Question

Given the code :

void transpose2(array dst,array src)
{
  int i,j;
  for ( i=0; i<4; i++) {
    for ( j=0; j<4; j++) {
     dst[i][j] = src[j][i];
    }
  }
}

Assumptions :

• int is 4 bytes

• src array starts at address 0 , dst starts at address 64

• the size of the cache is 32 bytes , at the beginning the cache is empty

Assuming that I have a cache with size of 32 bytes , under write through ,write allocate & LRU , using 2way set associative method , where each block is 8 bytes :

When I read from the memory , how many bytes do I take each iteration from the memory ?

is it 4 or 8 ?

What I’m quite sure about is that the cache has 4 cells , or rows , and each row has 8 bytes .Is this correct ?

What is a little confusing is the 2way part , I think that each way has 4 bytes , right ? please correct me if I’m wrong …

Then when I “take” a block from the memory , I just don’t exactly understand how many bytes !!?

Thanks in advance

Ron

Answer 1

The cache way (aka its associativity) does not affect the amount of data that’s transferred when a transfer occurs; the block size is the block size.

Associativity is simply a measure how many possible locations there are in the cache that a given block from memory could be stored. So:

• For a direct-mapped cache (associativity=1), memory address xyz will always map to the same cache location.
• For a two-way cache, xyz could map to either of two cache locations.
• For a fully-associative cache, xyz could map to anywhere in cache.

I’m really not saying anything here which isn’t already explained at e.g. Wikipedia: http://en.wikipedia.org/wiki/CPU_cache#Associativity.