smithisize

In the book, they were dealing with word-addressable blocks. That difference is the only reason I can see where the +2 might come from. I don't think we need it, because it makes sense to me that to select 1 of 512 blocks, I need 9 bits, and within those blocks, to select 1 of 64 bytes, I need 6 bits. The books equation just doesn't make sense, and isn't well explained.

Thanks for the reply, It turns out that you're correct, my first attempt was right. And, you're correct in saying that the 32 lines piece of information had nothing to do with it. For B, I've been working on it. So, in order to navigate the blocks, yes, my index needs to be 9 bits to select 1 of 512 blocks. These blocks are byte addressable. So, in order to access 1 byte out of 64, I need an additional 6 bits to select the byte being requested. Therefore, assuming a 32 address coming from the CPU, the number of bits allocated to my tag should be 32 - 9 - 6 or 32-15 which is 17 bits. Does that make more sense? It kind of makes sense to me, but I'd love someone else's opinion.

Hi everyone. So, I've got some homework that looks like this: And I'm searching through my book and cannot find the differentiation between address lines and blocks. But, regardless, I have two attempts at solving part A): Attempt 1: the cache is 32kB in size, and has a block size of 64B. So 32kB/64B = 512 Blocks Attempt 2: I found a PDF online that says that if there are 32 address lines, that corresponds to 2^32 words of memory. If the block size/length is 64B and word size is 32bits, then each block holds 16 words. So 2^32 words/16 words per block = 268435456 blocks. PDF http://umcs.maine.edu/~cmeadow/courses/cos335/COA04.pdf Those are my thought processes. Which, if either, are correct? How do I find that? Because A block in a cache is one "row" if we're visualizing things, hence my 1st attempt. But, the PDF says that bit of information about address lines, which confuses me because I don't know (and can't find) where they are. For part b, here's my though process. Assuming my first attempt at part A is correct, the cache size is 2^n=512 blocks, meaning we need an n bit index, in this case a 9 bit index. So, if 9 of the block size is 64 bytes, that is 512 bits. Meaning, 512-9 = 503 bit tag. Once more I'm thrown for a loop by something else though. My book says: Which leads me to think I'm completely wrong. Hopefully someone can help clarify this all! Thanks