Embed Size (px)
Citation preview
Senem Kumova Metin
STRUCTURESCHAPTER 9 in A
Book in C
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
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
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
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]; };
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
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
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
ID name
ID name
ID name
ID name
group[0]
group[1]
group[2]
group[3]
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 ofobjects in
array
number of objects in
array(array size)
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=40
width=15
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=40
width=15
Senem Kumova Metin
Structures : Accessing members of structures
// Defines structurestruct 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=200710
name= “doddy”
ID=200711
name =??
group[0]
group[1]
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;
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 */
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;
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;
Senem Kumova Metin
Structures in Structures#include<stdio.h>
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; }
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)
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); }
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 1222222customer 2 : Claire Bennet 1234567customer 3 : Hiro Nakamura 2222222customer 4 : Nathan Petrelli 1234569 customer 5 : Niki Sanders 2333333
Senem Kumova Metin
Exercise 2 on Structures struct customer{ char name[100];
char surname[100];int phone; };
main(){ struct customer heroes[5]; // Create
input objectsint 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);
}
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
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 );
}
Senem Kumova Metin
Structures as Function Arguments
Call by valueCall by reference
Senem Kumova Metin
// A function to add two integersint add(int a, int b){ int result= a+b;
return result; }
struct complex { int r; int i; };
// A function to add two complex numbersstruct 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
Senem Kumova Metin
// A function to print an integersvoid print(int a){ printf(“%d \n”,a); }
struct complex { int r; int i; };
// A function to print a complex numberstruct complex print ( struct complex a){ printf(“%d + i %d \n”, a.r,a.i); }
Exercise 1 : Define a function to print a
complex number
Senem Kumova Metin
Exercise 1 on Call by Value : Define main()
// Structure definitionstruct complex { int r; int i; } ;
// function prototypesstruct 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); }
Senem Kumova Metin
Exercise 2 on Call by Valuestruct 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.dreturn result;
}
Senem Kumova Metin
Exercise 3 on Call by Value
struct rectangular_prism{ int height;
int width;int length; };
……. volume(……){…….}
…… area (……){…….}
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;
}
