Computer uses a fully associative write-back data cache
Block size is 64 bytes,
Consider following code,
for(i=1;i<1024;i++) {
x[i] = b[i] * y; // S1
c[i] = x[i] + z; // S2
}
Assume x[1024], b[1024] and c[1024] are all double precision floating point arrays with arrays b and c already in cache,
moving x into the cache does not conflict with b or c
x[0] is stored at the beginning of a block
1)how many misses arise with respect to accessing x if the cache uses a no-write allocate policy?(Ans :256)
2)how many misses arise with respect to accessing x if the cache uses a write allocate policy?(Ans :128)
Are the answers correct?
Since the cache is write back first x[i] (S1) is write miss and second read miss will happen because of no write-allocate policy.
But on read miss ,block will be brought in, for other 7 elements it will be hit,so 2 misses for 1 block
Number of blocks=(1024*8)/64=128 hence 2*128=256 misses.
Am I wrong somewhere?
