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Why can't OS use entire 64-bits for addressing? Why only the 48-bits?

Asked 2010-07-10T14:32:19.893
20

I'm reading "Understanding Linux Kernel".

Paging for 64-bit Architectures

As we have seen in the previous sections, two-level paging is commonly used by 32-bit microprocessors. Two-level paging, however, is not suitable for computers that adopt a 64-bit architecture. Let's use a thought experiment to explain why:

Start by assuming a standard page size of 4 KB. Because 1 KB covers a range of 210 addresses, 4 KB covers 212 addresses, so the Offset field is 12 bits. This leaves up to 52 bits of the linear address to be distributed between the Table and the Directory fields. If we now decide to use only 48 of the 64 bits for addressing (this restriction leaves us with a comfortable 256 TB address space!), the remaining 48-12 = 36 bits will have to be split among Table and the Directory fields. If we now decide to reserve 18 bits for each of these two fields, both the Page Directory and the Page Tables of each process should include 218 entries that is, more than 256,000 entries.

  1. "If we now decide to use only 48 of the 64 bits for addressing". Why? & Why only 48 bits? Why not some other number?

  2. Well, I'm just a regular PC user & programmer. Its just hard to believe for me that 32-bit addressing i.e. 4GB (2GB/3GB to be more correct) address space per process is a limit. If you really encountered this limit. Please give me example.

  3. What is this limit for windows?

  4. I know that virtual memory != physical memory & processor address pins have nothing to do with virtual memory. This is a completely different question. How to know the number of address pins (= size of address bus) for a processor. http://ark.intel.com specifications of a processor doesn't inc

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2 Answers

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None of these answers are right, the reason that OSs don't use the full 64-bits is because the page tables would be far larger (64-bit is already up to 3 levels of page tables), and there's no reason to pay the extra indirection needed, 48 bits is enough. 48-bits is also convenient because you get some extra bits to store flags in (pointer tagging)

answered 2010-07-10T19:58:03.683
6

"If we now decide to use only 48 of the 64 bits for addressing". Why? & Why only 48bits? Why not some other number?

System architects make tradeoffs. 256TB seems like more than enough room for 1 process's address space. Remember virtual address != physical address, and generally speaking, each process has its own address space.

As long as pointers are 64 bits, this is more of a performance capability issue than anything else. If & when 48 bits becomes a limitation, the OS could be tweaked to use more bits of the 64-bit address space without breaking application incompatibility. For now, the architects are just buying themselves a very comfortable amount of time.

It may have to do with processor-side virtual addressing capabilities, as many processors now have memory management units to handle the virtual -> physical memory mapping.

How to know the number of address pins (= size of address bus) for a processor. http://ark.intel.com specifications of a processor doesn't include this spec.

This is for the most part irrelevant. It's a way for a processor to implement various physical addressing schemes. A 64-bit processor could achieve external address/data buses for its complete address space with 64, 32, 16, 8, 4, 2, or 1 address pin if the bus is synchronous and the address bits get multiplexed in time. Again, virtual address != physical address; 64-bit virtual addressing could be implemented with 48-bit or 32-bit physical addresses (just that you would be limited to 248 or 232 words of memory).

update: if you really want to know, you have to look at the datasheet of each processor in question. E.g. answered 2010-07-10T14:39:26.990

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