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J. Glenn Brookshear

Chapter 8. C H A P T E R 3. Data Abstractions. J. Glenn Brookshear 蔡 文 能. J. Glenn Brookshear. Chapter 8: Data Abstractions. 8.1 Data Structure Fundamentals 8.2 Implementing Data Structures 8.3 A Short Case Study 8.4 Customized Data Types 8.5 Classes and Objects

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J. Glenn Brookshear

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  1. Chapter8 C H A P T E R 3 Data Abstractions J. Glenn Brookshear 蔡 文 能 J. Glenn Brookshear

  2. Chapter 8: Data Abstractions • 8.1 Data Structure Fundamentals • 8.2 Implementing Data Structures • 8.3 A Short Case Study • 8.4 Customized Data Types • 8.5 Classes and Objects • 8.6 Pointers in Machine Language • See the Assembly code: tcc -S file.c gcc -S file.cg++ -S file.cpp

  3. Basic Data Structures • Homogeneous array • Row major: Pascal, C/C++, Java, C# • Column major: FORTRAN • Heterogeneous array: struct/class [, union] • Sequence • List : ArrayList, LinkedList • Stack (in Java, a Stack is a Vector) • Queue • Tree Data Structures: How to arrange data in memory

  4. Arrays vs. Struct (including class) • Homogeneous arrays • Row-major order versus column major order • Address polynomial for each of them • Heterogeneous arrays (class, struct) • Components can be stored one after the other in a contiguous block • Components can be stored in separate locations identified by pointers • Give a formula for finding the entry in the i-throw and the j-thcolumn of a two-dimensional array if it is stored in column major order rather than row major order.

  5. FORTRAN 用小括號 m(2,4) 在 何位址? m[1][3]在 何位址? Array allocation: Row major vs. Column major • Column major (column by column) • FORTRAN: integer*4 m(3,6) • Row major (row by row) • Pascal, C/C++,Java • int m[3][6]; • Subscript 從 0 開始 在 2010 1 2 3 4 5 6 在 2014 1 2 3 0 1 2

  6. M(1,1) 5 6 M(2,1) 1 2 3 4 7 M(1,2) 1 2 Column-major order (FORTRAN) The array is stored as a sequence of arrays consisting of columns (column by column) Integer*4 M(2,7) M(1,2) x(2,4)

  7. 須知道每列(row)有幾欄(column)才能算出下列從何處開始安排須知道每列(row)有幾欄(column)才能算出下列從何處開始安排 • Give a formula for finding the entry in the i-throw and the j-thcolumn of a two-dimensional array if it is stored in column major order rather than row major order. Figure 8.6 A two-dimensional array with four rows and five columns stored in row major order

  8. X[0][0] 4 5 X[0][1] 0 1 2 3 6 X[0][2] X[0][1] 0 X[1][3] 1 Row-major order: Pascal, C/C++, Java,C# The array is stored as a sequence of arrays consisting of rows (row by row) int x[2][7]; // declare as 2 rows x 7 columns x * 在我之前有幾個元素 ?* 每個元素佔幾個 address? (現代電腦大多一byte一個address)

  9. Accessing elements in 2D arrays (1/2) m, n should be constant in most languages int x[m][n]; • m=number of rows • n=number of columns • b=address of x[u][v] • Address of x[i][j] (row major order) is b + ((i-u)*n + j-v) * (size of entry) 通常用u=0, v=0 寫個C/C++小程式然後 gcc -S file.c 或 g++ -S file.cpp編譯成 Assembly code 研究 Brookshear page 306

  10. Accessing elements in 2D arrays (2/2) The formula for finding the address of an element is i = b + s*n s*n = s*e*k + s*n’ = s*(e*k + n’) e = number of elements in a row or column k = number of rows or columns to be skipped n’ = the number of elements between the target element and the beginning of a row or column where b = base address s = size of each element n = the number of elements between the target element and the base address * 在我之前有幾個元素 ?* 每個元素佔幾個 address? (現代電腦大多一byte一個address)

  11. 0 1 2 3 0 1 2 target element Example e*k + n’ = 6 skip 6 elements 在我之前有6個 Row-major order: e = 4, k = 1, n’ = 2 Column-major order: e = 3, k =2, n’ = 1 e*k + n’ = 7 skip 7 elements 在我之前有 7 個 Base address

  12. Dynamic Array ? • Most Languages do NOT support dynamic array • Use pointer in C/C++ int *p; p = new int[ numberOfElementsNeeded ]; • In Java, array is an object; create it dynamically! int p[ ]; // p is only a reference p = new int[ numberOfElementsNeeded ]; Q: C++ STL 的 vector 如何“自動長大” ?

