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Senem Kumova Metin STRUCTURES CHAPTER 9 in A Book in C.

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Presentation on theme: "Senem Kumova Metin STRUCTURES CHAPTER 9 in A Book in C."— Presentation transcript:

1 Senem Kumova Metin STRUCTURES CHAPTER 9 in A Book in C

2 Senem Kumova Metin Structures in C A structure is –a convenient way of grouping several pieces of related information together –a collection of variables under a single name Examples : real number && imaginary number  complex number ( 3+5i ) height && width && length  rectangular prism

3 Senem Kumova Metin Structures in C The variables in structures can be of different types, and each has a name which is used to select it from the structure Example : ID (integer) && Name (char array)  A Student record A structure is a new named type that may involve several different typed variables

4 Senem Kumova Metin Defining Structures 1 struct complex_number { int real_part; int imaginary_part; }; “complex_number” is the tag name “struct complex_number” is the new type

5 Senem Kumova Metin Defining Structures 2 /* DEFINITION OF RECTANGULAR PRISM */ struct rectangular_prism {int height; int width; int length;}; // name of new type?? /* DEFINITION OF STUDENT RECORD */ struct student_record { int ID; char name[100]; };

6 Senem Kumova Metin Structures : Creating objects struct complex_number { int real_part; int imaginary_part;}; // Below the definition of structure you can create // objects from it  “s1” and “s2” struct complex_number s1; struct complex_number s2; s2 real_part imaginary_part s1 real_part imaginary_part

7 Senem Kumova Metin Structures : Creating objects struct rectangular_prism {int height; int width; int length;}; // Below the definition of structure you can create // objects from it  “obj” struct rectangular_prism obj; obj height lengthwidth

8 Senem Kumova Metin Structures : Creating static arrays of objects struct student_record { int ID; char name[100]; }; // Creates an array of student records  “group” struct student_record group[4]; group IDname IDname IDname IDname group[0]  group[1]  group[2]  group[3] 

9 Senem Kumova Metin Structures : Creating dynamic arrays of objects struct rec { int ID; char name[100]; }; // Creating a dynamic array // of student records  “group” struct rec * group; // DECLARES POINTER group = ( struct rec * ) malloc ( sizeof (struct rec) * 4 ); size of objects in array number of objects in array (array size)

10 Senem Kumova Metin Structures : Accessing members of structures struct rectangular_prism {int height; int width; int length;}; // Create an object from the structure defined above  “obj” struct rectangular_prism obj; // Members of the objects can be accessed by putting a dot // following the object name obj.height=10; obj.width=15; obj.length=40; obj height=10 length=40width=15

11 Senem Kumova Metin Structures : Accessing members of structures struct rectangular_prism {int height; int width; int length;}; struct rectangular_prism obj; // Create a pointer to point object “obj” struct rectangular_prism *p = &obj; (*p).height=10 // or obj.height or p->height=10 p->width=15; p->length=40; obj == *p height=10 length=40width=15

12 Senem Kumova Metin Structures : Accessing members of structures // Defines structure struct student_record { int ID; char name[100]; }; // Creates an array of student records  “group” struct student_record group[2]; group[0].ID=200710; strcpy(group[0].name, “doddy”); group[1].ID=200711; strcpy(group[1].name,group[0].name); group ID=200710name= “doddy” ID=200711name =?? group[0]  group[1] 

13 Senem Kumova Metin Structures : DECLARATION ALTERNATIVES Declaration 1 : struct record { int ID; char * name; char grade; }; struct record s1; struct record s2; struct record s3; Declaration 2 : struct record { int ID; char * name; char grade; } s1, s2; struct record s3;

14 Senem Kumova Metin Structures : DECLARATION ALTERNATIVES Declaration 1 : struct record { int ID; char * name; char grade; }; struct record s1; struct record s2; Declaration 3 : struct { int ID; char * name; char grade; } s1, s2; /* no tag name */ /* no permission to declare other variables of this type */

15 Senem Kumova Metin Structures : DECLARATION ALTERNATIVES Declaration 4 : struct record { int ID; char * name; char grade; }; typedef struct record rec; rec s1; struct record s2; Declaration 5 : /* high degree of modularity and portability */ typedef struct { int ID; char * name; char grade; } rec; rec s1; rec s2;

16 Senem Kumova Metin Initialization of Structure Objects 1.struct names { char name[10]; int length; int weigth;} man[3]= { “Tom”, 180, 65, “George”, 170, 68, “Bob”, 190, 100 }; 2.struct names woman[2]={{“Mary”, 170, 55}, {“Sue”, 160,67}}; 3.struct names your; your. name=“Jane”; your.length=160; your.weigth=50;

