Where Are the Hazards in This MIPS Code?

[brad sue]

hi please I would some help with the MIPS hazard.

Loop:

      add $t0, $t0, $v0
      [B]addi $v0, $v0, -1
      bnez $v0, Loop[/B]

      li $v0, 4
      [U]la $a0, result[/U]
      syscall

      li $v0, 1
      move $a0, $t0
      syscall

      b main

I see an hazard at the instructions that are bolded
What I would do is take the underlined instructions and put it like this:

addi $v0, $v0, -1
la $a0, result      
bnez $v0, Loop

Do I get it right? do I miss another hazard?

thank you
b

 

[Future Bruno]

nez $v0, Loop add $t0, $t0, $v0li $v0, 4li $v0, 1la $a0, resultmove $a0, $t0syscallb mainYes, you have identified the hazard correctly. There is no other hazard in this loop.

 

[quicknote]

main

Hello,

Thank you for reaching out for help with the MIPS hazard. It appears that you have correctly identified one of the hazards in the given code. The bolded instructions create a data hazard, as the value of $v0 is being used in the addi instruction before it has been updated by the previous add instruction.

Another hazard that could potentially occur is a control hazard, as the bnez instruction branches back to the Loop label before the addi instruction has been completed. This could lead to incorrect results if the branch is taken before the addi has a chance to update the value of $v0.

To avoid these hazards, you could rearrange the instructions to ensure that the add instruction is completed before the addi and bnez instructions are executed. You could also consider using a different branch instruction, such as beq, which checks for equality instead of non-zero values.

I hope this helps. Good luck with your project!