  13. Represent the data and Store the data Primitive data: (大部份電腦用硬體就可處理的) • bit, byte, word, double word, quad word • C: char, short, int, long, float, double,[long double], and pointer of any data type • C++: bool, + ALL types in C • Java: boolean, byte, char(16 bits),short(16 bits) • int(32 bits), long(64 bits), float(32 bits), double(64 bits) Abstract data (User defined data type): List, Deque, Stack, Queue, Tree, …

  14. 結構與型別 (如何自定資料結構?)solution: struct • 只能用前述基本 data type 嗎? • User defined data type? • 考慮寫程式處理全班成績資料, 包括加總平均並排序(Sort)然後印出一份照名次排序的以及一份照學號排序的全班資料 • 如何做 Sort (排序) ? • Sort 時兩學生資料要對調, 要如何對調? • 有很多 array ? 存學號的 array, 存姓名的 array? 存成績的 array? … Bubble sort, Insertion sort, Selection sort

  15. So, Why using struct in C ? Why using class in C++ ? • Struct 可以把相關資料 group 在一起 struct student x[99], tmp; /* … */ tmp = x[i]; x[i] = x[k]; x[k] = tmp; • 增加程式可讀性 • 程式更容易維護 • 其實都不用到struct也能寫出所有程式! • How to define / use struct ? • Next slides User defined data type is a template for a heterogeneous structure

  16. User-defined Data Type 考慮寫個程式處理學生的成績資料, 如何表示一個學生的資料? 想一想. . . #include <stdio.h> struct Student { long sid; char name[9]; /*可存四個Big5中文 */ float score[13]; /*每人最多修13 科 */ }; /*注意struct與class之 分號不能省掉*/ int main( ) { structStudent x; /* C++和 C99不用寫 struct*/ x.sid = 123; /* dot notation */ strcpy(x.name, "張大千") ; /*注意字串不能= */ /* 用 loop 把成績讀入 x.score[?] */ } 習慣上第一個字母大寫 //C 的字串不能直接用 = 複製! 也可用 memcpy( )

  17. x pointer 6087 The pointer points to memory cell x which contains the value 6087. Dynamic Data Structures • Static Data Structures: Size and shape of data structure does not change • Dynamic Data Structures: Size and shape of data structure can change at Run time (How?) • Pointers: An integer which is a memory address points to some data unit. For example: the program counter in the CPU. (Intel CPU uses CS:IP)

  18. Sequence • A group of elements in a specified order is called a sequence. • A tuple is a finite sequence. • Ordered pair (x, y), triple (x, y, z), quadruple, and quintuple • A k-tuple is a tuple of k elements. • List, Stack, Queue , Deque, Vector all are sequence data structures that their size may vary.

  19. Figure 8.1 Lists, stacks, and queues C++ STL, Java 都有提供 iterator 可走遍整個資料結構 An iteratoris an object that enables a user to loop througha collection withoutaccessing the collection’s fields.

  20. Lists • A list is a collection of entries that appear in sequential order. • Can be an ArrayList or be a LinkedList. • ArrayList: List stored in a homogeneous array; also known as Contiguous list • LinkedList: List in which each entries are linked by pointers • Head pointer: Pointer to first entry in list • NIL pointer: A “non-pointer” value used to indicate end of list (NULL or 0) • List in C++ STL vs. List in Java • In C++ STL, list is a template class which implements as a linkedlist • In Java, java.util.List is a interface which extends from java.util.Collection interface; java.util.ArrayList and java.util.LinkedList both implements List interface and thus they are both Lists.

