How many bits is the index for a fully associative cache?

How many bits is the index for a fully associative cache?

Consider a main memory of 16 kilobytes, which is organized as 4-byte blocks, and a fully associative cache of 256 bytes and a block size of 4 bytes. Because the main memory is 16kB, we need a minimum of 14 bits to uniquely represent a memory address.

What does fully associative cache mean?

A fully associative cache permits data to be stored in any cache block, instead of forcing each memory address into one particular block. — When data is fetched from memory, it can be placed in any unused block of the cache.

Why are cache lines 64 bytes?

A cache line of 64 bytes for instance means that the memory is divided in distinct (non-overlapping) blocks of memory being 64bytes in size. 64bytes mean the start address of each block has the lowest six address bits to be always zeros.

Are there index bits in a fully associative cache?

There are no index bits in a fully associative cache. The index bits are used to uniquely identify which set the block belongs. In a fully associative cache, all blocks are essentially part of the same set.

How are the index and offset bits used?

Outside of DSPs, almost all recent processors use byte addressing, so it would be safe to assume byte addressing (and five offset bits). There are no index bits in a fully associative cache. The index bits are used to uniquely identify which set the block belongs.

Which is the best definition of fully associative cache memory?

Fully Associative Cache Memory: This cache is most flexible cache architecture where data blocks from main memory can be paced in any location in cache memory. In other words, this cache has ā€˜P’ locations to store data from Main Memory.

How to calculate the number of cache blocks?

Let’s assume the system is byte addressable. Then each cache block contains 8 words* (4 bytes/word)=32=2 5 bytes, so the offset is 5 bits. The index for a direct mapped cache is the number of blocks in the cache (12 bits in this case, because 2 12 =4096.) Then the tag is all the bits that are left, as you have indicated.