A byte addressable direct-mapped cache with 1024 blocks/lines, and with each block are having 8 32-bit words. How many bits are required for block offset, assuming a 32-bit address?
Answers
"""The inquiry as expressed isn't exactly responsible. A word has been characterized to be 32-bits. We have to know whether the framework is ""byte-addressable"" (you can get to an 8-bit piece of information) or "" word-addressable"" (littlest open lump is 32-bits) or even "" half-word addressable"" (the little lump of information you can get to be 16-bits.)
You have to realize this to recognize what the most reduced request bit of a location is letting you know.
At that point, you work from the base up. We should accept the framework is byte addressable. At that point each reserve square contains 8 words* (4 bytes/word) =32=25 bytes, so the counterbalance is 5 bits.
The record in a direct mapped store is the number of squares in reserve (12 bits for this situation, because of 212=4096.) At that point, the tag is every one of the bits that are left, as you have shown.
As the reserve gets more cooperative,. However, remains a similar size. there are fewer list bits and more label bits.""
"