How long is a virtual page number in a 48-bits system with 4KB page size

Question

In a 32-bits system with 4KB page size, the page number PN from its address VA is 20 bits long obtained with PN = VA / 4KB.
But if we have a 48-bits system, is still the page number 20 bits long?

Concretely, I have a virtual address of the form 0x5641ba0c7000 and I want to know if the virtual page number is 5641ba0c7 (= 5641ba0c7000 / 4KB) or 5641b (20 bits)

score 1 · Accepted Answer · answered Oct 14 '21 at 15:33

1

Normally the virtual page number would be everything except the low bits of the address that give the offset within the page. With 4 KB (= 2**12 byte) pages, the offset is the low 12 bits, so the virtual page number is the remaining 48-12=36 bits. That's 0x5641ba0c7 in your example. (This number is likely to be broken up further as indexes into several levels of page tables.)

There's nothing universal about 20 bits; it's just that if you have 32-bit virtual addresses then 32-12=20, so the high 20 bits give the page number in that case.

answered Oct 14 '21 at 15:33

Nate Eldredge

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thanks for the explanation So I'm not wrong, we can talk about page number both for virtual and physical memory? – S Bitchebe Oct 14 '21 at 16:48
@StellaB: Sure, we can. – Nate Eldredge Oct 14 '21 at 16:53

How long is a virtual page number in a 48-bits system with 4KB page size

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