This is part of a a homework assignment, I want to make sure that I am understanding correctly. In class, my professor said that in direct mapping using 16 bit addresses, we have 4 tag bits, 4 set bits and 8 offset bits. The size of the block is 256 (2^8).
For the homework, we have a direct mapped cache, using 16 bit addresses. the right most four bits are the offset and there is 4 sets. one of the questions is what is the size of the block? Is 2^4 or 16 correct?
Yes. But that would imply that you have 8 bits as tag values. Unless the tag values do some special line selection within a block, I guess the block size would be 2^4.