Assume that a system has a 32-bit virtual address with a 4-KBpage size. Given th

Question

Assume that a system has a 32-bit virtual address with a 4-KBpage size. Given the virtual address 19968
Find the page number and offset

The answer is:
The address 19986 contains:
page number = 4
offset = 3602

I'm not real sure how this is done. Someone provided theexplanation below but for some reason I am still confused as to howthis works. Primarily the part in red. How do they getthis number? Any help would be greatly appreciated

The virtual address is divided into a pageoffset and a virtual page number:

The virtual page number indicates whichpage of virtual memory the data that the CPU needs isin.

Physical page number of a virtual page isobtained from the virtual page table, if that page is inmemory.

Here, Given

The virtual address size = 32bit

Explanation / Answer

Dear.., In general, on a 32-bit machine can have 32-bitvirtual addresses. So, with 4KB pages need 12 bits for offset and20 bits for page address. thats the reason 12-bits for offset. pagesize is 32-bits address = 232/212 =220 entries ~ 4bytes each..... Here we will use page table to fing out the page numberaccording to the offset. I Hope this will helps you. pagesize is 32-bits address = 232/212 =220 entries ~ 4bytes each..... Here we will use page table to fing out the page numberaccording to the offset. I Hope this will helps you.

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