Question

for the valid and tag bits, write your answers in binary without the 0b prefix. write the complete tag bits per block (including leading zeros, if applicable) given the length of the memory address. for the data field, write your answers in hexadecimal without the 0x prefix. as an example, let's say we have a 16-byte, 8-bytes per block, direct-mapped, write-back cache. when we execute sh x0 0x61c(x0), the cache (which started out empty) would look like the following: (this is already a hint. check the memory contents provided earlier and check for

Answer 1

The correct answer involves explaining cache operation, specifically for a store halfword instruction to a direct-mapped write-back cache.

Step-by-step explanation:

The correct answer involves describing how a direct-mapped write-back cache functions when a store halfword instruction is executed. A memory address is split into three parts when dealing with caches: the tag, the index, and the block offset.

For a 16-byte cache with 8-byte blocks, we would have 2 blocks, which means 1 bit for the index (since 21 = 2). With the address being 0x61c, we typically ignore the least significant 3 bits because an 8-byte block would align with addresses on 8-byte boundaries, and these least significant bits represent the block offset.

The remaining higher order bits would be used for the tag. The valid bit would be set to 1 since data is being written to the cache. The data in question would be written in hexadecimal but will depend on the actual bytes being stored.

The memory address must be divided into the tag, index, and block offset, while the valid bit is set, and data is written in hexadecimal form.