2.30
sll $a2, $a2, 2 # max i= 2500 * 4
sll $a3, $a3, 2 # max j= 2500 * 4
add $v0, $zero, $zero # $v0 = 0
add $t0, $zero, $zero # i = 0
outer: add $t4, $a0, $t0 # $t4 = address of array 1[i]
lw $t4, 0($t4) # $t4 = array 1[i]
add $t1, $zero, $zero # j = 0
inner: add $t3, $a1, $t1 # $t3 = address of array 2[j]
lw $t3, 0($t3) # $t3 = array 2[j]
bne $t3, $t4, skip # if (array 1[i] != array 2[j]) skip $v0++
addi $v0, $v0, 1 # $v0++
skip addi $t1, $t1, 4 # j++
bne $t1, $a3, inner # loop if j != 2500 * 4
addi $t0, $t0, 4 # i++
bne $t0, $a2, outer # loop if i != 2500 * 4
The code determines the number of matching elements between the two arrays
and returns this number in register $v0.
2.31 Ignoring the four instructions before the loops, we see that the outer loop
(which iterates 2500 times) has three instructions before the inner loop and two
after. The cycles needed to execute these are 1 + 2 + 1 = 4 cycles and 1 + 2 = 3
cycles, for a total of 7 cycles per iteration, or 2500 × 7 cycles. The inner loop
requires 1 + 2 + 2 + 1 + 1 + 2 = 9 cycles per iteration and it repeats 2500 × 2500
times, for a total of 9 × 2500 × 2500 cycles. The total number of cycles executed is
therefore (2500 × 7) + (9 × 2500 × 2500) = 56,267,500. The overall execution time
is therefore (56,267,500) / (2 × 109) = 28 ms. Note that the execution time for the
inner loop is really the only code of significance.
