Question

1. In the following C code, elements of each row are stored contiguously in memory.

Assume each element in an array is a 32-bit integer.

p = 0;

for (j=0; j<8; ++j)

for(i=500; i>0; --i )

{    

A[ i] = 2 * B[j];

++p;

}

a. How many 32-bit integers can be stored in a 32-byte cache block?

b. Which variable references exhibit temporal locality?

c. Which variable references exhibit spatial locality?

Answer 1

a. 1 byte= 8 bits

This implies 32 bits = 32/8 = 4 bytes

No: of 32 bit (4 bytes) integers than can be stored in a 32 - byte cache block = 32/4 = 8

Hence 8 32 bit integers can be stored in a 32 byte cache block

b. p is referenced by the instruction ++p frequently. Hence we can say that the variable p exhibit temporal locality. Similary B[j] (j being any integer) is referenced by the instruction A[i] = 2* B[j] frequently. Hence B[j] also exhibit temporal locality

c. In the inner loop, A[i], i being the variable of the loop, is executed continuously. Hence A[i[ and A[i+] will be saved next to each other in memory. This implies that A[i] exhibits spatial locality.