Question

Consider the following fragment of C code:

for (i=0; i<=100; i++)
{
a[ i ] = b[ i ] + C;
}

Assume that a and b are arrays of words and the base address of a is in $a0 and the base address of b is in $a1. Register $t0 holds the variable i and register $s0 the constant C. Write the code for MIPS. How many instructions are executed during the running of your code? How many memory data references will be made during execution?

Answer 1

711; 202

Step-by-step explanation:

MIPS code for the C fragment:

addi $t6, $zero, 101 # the loop termination value

add $t0, $zero, $zero # i = 0

addi $t2, $a0, 0 # ptr to current A[i]

addi $t3, $a1, 0 # ptr to current B[i]

loop: lw $t4, 0($t3) # load B[i]

add $t4, $t4, $s0 # B[i] + c

sw $t4, 0($t2) # store in A[i]

addi $t0, $t0, 1 # i++

addi $t2, $t2, 4 # ptr to next A[i]

addi $t3, $t3, 4 # ptr to next B[i]

bne $t0, $t6, loop # if i < 101, goto loop

The loop is executed 101 times and the loop contains 7 statements. There are 4 statements outside the loop. Therefore, the total number of instructions executed is 4 + (7*101) = 711.

The total number of memory data reference: 101 * 2 = 202.