Timeline for Total number of bits of storage for direct mapped cache
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 4, 2016 at 20:30 | vote | accept | Carousser | ||
| Dec 4, 2016 at 0:27 | comment | added | Carousser | Thanks. One last question - would that work to have the total bits for storage be 38bits, even though the address size is 32bits? I was under the impression when you computed the bits for storage, it would equal the address size as the address contains the tag, valid bit, and the data? | |
| Dec 4, 2016 at 0:25 | comment | added | Surt | It sounds more correct to me :) how many bits your using exactly for the 16 bytes (256 bits) I'll leave up to you. | |
| Dec 4, 2016 at 0:18 | comment | added | Carousser | So let's say the cache is full. Using 1024 "cache lines", which equates to 1024 blocks of data, which equates to 1024 frames, and using the formula from the main post, that would give us a total of 38 bits required for storage, and give us an overhead of ~18.75% for the tags and valid bits. To me, that seems way too low of an overhead for the tags, considering the tags and valid bit would consume 24 of the 38 bits | |
| Dec 4, 2016 at 0:01 | comment | added | Surt | The 1024 sets (cache lines, set-associativity is not the same) can be in the cache at a time, that there are more different sets that could be loaded at some time doesn't impact what is in the cache, but the tag needs to cover it. | |
| Dec 3, 2016 at 23:55 | comment | added | Carousser | Or would it be 16kb/16b, giving us 1024 sets? | |
| Dec 3, 2016 at 23:49 | comment | added | Carousser | As for my understanding of a direct mapped cache, a frame/set consists of: 1 block of data, 1 valid bit, and a tag. Since the cache is 16kb, using 16byte block sizes, wouldn't that mean there are 65536 blocks of data, giving us 65536 frames/sets? | |
| Dec 3, 2016 at 23:41 | comment | added | Surt | In the cache these can only be 16K / 16 bytes sets ie. cache lines. | |
| Dec 3, 2016 at 23:05 | comment | added | Carousser | Thanks for the input. However, the way I described above is how my professor wants it, and I guess I should refine my question to: Am I right on the number of sets/frames being 65536? | |
| Dec 3, 2016 at 22:27 | review | First posts | |||
| Dec 3, 2016 at 23:44 | |||||
| Dec 3, 2016 at 22:26 | history | answered | Surt | CC BY-SA 3.0 |