17 Senem Kumova Metin Structures in Structures #include struct physical_info { int length; int weigth; } ; struct record { int salary; int working_hour; struct physical_info man; } ; main() {struct record s1; s1.salary=10000; s1.working_hour= 6; s1.man.length=180; s1.man.weigth=78; }

18 Senem Kumova Metin Exercise 1 on Structures Declare a structure to represent fractions Create 2 objects of fractions Ask user to fill the objects Calculate and print out the multiplication value  Fraction x / y  x over y (3/7, 8/5)  Need 2 numbers as numerator and denominator  User has to give 2 numbers for each object  Multiplication rule  a/b * c/d = (a*c) / (b*d)

19 Senem Kumova Metin Exercise 1 on Structures struct fraction {int n; // an integer to represent numerator int d; // an integer to represent denominator }; main() { struct fraction obj1, obj2; // Create input objects struct fraction result; // Create an object to store the result of // multiplication printf(“please enter values for fractions”); scanf(“%d%d”, &obj1.n, &obj1.d); scanf(“%d%d”, &obj2.n, &obj2.d); result.n = obj1.n * obj2.n; result.d = obj1. d * obj2. d; printf(“result is %d / %d”, result.n, result.d); }

20 Senem Kumova Metin Exercise 2 on Structures Declare a structure for a customer record (name, surname, phone number) Create 5 customer objects Ask user to fill the objects Print out the information of all customers as in the following output customer 1 : Gabriel Gray customer 2 : Claire Bennet customer 3 : Hiro Nakamura customer 4 : Nathan Petrelli customer 5 : Niki Sanders

21 Senem Kumova Metin Exercise 2 on Structures struct customer {char name[100]; char surname[100]; int phone; }; main() { struct customer heroes[5]; // Create input objects int i; printf("please fill customer records"); for(i=0;i<5;i++) scanf("%s%s%d", heroes[i].name, heroes[i].surname, &heroes[i].phone ); for(i=0;i<5;i++) printf(“customer %d : %s%s%d\n", i+1, heroes[i].name, heroes[i].surname, heroes[i].phone); }

22 Senem Kumova Metin Exercise 3 on Structures Declare a structure for complex numbers (real and imaginary part) Create 1 dynamic object (use pointers) Ask user to fill the object Print out the complex number as given below output Please give the values for complex number : 5 6 Complex number : 5 + 6i

23 Senem Kumova Metin Exercise 3 on Structures struct complex {int real; int imaginary; }; main() { struct complex * p; // Declare pointer for object // Memory allocation for object p=(struct complex *) malloc(sizeof(struct complex)); printf( " Please give the values for complex number : " ); scanf( "%d%d", &(p->real), & (p->imaginary) ); printf( "Complex number : %d + %d i", p->real, p->imaginary ); }

24 Senem Kumova Metin Structures as Function Arguments Call by value Call by reference

25 Senem Kumova Metin // A function to add two integers int add(int a, int b) { int result= a+b; return result;} struct complex { int r; int i; }; // A function to add two complex numbers struct complex add ( struct complex a, struct complex b ) { struct complex result; result.r=a.r+b.r; result.i=a.i+b.i; return result;} Exercise 1 : Define a function to add two complex numbers

26 Senem Kumova Metin // A function to print an integers void print(int a) { printf(“%d \n”,a);} struct complex { int r; int i; }; // A function to print a complex number struct complex print ( struct complex a) { printf(“%d + i %d \n”, a.r,a.i);} Exercise 1 : Define a function to print a complex number

27 Senem Kumova Metin Exercise 1 on Call by Value : Define main() // Structure definition struct complex { int r; int i; } ; // function prototypes struct complex add ( struct complex a, struct complex b ); void print ( struct complex a); main() { struct complex e1={2,3}; struct complex e2 ={1,2}; struct complex e3; print(e1); print(e2); e3=add(e1, e2); print(e3);}

28 Senem Kumova Metin Exercise 2 on Call by Value struct fraction {int n; // an integer to represent numerator int d; // an integer to represent denominator }; struct fraction sum(struct fraction x, struct fraction y) { struct fraction result; result.d=x.d*y.d; result.n=x.n*y.d+y.n*x.d; return result; } struct fraction divide(struct fraction x, struct fraction y) { struct fraction result; result.d=x.d*y.n; result.n=x.n*y.d return result; }

29 Senem Kumova Metin Exercise 3 on Call by Value struct rectangular_prism {int height; int width; int length;}; ……. volume(……) {…….} …… area (……) {…….}

30 Senem Kumova Metin Exercise 3 on Call by Value struct rectangular_prism {int height; int width; int length;}; int volume(struct rectangular_prism x) {return x.height*x.width*x.length; } int area (struct rectangular_prism x) { return 2*x.width*x.lenght + 2*x.lenght*x.height + 2*x.width*x.lenght; }


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