CACHE direct mapping

Question

Comments

ok. sorry for that. will keep in my mind next time :)

---

don't be sorry ! its just a suggestion ,All the best :)

---

Oh! that's nothing compared to what happened with me, I just mentioned a certain person's name and my comment was first flagged and then I got banned.....
Site is great and all, but at least a warning is to be issued. It's like the exam itself, if you are wrong you are wrong... Why you are wrong? what was your mistake? they don't care! we are just wrong and ultimately we suffer loss!

Answer 1

Given, memory address is of 16-bits.

Each cache block contains 16 bytes(i.e. = $2^4$ bytes), so the WORD offset field will be of 4 bits (since memory is byte addressable).

Then we have total 16 blocks (i.e = $2^4$ blocks) in the cache. So BLOCK field will be of 4 bytes.

Then the remaining 16 - 4 - 4 = 8 bits are for TAG field.

            TAG(8 bits)            |       BLOCK(4 bits)     |        OFFSET(4 bits)

Memory address given: $(9A81)_{16}$

$1001-1010-1000-0001$

So, it maps to block $(8)_{16}$