Consider the following MIPS loop: *(show work) LOOP: slt $t2, $0, $t1 beq $t2, $

Question

Consider the following MIPS loop: *(show work)

LOOP:    slt           $t2,        $0,         $t1

               beq        $t2,        $0,         DONE

               subi        $t1,        $t1,        1

               addi       $s2,        $s2,        2

               j              LOOP

DONE:

a) Assume that the register $t1 is initialized to the value 10. What is the value in register $s2 assuming $s2 is initially zero?

b) For each of the loops above, write the equivalent C code routine. Assume that the registers $s1, $s2, $t1, and $t2 are integers A,B, I, and temp, respectively.

c) For the loops written in MIP assembly above, assume that the register $t1 is initialized to the value N. How many MIPS instructions are executed?

Explanation / Answer

The loop will run for condition ($0 < $t1)

It is similar to c code:

$t1 = 10; $s2 = 0;

while($0 < $t1) {

$t1 = $t1 - 1;

$s2 = $s2 + 2;

}

Therefore

(a) $s2 = 20

// loop will run 10 times

(b)

I = 10; B = 0;

while((temp = (0 < I)) != 0) {

I = I - 1;

B = B + 2;

}

// loop will run 10 times

(c)

As seen above from c code

For $t1 = N

No of executions of loop = N

For each loop execution --> 5 MIPS instruction

Therefore total MIPS instruction = 5N

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