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Answer:


        # create the second node 
        li      $v0,9             # allocate memory
        li      $a0,8             # 8 bytes
        syscall                   # $v0 <-- address 
        
        # link the second node to the first
        sw      $v0,4($s1)        # copy address of second node
                                  # into the first node
        
        # make the new node the current node
        move    $s1,$v0           # $s1 = &(second node)

        # initialize the second node
        li      $t0,2             # store 2
        sw      $t0,0($s1)        # at displacement 0

$s1 as the Current node

Three Node List

After the above code has finished, the first two nodes have been created. Register $s1 is now pointing to the second node. The first node has not been lost, because the field first points to it. When a register is used to point at the node were work is being done (as does $s1) it is sometimes called a pointer to the current node.

Now let us work on the third node. Again, $s1 is used as the pointer to the current node.

        # create the third node 
        li      $v0,9             # allocate memory
        li      $a0,8             # 8 bytes
        syscall                   # $v0 <-- address
        
        # link the third node to the second
        sw      $v0,4($s1)        # copy address of third node
                                  # into the second node
        
        # make the new node the current node
        move    $s1,$v0           # $s1 = &(third node)

        # initialize the third node
        li      $t0,3             # store 3
        sw      $t0,0($s1)        # at displacement 0

        # end the list
        sw      $0,4($s1)         # put null in the link field

Linked lists can be aggravating. Look at the line of code where the current pointer is changed from the second node to the third:

        # make the new node the current node
        move    $s1,$v0           # $s1 = &(third node)

This "change in meaning" of $s1 must be done in the correct sequence. The previous value of $s1 must already be saved in the second node.


QUESTION 10:

Register $s1 no longer points at the second node. Has the second node been lost?