  21. ArrayList -- Contiguous Storage of a List • List entries are stored consecutively in memory : ArrayList • Advantage: simple, quick random access • Disadvantage: insertion and deletion difficult. Java.util.ArrayList is NOT a Queue because it does NOT implement java.util.Queue interface; However, java.util.LinkedList is a Queue and thus the following statement is correct: Queue<Integer> gg = new LinkedList<Integer> ( );

  22. Linked Lists • Each list entry contains a pointer to the next entry • List entries can be stored in any part of memory. • There is a special head pointer for the first entry and a Nil pointer in the last entry. java.util.LinkedList is a Queue and thus the following statement is correct: Queue<Integer> gg = new LinkedList<Integer> ( ); However, the following is WRONG ! Queue<Integer> gg = new ArrayList<Integer> ( ); // wrong In JDK 1.5, LinkedList implements Queue; In JDK 1.6, it implements Deque which extends Queue

  23. Figure 8.9 The structure of a linked list Singly LinkedList Java 的 LinkedList 與 C++ STL 的 list 都是doubly linked list

  24. Figure 8.10 Deleting an entry from a linked list Singly LinkedList Figure 8.11 Inserting an entry into a linked list

  25. head p Add before this node head tmp p Inserting an entry into a doubly linked list 共有四個 Link 要改 ! What about delete ?

  26. Stacks • A stack is a list in which all deletions and insertions occur at the same end. • Alternative name: LIFO structure • Last In First Out • Top: place where insertions and deletions occur. • Base: opposite end to the top. • Insertion=push, deletion = pop. A stack in memory

  27. Stacks and Procedures/Functions Call • When a procedure P is called a pointer to the return location is pushed onto a stack. • If a second procedure Q is called from P then a pointer to the return location for Q is pushed onto the same stack. • When a procedure finishes, the return location is popped from the stack. 注意gcc/g++ 翻譯出的 組合語言格式與 Intel 公佈的或是Microsoft MASM 格式不同!除了指令名稱略不同外,指令中運算元(operands)的 左右順序相反! tcc –S filename.c tcc –S filename.cpp gcc –S filename.c g++ –S filename.cpp

  28. C++ STL stack contains a deque [, vector, list ] template <class T, class T38=deque<T>> class stack { protected: T38 c; // the actual data is here public: typedef typename T38::value_typevalue_type; typedef typename T38::referencereference; typedef typename T38::const_referenceconst_reference; stack( ):c( ) { } explicit stack(const T38& seq) : c(seq) { } bool empty( ) const { return c.empty( ); } int size( ) const { return c.size ( ); } reference top( ) { return c.back( ); } const_reference top( ) const { return c.back( ); } void push(const value_type& x) { c.push_back(x); } void pop( ) { c.pop_back( ); } }; // class stack stack<int> gg; stack<int, vector<int> > yy;

  29. Stack in java.util.*; package java.util; public class Stack<E> extends Vector<E> { public Stack() { } public boolean empty() { return size( ) == 0; } public E push(E item) { addElement(item); return item; } public synchronized E pop( ) { E obj; int len = size( ); obj = peek( ); removeElementAt(len - 1); return obj; } public synchronized E peek( ) { int len = size( ); if (len == 0) throw new EmptyStackException(); return elementAt(len - 1); } public synchronized int search(Object o) { int i = lastIndexOf(o); if (i >= 0) { return size( ) - i; } return -1; } private static final long serialVersionUID = 1224463164541339165L; } // class Stack

  30. Vector in java.util.*; package java.util; public class Vector<E> extends AbstractList<E> implementsList<E>, RandomAccess, Cloneable, java.io.Serializable { protected Object[ ] elementData; // Java 的參考是類似C 的指標 protected int elementCount; protected int capacityIncrement; // if 0, double the capacity (array length) private static final long serialVersionUID = -2767605614048989439L; public Vector(int initialCapacity, int capacityIncrement) { super( ); // invoke constructor of my father if (initialCapacity < 0) throw new IllegalArgumentException("Illegal Capacity: "+ initialCapacity); this.elementData = new Object[initialCapacity]; this.capacityIncrement = capacityIncrement; } public Vector(int initialCapacity) { // default increment 0 will double capacity this(initialCapacity, 0); } public Vector( ) { this(10); } // Default capacity is 10 public synchronized int size( ) { return elementCount; } // . . .

  31. Stack 應用 again • 使用堆疊把 infix  postfix • Stack 內放運算符號和左括號, 注意優先權 • Operand 直接輸出 • 使用堆疊計算 postfix 的答案 • Stack 內放 operand 值 (value, 整數或實數) • 遇到 operator 就把 operand pop出來算 • 算出的 value 再 push 進去

  32. p long x next Handling Pointers struct Student { long x; struct Student * next; }; • Assign the value 4 to x: p->x = 4; // C, C++ p.x = 4; // Java, C# • Initialise a pointer q to point to the next item: q = p->next; // C, C++ q = p.next; // Java struct Student * p;

  33. Figure 8.19 A procedure for printing a linked list

  34. Print a Linked List in Reverse Order Procedure reverseprint(L) p = head(L) While p<>Nil do push(p.data, stack) p=p.next EndWhile While stack not empty Print(pop(stack)) EndWhile End Procedure

  35. Queue • A queue is a list in which insertions are performed at one end (tail) and deletions are performed at the opposite end (head). • FIFO: First In, First Out tail: insertions tail pointer: to next unused location head: deletions head pointer: to head item

  36. Circular Queue empty vs. full EMPTY QUEUE [3] [2] [3] [2] J2 J3 [1] [4] [1] [4] J1 [0] [5] [0] [5] front = 0 front = 0 rear = 0 rear = 3 What if Queue is full ?

  37. Leaveone empty space when Queue is full Why? FULL QUEUE FULL QUEUE [2] [3] [2] [3] J8 J9 J2 J3 J7 [1] [4][1] [4] J1 J4 J6 J5 J5 [0] [5] [0] [5] front =0 rear = 5 front =4 rear =3 How to test when queue is empty? How to test when queue is full?

  38. void enqueue(int front, int *rear, element item){/* add an item to the queue */ *rear = (*rear +1) % MAX_QUEUE_SIZE; if (front == *rear) /* reset rear and print error */ return; } queue[*rear] = item; } Enqueue in a Circular Queue

  39. element dequeue(int* front, int rear){ element item; /* remove front element from the queue and put it in item */ if (*front == rear) return queue_empty( ); /* queue_empty returns an error key */ *front = (*front+1) % MAX_QUEUE_SIZE; return queue[*front];} Dequeue from Circular Queue

  40. Trees Earth Europe N. America Africa Asia S. America China Chad UK France Peru USA India Beijing Paris London Lima

  41. Figure 8.2 An example of an organization chart

  42. Definition of a Tree • A tree is a connected undirected graph in which there are no circuits. • A rooted tree is a tree in which one node is selected as the root. • In computing, trees are usually rooted.

  43. Terminology • Node, root, leaf, terminal node, parent, child. • Ancestor: Parent, parent of parent, etc. • Descendent: Child, child of child, etc. • Siblings: Nodes sharing a common parent • Subtree: a node together with all the nodes below it. • Depth: number of nodes on longest path from root to a leaf. • Binary tree: each node has at most two children.

  44. Figure 8.3 Tree terminology

  45. Binary Trees • A binary tree is a finite set of nodes that is either empty or consists of a root and two disjoint binary trees called the left subtreeand the right subtree. • Any tree can be transformed into binary tree. • by left child-right sibling representation • The left subtree and the right subtree are distinguished.

  46. A A A B B B C C F G D E D E H I Samples of Binary Trees Complete Binary Tree 1 2 Skewed Binary Tree 歪斜樹 3 4 4 5 What is a Full Binary Tree ?

  47. Figure 8.16 The conceptual and actual organization of a binary tree using a linked storage system

  48. Figure 8.17: A tree stored without pointers 適合大部份 node 都存在的 tree : complete tree, full tree

  49. Figure 8.18: A sparse, unbalanced tree shown in its conceptual form and as it would be stored without pointers 大部份空間都浪費掉了!

  50. Recursive Printing of a Binary Tree Procedure printTree(tree) If (tree not empty) printTree(left subtree) print(root) printTree(right subtree) EndIf EndProcedure In-order, pre-order, post-order traversal

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