help2002

It's UWAweek 44 (2nd semester, 1st exam week)

help2002

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UWA week 43 (2nd semester, study break) ↓

final exam calculating page size (all 5)

6:25pm Fri 25th Oct, Chenjun H.

Hi, just a quick question about calculating the page size from the offset. if there are 10 bits for the offset in a 32 bit system, would the page size be 2^10 = 1kB bytes, where each offset contains 2 bits to determine which byte of the 4 bytes in the address to read, or would it be 2^10 * 4 = 4kB bytes, since each address is 4 bytes (32 / 8 = 4)?

final exam calculating page size (all 5)

7:25pm Fri 25th Oct, Amitava D.

"Chenjun Hou" <23*6*8*9@s*u*e*t*u*a*e*u*a*> wrote:

> Hi, just a quick question about calculating the page size from the offset. > > if there are 10 bits for the offset in a 32 bit system, would the page size be 2^10 = 1kB bytes, where each offset contains 2 bits to determine which byte of the 4 bytes in the address to read, or would it be 2^10 * 4 = 4kB bytes, since each address is 4 bytes (32 / 8 = 4)?

The page size is 2^10, but what you have written after that does not make any sense. I guess you haven't spent time to understand the lectures. It is impossible to explain here.

final exam calculating page size (all 5)

7:40pm Fri 25th Oct, Chenjun H.

Sorry for the confusion, when the offset is 10 bits, does it mean there are 2^10 bytes, or 2^10 32-bit addresses?

final exam calculating page size (all 5)

8:19pm Fri 25th Oct, ANONYMOUS

an off set of 10 bits means the page size is 2^10 Bytes which is 1 KB. the amount of pages is determined by the maximum page number. Also, each digit in a memory address represents a byte so a 32 bit computer would have 2^32 unique addresses which is equal to 4GB

final exam calculating page size (all 5)

8:23pm Fri 25th Oct, ANONYMOUS

meant to say the amount of frames is determined by the maximum frame number which is everything left of the offset. so in a 32 bit computer if 10 were for offset then it means that there are up to 2^22 pages and if each of those pages are 2^10 bytes in size, it would occupy 2^32 bytes which again is 4GB