# 10.3 Implementing pointers and objects

## 10.3-1

Draw a picture of the sequence $\langle 13, 4, 8, 19, 5, 11 \rangle$ stored as a doubly linked list using the multiple-array representation. Do the same for the single-array representation.

• A multiple-array representation with $L = 2$,

$$\begin{array}{|r|c|c|c|c|c|c|c|} \hline index & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline next & & 3 & 4 & 5 & 6 & 7 & \diagup \\ \hline key & & 13 & 4 & 8 & 19 & 5 & 11 \\ \hline prev & & \diagup & 2 & 3 & 4 & 5 & 6 \\ \hline \end{array}$$

• A single-array version with $L = 1$,

$$\begin{array}{|r|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline index & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 \\ \hline key & 13 & 4 & \diagup & 4 & 7 & 1 & 8 & 10 & 4 & 19 & 13 & 7 & 5 & 16 & 10 & 11 & \diagup & 13 \\ \hline \end{array}$$

## 10.3-2

Write the procedures $\text{ALLOCATE-OBJECT}$ and $\text{FREE-OBJECT}$ for a homogeneous collection of objects implemented by the single-array representation.

ALLOCATE-OBJECT()
if free == NIL
error "out of space"
else x = free
free = A[x + 1]
return x

FREE-OBJECT(x)
A[x + 1] = free
free = x


## 10.3-3

Why don't we need to set or reset the $prev$ attributes of objects in the implementation of the $\text{ALLOCATE-OBJECT}$ and $\text{FREE-OBJECT}$ procedures?

We implement $\text{ALLOCATE-OBJECT}$ and $\text{FREE-OBJECT}$ in the hope of managing the storage of currently non-used object in the free list so that one can be allocated for reusing. As the free list acts like a stack, to maintain this stack-like collection, we merely remember its first pointer and set the $next$ attribute of objects. There is no need to worry the $prev$ attribute, for it hardly has any impact on the resulting free list.

## 10.3-4

It is often desirable to keep all elements of a doubly linked list compact in storage, using, for example, the first $m$ index locations in the multiple-array representation. (This is the case in a paged, virtual-memory computing environment.) Explain how to implement the procedures $\text{ALLOCATE-OBJECT}$ and $\text{FREE-OBJECT}$ so that the representation is compact. Assume that there are no pointers to elements of the linked list outside the list itself. ($\textit{Hint:}$ Use the array implementation of a stack.)

ALLOCATE-OBJECT()
if STACK-EMPTY(F)
error "out of space"
else x = POP(F)
return x

FREE-OBJECT(x)
p = F.top - 1
p.prev.next = x
p.next.prev = x
x.key = p.key
x.prev = p.prev
x.next = p.next
PUSH(F, p)


## 10.3-5

Let $L$ be a doubly linked list of length $n$ stored in arrays $key$, $prev$, and $next$ of length $m$. Suppose that these arrays are managed by $\text{ALLOCATE-OBJECT}$ and $\text{FREE-OBJECT}$ procedures that keep a doubly linked free list $F$. Suppose further that of the $m$ items, exactly $n$ are on list $L$ and $m - n$ are on the free list. Write a procedure $\text{COMPACTIFY-LIST}(L, F)$ that, given the list $L$ and the free list $F$, moves the items in $L$ so that they occupy array positions $1, 2, \ldots, n$ and adjusts the free list $F$ so that it remains correct, occupying array positions $n + 1, n + 2, \ldots, m$. The running time of your procedure should be $\Theta(n)$, and it should use only a constant amount of extra space. Argue that your procedure is correct.

We represent the combination of arrays $key$, $prev$, and $next$ by a multible-array $A$. Each object of $A$'s is either in list $L$ or in the free list $F$, but not in both. The procedure $\text{COMPACTIFY-LIST}$ transposes the first object in $L$ with the first object in $A$, the second objects until the list $L$ is exhausted.

COMPACTIFY-LIST(L, F)
i = 2
while l != NIL
TRANSPOSE(A[l], A[i])
if F == i
F = l
l = A[l].next
i = i + 1

TRANSPOSE(a, b)
SWAP(a.prev.next, b.prev.next)
SWAP(a.prev, b.prev)
SWAP(a.next.prev, b.next.prev)
SWAP(a.next, b.next)