Concept of "block size" in a cache

hektor picture hektor · Nov 12, 2011 · Viewed 44.6k times · Source

I am just beginning to learn the concept of Direct mapped and Set Associative Caches. I have some very elementary doubts . Here goes.

Supposing addresses are 32 bits long, and i have a 32KB cache with 64Byte block size and 512 frames, how much data is actually stored inside the "block"? If i have an instruction which loads from a value from a memory location and if that value is a 16-bit integer, is it that one of the 64Byte blocks now stores only a 16 bit(2Bytes) integer value. What of the other 62 bytes within the block? If i now have another load instruction which also loads a 16bit integer value, this value now goes into another block of another frame depending on the load address(If the address maps to the same frame of the previous instruction, then the previous value is evicted and the block again stores only 2bytes in 64 bytes). Correct?

Please forgive me if this seems like a very stupid doubt, its just that i want to get my concepts correctly.

Answer

user541686 picture user541686 · Nov 13, 2011

I typed up this email for someone to explain caches, but I think you might find it useful as well.

You have 32-bit addresses that can refer to bytes in RAM. You want to be able to cache the data that you access, to use them later.

Let's say you want a 1-MiB (220 bytes) cache.

What do you do?

You have 2 restrictions you need to meet:

  1. Caching should be as uniform as possible across all addresses. i.e. you don't want to bias toward any particular kind of address.
    • How do you do this? Use remainder! With mod, you can evenly distribute any integer over whatever range you want.
  2. You want to help minimize bookkeeping costs. That means e.g. if you're caching in blocks of 1 byte, you don't want to store 4 bytes of data just to keep track of where 1 byte belongs to.
    • How do you do that? You store blocks that are bigger than just 1 byte.

Let's say you choose 16-byte (24-byte) blocks. That means you can cache 220 / 24 = 216 = 65,536 blocks of data.

You now have a few options:

  • You can design the cache so that data from any memory block could be stored in any of the cache blocks. This would be called a fully-associative cache.
  • The benefit is that it's the "fairest" kind of cache: all blocks are treated completely equally.
  • The tradeoff is speed: To find where to put the memory block, you have to search every cache block for a free space. This is really slow.
  • You can design the cache so that data from any memory block could only be stored in a single cache block. This would be called a direct-mapped cache.
  • The benefit is that it's the fastest kind of cache: you do only 1 check to see if the item is in the cache or not.
  • The tradeoff is that, now, if you happen to have a bad memory access pattern, you can have 2 blocks kicking each other out successively, with unused blocks still remaining in the cache.
  • You can do a mixture of both: map a single memory block into multiple blocks. This is what real processors do -- they have N-way set associative caches.

Direct-mapped cache:

Now you have 65,536 blocks of data, each block being of 16 bytes.
You store it as 65,536 "rows" inside your cache, with each "row" consisting of the data itself, along with the metadata (regarding where the block belongs, whether it's valid, whether it's been written to, etc.).

Question: How does each block in memory get mapped to each block in the cache?

Answer: Well, you're using a direct-mapped cache, using mod. That means addresses 0 to 15 will be mapped to block 0 in the cache; 16-31 get mapped to block 2, etc... and it wraps around as you reach the 1-MiB mark.

So, given memory address M, how do you find the row number N? Easy: N = M % 220 / 24.
But that only tells you where to store the data, not how to retrieve it. Once you've stored it, and try to access it again, you have to know which 1-MB portion of memory was stored here, right?

So that's one piece of metadata: the tag bits. If it's in row N, all you need to know is what the quotient was, during the mod operation. Which, for a 32-bit address, is 12 bits big (since the remainder is 20 bits).

So your tag becomes 12 bits long -- specifically, the topmost 12 bits of any memory address.
And you already knew that the lowermost 4 bits are used for the offset within a block (since memory is byte-addressed, and a block is 16 bytes).
That leaves 16 bits for the "index" bits of a memory address, which can be used to find which row the address belongs to. (It's just a division + remainder operation, but in binary.)

You also need other bits: e.g. you need to know whether a block is in fact valid or not, because when the CPU is turned on, it contains invalid data. So you add 1 bit of metadata: the Valid bit.

There's other bits you'll learn about, used for optimization, synchronization, etc... but these are the basic ones. :)