6. Consider the following MIPS machine code. bne $0, $0, 2 addi $t0, $0, 5 addi

Question

6.       Consider the following MIPS machine code.

         bne $0, $0, 2

         addi $t0, $0, 5

         addi $t1, $0, 10

What will be in $t0 and $t1 after the above program is executed?

a.       $t0 = 5, $t1 is unknown because 3rd statement will not be executed

b.      $t1 = 10, $t0 is unknown because 2nd statement will not be executed

c.       $t0 = 5, $t1 = 10

d.      $t0 = 0, $t1 = 0

BONUS (5 points)

7.      

       addi $t0, $0, 3
       add $s0, $0, 0
doLoop:
       addi $t0, $t0, -1
       add $s0, $s0, $t0
       bne $t0, $0, doLoop

       addi $t0, $0, 3
       add $s0, $0, 0
       addi $t0, $t0, -1
       add $s0, $s0, $t0
       bne $t0, $0, -2

Consider the following code: What will be in $s0 after the program is executed?

       OR

a.       1

b.      3

c.       6

d.      Program will not terminate- it will loop forever.

       addi $t0, $0, 3
       add $s0, $0, 0
doLoop:
       addi $t0, $t0, -1
       add $s0, $s0, $t0
       bne $t0, $0, doLoop

Explanation / Answer

6. Option c: $t0 = 5, $t1 = 10
Explanation:
since $0 =$0; the branch will not take place
in the 2nd statement t0 will get 5 and in the 3rd statement t1 will become 10

7. Option c: 3
$s0 is incremented 3 times.

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