Question 9387 – Cache
November 12, 2023Question 9394 – Cache
November 12, 2023Question 9393 – Cache
A CPU has a 32 KB direct mapped cache with 128-byte block size. Suppose A is a two-dimensional array of size 512×512 with elements that occupy 8-bytes each. Consider the following two C code segments, P1 and P2.
P1: for (i=0; i<512; i++) { for (j=0; j<512; j++) { x += A[i][j]; } } P2: for (i=0; i<512; i++) { for (j=0; j<512; j++) { x += A[j][i]; } }
P1 and P2 are executed independently with the same initial state, namely, the array A is not in the cache and i, j, x are in registers. Let the number of cache misses experienced by P1 be M1 and that for P2 be M2.
The value of M1 is:
Correct Answer: C
Question 14 Explanation:
[P1] Access A in row major order.
[P2] Access A in column major order.
No. of cache blocks=Cache size/Block size = 32KB / 128B = 32×210B / 128B = 215 / 27 = 256
No. of array elements in each block = Block size / Element size = 128B / 8B = 16
Total misses for (P1) = Array size * No. of array elements / No. of cache blocks = (512×512) * 16 / 256 = 16384
[P2] Access A in column major order.
No. of cache blocks=Cache size/Block size = 32KB / 128B = 32×210B / 128B = 215 / 27 = 256
No. of array elements in each block = Block size / Element size = 128B / 8B = 16
Total misses for (P1) = Array size * No. of array elements / No. of cache blocks = (512×512) * 16 / 256 = 16384
0
2048
16384
262144
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