Final answer:
For a directly mapped cache with 256 entries and 32 byte cache blocks, operating with a 32-bit address system, the size of the tag would be 19 bits. This is determined by subtracting the index bits (8 bits) and the block offset bits (5 bits) from the total address size (32 bits).
Step-by-step explanation:
When working with a directly mapped cache with 256 entries, 32 byte cache blocks, and an address size of 32 bits, calculating the size of the tag requires understanding how a cache is organized. The key components of the cache address include the tag, index, and block offset.
In this scenario, the number of cache entries, also referred to as lines, is 256. Since cache addresses are binary, we consider the index bits as the log base 2 of the number of entries. Therefore, we have log2(256) = 8 bits for the index.
The size of each cache block is 32 bytes, which means the number of block offset bits is log2(32) = 5 bits. We now have accounted for 5 bits for the offset and 8 bits for the index.
Given a 32-bit address, we can calculate the size of the tag by subtracting the index and offset bits from the total address bits. Thus, 32 - 8 (index bits) - 5 (block offset bits) gives us 19 bits for the tag. The tag size in this cache configuration is therefore 19 bits.