Answer:

A null (zero) is placed in the link field of the last node. (Actually, for SPIM, a zero is already there. But with a real OS where dynamic memory is both allocated and deallocated you must be sure to zero the link.)

        
done:
        # end the list
        sw      $0,4($s1)         # put null in the link field
                                  # of the current node, which
                                  # is the last node.

Traversing the List

Now the linked list has been constructed. The field first points to the first node. Here is some code from the previous chapter. The code traverses a linked list, printing out the data found at each node.

Our new linked list has the same form as the previous chapter's linked list. This code will work fine with it.

          # assume that the linked list has been built,
          # (as above) and that the field "first" points 
          # to the first node.
          
             ,     # get a pointer to the first node
                                # put it in $s0
          
lp:       beqz   $s0,endlp      # while the pointer is not null
          lw     $a0,0($s0)     #   get the data of this node

                                #    
          li     $v0,1          #   print it
          syscall               #
          la     $a0,sep        #   print separator
          li     $v0,4          #
          syscall               #
          
          lw     $s0,4($s0)     #   get the pointer to the next node
          b      lp
          
endlp:     . . . 

          .data
first:    .word  0
sep:      .asciiz "  "

Answer:

Traversing the List

QUESTION 17: