Les pointeurs 1. 00A0 00A1 00A2 00A3 00A4 00A5 00A6 00A7 00A8 00A9 00AA 00AB 00AC 00AD 00AE 00AF 00B0 2 Code du programme BSSTas void main(){ node exemple(0);

Slides:



Advertisements
Similar presentations
Complete Structure class Date {class Date { private :private : // private data and functions// private data and functions public :public : // public data.
Advertisements

LINKED LIST, STACKS AND QUEUES Saras M Srivastava, PGT – Comp. Sc. Kendriya Vidyalaya TengaValley.
FACTORISATION à laide dun facteur commun Exemple : Exemple : Factoriser lexpression suivante : (3x + 2)(x – 1) + (2x + 4)( 3x + 2)
For(int i = 1; i
Inheritance (2).
Review of Inheritance. 2 Several Levels of Inheritance Base Class B Derived class D Derived class D1.
1 Class Vehicle #include #define N 10../.. 2 Class Vehicle class vehicle { public: float speed; char colour[N+1]; char make[N+1];
S. Sudarshan Based partly on material from Fawzi Emad & Chau-Wen Tseng
Les questions orales.  ____________________________________________  ferai  _____________________________________  ________________________________________________.
AB 11 22 33 44 55 66 77 88 99 10  20  19  18  17  16  15  14  13  12  11  21  22  23  24  25  26  27  28.
Deux applications. Freelists Les appels système new et delete sont lents: // Singly-linked list node with freelist template class Link { private: static.
DictionaryADT and Trees. Overview What is the DictionaryADT? What are trees? Implementing DictionaryADT with binary trees Balanced trees DictionaryADT.
Recursion practice. Problem 0 Using recursion (and no arrays), write the code to read in a series of numbers (until EOF) and then print them backwards.
Passing arguments by value void func (int x) { x = 4; }... int a = 10;... func(a); cout
計算機概論實習 Claim a Parameter int a; a = 5; typenameaddress inta0x0001 addressvalue 0x0001 0x0002 0x0003 0x0004 0x0005 0x0006 0x0007 0x0008.
Copyright 2002, Pearson Education Canada1 How to Read and Understand Graphs Appendix to Chapter 1.
Insertion into a B+ Tree Null Tree Ptr Data Pointer * Tree Node Ptr After Adding 8 and then 5… 85 Insert 1 : causes overflow – add a new level * 5 * 158.
Chapter 17 Linked List.
1 Chapter 3 Data Representation Part 1. 2 Goals Introduce the different ways in which data may be represented Concepts –Abstract data types –Formula-based,
1 Chapter 16-1 Linked Structures Dale/Weems. 2 Chapter 16 Topics l Meaning of a Linked List l Meaning of a Dynamic Linked List l Traversal, Insertion.
Chapter 4 sections 1 and 2.  Fig. 1  Not connected  All vertices are even.  Fig. 2  Connected  All vertices are even.
CS162 - Topic #11 Lecture: Recursion –Problem solving with recursion –Work through examples to get used to the recursive process Programming Project –Any.
Chair to Horse Proportion Lines of Superficies The marks are not equally spaced.
1 AA AB AC AD AE AF AG AH AG39 006AG Origin Point 180 E/W & 90S 006 AG Cell to Quadrant to.
 Trees Data Structures Trees Data Structures  Trees Trees  Binary Search Trees Binary Search Trees  Binary Tree Implementation Binary Tree Implementation.
CS342 Data Structures End-of-semester Review S2002.
An Introduction to C++ A First Look. void Functions #include #include void main( ) { cout
Adding SubtractingMultiplyingDividingMiscellaneous.
Stack Implementations Chapter 6 Copyright ©2012 by Pearson Education, Inc. All rights reserved.
Algorithms and Data Structures
LINKED LISTS Midwestern State University CMPS 1053 Dr. Ranette Halverson 1.
10.5 Tangents Tangent: a line that shares only one point with a circle and is perpendicular to the radius or diameter at that point. Point of tangency:
1 For Loops l From Chapter 9 l A shorthand way of coding count loops.
PROPERTIES OF REAL NUMBERS. COMMUTATIVE PROPERTY OF ADDITION What it means We can add numbers in any order Numeric Example Algebraic Example
Determine the sequence of genes along a chromosome based on the following recombination frequencies A-C 20% A-D 10% B-C 15% B-D 5%
 Memory from the heap  Dynamic memory allocation using the new operator  Build a dynamic linked list  Another way of traversing a linked list 
KENTUCKY CLASSIC CDE Marathon Obstacles. OBSTACLE 1 Course Designer Richard Nicoll IN/OUT A B D C E F FEI A-F INT A-E PR A-D.
NESTED CLASS. Apa itu nested class ? Nested class is a class defined inside a class, that can be used within the scope of the class in which it is defined.
Programming Circular Linked List.
UNIT – I Linked Lists.
Linked List.
Abstract Data Types Polynomials CSCI 240
Doubly Linked List Review - We are writing this code
Stack Lesson xx   This module shows you the basic elements of a type of linked list called a stack.
Get IEEE 802® Update IEEE 802 Plenary; 802 March, 2015.
Recursion & Linked Lists
تصنيف التفاعلات الكيميائية
Popping Items Off a Stack Lesson xx
Geometry CP Mrs. Mongold
Dynamic Memory A whole heap of fun….
Numerics for PDE using EXCEL
Get IEEE 802® Update IEEE 802 Plenary; 802 EC November 2014.
1-6 Midpoint Find the midpoint of: DE: DC: AB: AE: (2+5) / 2 = 3.5
3-bit calculator using 7-segment LEDs
Week 10, Finite State Machine
Further Mathematics Support Programme
AB AC AD AE AF 5 ways If you used AB, then, there would be 4 remaining ODD vertices (C, D, E and F) CD CE CF 3 ways If you used CD, then, there.
-.&- ·Af& Q 0 "i'/
Dynamic Memory.
Adding with 9’s.
Adding with 10’s.
Copyright © 2013 Elsevier Inc. All rights reserved.
If AD = 10, DC =6, and ED = 15, find DB.
Adding ____ + 10.
BETONLINEBETONLINE A·+A·+
Copyright © 2013 Elsevier Inc. All rights reserved.
Get IEEE 802® Update IEEE 802 Plenary; 802 EC November 2014.
Mobile genetic elements associated with bla OXA genes.
Linked Lists.
Properties of Numbers Review Problems.
Presentation transcript:

Les pointeurs 1

00A0 00A1 00A2 00A3 00A4 00A5 00A6 00A7 00A8 00A9 00AA 00AB 00AC 00AD 00AE 00AF 00B0 2 Code du programme BSSTas void main(){ node exemple(0); exemple.valeur=0; exemple.suivant=new node(5); node *iter; iter=&exemple; iter=iter->suivant; node *iter2=iter; iter2->suivant=new node(10); iter2=iter2->suivant; iter2->suivant=new node(20); iter2=null; iter=null; delete iter; iter=null; } class node{ int valeur; node *suivant; }

00A0 00A1 00A2 00A3 00A4 00A5 00A6 00A7 00A8 00A9 00AA 00AB 00AC 00AD 00AE 00AF 00B0 3 Code du programme BSSTas void main(){ node *iter=new node(-1); for(int i=0;i<5;i++){ iter->suivant=new node(i); iter=iter->suivant(); }; node *iter1=iter; iter=iter1; for(int i=0;i<5;i++){ cout valeur; iter=iter->suivant(); } class node{ int valeur; node *suivant; }

00A0 00A1 00A2 00A3 00A4 00A5 00A6 00A7 00A8 00A9 00AA 00AB 00AC 00AD 00AE 00AF 00B0 4 Code du programme BSSTas void main(){ node *iter=new node(-1); node *iter1=iter; for(int i=0;i<5;i++){ iter->suivant=new node(i); iter=iter->suivant(); } iter=iter1; while (iter != null) { cout valeur; iter=iter->suivant(); } class node{ int valeur; node *suivant; }