Need help solving this loop in MIPS(assembly)

Question

I have a loop in MIPS(assembly) that looks like this:

LOOP: 
      slt $t2, $0, $t1
      beq $t2, $0, DONE
      subi $t1, $t1, 1
      addi $s2, $s2, 2
      j LOOP
DONE:

If the value of $t1 is set to 10 in the beginning, what is the value of $s2 if $s2 is set to ZERO in the beginning?

I tried to solve the answer to the question and I'm getting 20. But, the solutions manual says its 200. Can anyone tell me what im missing out here? Won't $s2 keep getting incremented by 2 until $t1 is 0?

Answer 1

I agree with @Joe Farrell, this loop is basically equivalent to this C-style loop:

while(t1 > 0)
{
  --t1;
  s2 += 2;
}

Looks like you're working out of the book "Computer Organization and Design: The Hardware/Software Interface" I only managed to find errata sheets for 3rd and 4th editions, and I didn't see this typo addressed in either. Which edition are you using?

Answer 2

答案将是22。因为如果u重复循环直到t1 = -1，则s2将为22，并且当循环再次开始时，t2变为0，这导致打印DONE（循环退出的地方）。