A Consider the following fragment of C code for i0 i

A. 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 this fragment in MIPS. How many total instructions are executed during the running of your code? How many memory data references will be made during execution? Typing your answer is recommended.

B. Add comments to each line of the following MIPS code. Then, describe what it computes (one or two sentences). Assume that $a0 is used for the input and initially contains n, a positive integer. Assume that $v0 is used for returning the output value.

begin: addi $t0, $zero, 0

addi $t1, $zero, 1

loop: slt $t2, $a0, $t1

bne $t2, $zero, finish

add $t0, $t0, $t1

addi $t1, $t1, 2

j loop

finish: add $v0, $t0, $zero

Solution

(1) One example of MIPS code:

clear $t0;

addi $s0, $zero, 100

loop: lw $t1, 0($a1)          # $t1 = b[i]

add $t1, $t1, $s0               # $t1 = a[i]

sw $t1, 0($a0)                  # store $t1 to address of a[i]

addi $a0, $a0, 4               # $a0 = address of a[i+1]

addi $a1, $a1, 4             # $a0 = address of a[i+1]

addi $t0, $t0, 1               # $t0 = $t0 + 1

beq $t0, $s0, finish      # if ($t0 = 100) finish

j loop

finish:   

b. instructions before loop are executed 1 time; 7 instructions between loop are executed 101 times; instruction

“j loop” executed 100 times. Therefore: Total instructions executed (in this case): 2*1 + 7*101 + 1*100 = 809.

c.

Memory data reference: 101 * 2 = 202.

B.

t=0;

if(b==0) go to finish

t = t+a, increase t by a

b = b-1,

go to loop

t = t+100

$v0 = t, output

t=0;

while(b!=0)

{

t = t + a;

b = b – 1;

}

t = t + 100;

v = t;

So, v = a*b+100