2 of the shelf 100002 (because my page start at shelf 10000) We can both refer to 32^2 different book, but they are not the same (and the library can have much more than that).

(Change have changed lots since I studied computer, so what I'm saying can be not 100 % accurate, but I think the idea is there)