Fundamental of Programming (C)

Lecturer: Omid Jafarinezhad

Sharif University of TechnologyDepartment of Computer Engineering 1

Fundamental of Programming (C)

Lecture 8Structures, Unions, Bit Manipulations and Enumerations

Structures, Unions, Bit Manipulations and Enumerations – Lecture 9


Outline• Structures– Be able to use compound data structures in

programs• Unions– Be able to share storage space of their members

• Bit fields Structures– Be able to do simple bit-vector manipulations

• Enumerations– Be able to use compound symbolic constants



User Defined Data Types (typedef)• The C language provides a facility called typedef for creating synonyms

for previously defined data type names. • For example, the declaration:

typedef int Length;

makes the name Length a synonym (or alias) for the data type int. • The data type name Length can now be used in declarations in exactly

the same way that the data type int can be used:

Length a, b, len ;Length numbers[10] ;

typedef char String[50];typedef int Array[10];String name;Array ages;



Structures (struct) • Structures—sometimes referred to as aggregates—are collections of

related variables under one name

• Structures may contain variables of many different data types—in contrast to arrays that contain only elements of the same data type

• Structures are commonly used to define records to be stored in files

• Pointers and structures facilitate the formation of more complex data structures such as linked lists, queues, stacks and trees

• Structures are derived data types—they are constructed using objects of other types



Declaring Structures (struct)• The name "employee" is called a structure tag• Variables declared within the braces of the structure

definition are the structure’s members

struct employee {

char firstName[ 20 ]; char lastName[ 20 ]; int age; char gender; double hourlySalary;

} Ali, Sara, empDTS[20];

struct employee Reza, *emp;

struct employee {


};struct employee Ali,

emp[10];

struct {


} Ali;



Declaring Structures (struct)• Often, typedef is used in combination with struct to declare a

synonym (or an alias) for a structure:

typedef struct {


} employee; /* The "alias"

employee Ali; /* Create a struct variable */

struct employee {


} Ali, Sara, empDTS[20];

struct employee Reza, *emp;



Declaring Structures (struct)• Members of the same structure type must have unique

names, but two different structure types may contain members of the same name without conflict

• Each structure definition must end with a semicolon

struct employee {

char Name[ 20 ]; char Name[ 20 ]; // Error!!! int age; char gender; double hourlySalary;

} Ali, Sara, empDTS[20]; struct employee Reza, *emp;

struct Student {

char Name[ 20 ]; // OK int age; char gender;

}; struct Student Ce40153[80];



Declaring Structures (struct)• A structure cannot contain an instance of itself • For example, a variable of type struct employee cannot be

declared in the definition for struct employee A pointer to struct employee, however, may be included

• A structure containing a member that is a pointer to the same

structure type is referred to as a self-referential structure

struct employee2 { // … double hourlySalary; struct employee2 person; /* ERROR */ struct employee2 *ePtr; /* pointer */};



Declaring Structures (struct)• The structure tag name is optional

• If a structure definition does not contain a structure tag name, variables of the structure type may be declared only in the structure definition—not in a separate declaration

struct {


} Ali;



Structure’s sizeof • Structure definitions do not reserve any space in

memory; rather, each definition creates a new data type that is used to define variables

sizeof(struct …) = sum of sizeof(members) + alignment padding (Processor- and compiler-specific)

struct employee { char firstName[ 20 ]; char lastName[ 20 ]; int age; char gender; double hourlySalary;};

struct employee Ali, emp[10];printf("%d", sizeof(Ali));printf("%d", sizeof(emp));printf("%d", sizeof(struct employee));



Memory layoutstruct COST {

int amount; char currency_type[2];

}struct PART {

char id[2]; struct COST cost; int num_avail; }

• Here, the system uses 4-byte alignment of integers, so amount and num_avail must be aligned Four bytes wasted for each structure!

id amount num_avail

cost

currency_type



Memory layoutstruct COST {

int amount; char currency_type[2];

}struct PART {

struct COST cost; char id[2];int num_avail;

}

• Implementation dependent!!!

idamount num_avail

cost

currency_type



Accessing Struct Members• Individual members of a struct variable may be accessed

using the structure member operator (the dot, "."): myEmp.firstName ;

employee. firstName; // Error• Or , if a pointer to the struct has been declared and

initialized employee *emp = &myEmp ;– by using the structure pointer operator :

emp -> firstName; // arrow operator– which could also be written as:

(* emp).firstName;

struct employee { char firstName[ 20 ]; // …

} myEmp;



An Example - Initialization

struct identity js = {"Joe", "Smith", 25}, *ptr = &js ;js.person.id = 123456789 ;js.person.gpa = 3.4 ;

printf ("%s %s %d %ld %f\n", js.FirstName, js.LastName, js.age, js.person.id, js.person.gpa) ;printf ("%s %s %d %ld %f\n", ptr->FirstName, ptr->LastName,ptr->age, ptr->person.id,

ptr->person.gpa) ;

struct identity{ char FirstName[30]; char LastName[30]; unsigned age; struct personal person;};

//Create a struct but don’t reserve spacestruct personal{

long id; // student ID float gpa; // grade point average};

js = {"Joe", "Smith", 25, 9, 10}js.personal.id Errorstrcpy(js.FirstName, "Joe");



An Example - Assignment

struct identity js = {"Joe", "Smith", 25}, oj ;js.person.id = 123456789 ;js.person.gpa = 3.4 ;

oj = js;

printf ("%s %s %d %ld %f\n", oj.FirstName, oj.LastName, oj.age, js.person.id, oj.person.gpa) ;printf ("%s %s %d %ld %f\n", ptr->FirstName, ptr->LastName,ptr->age, ptr->person.id,

ptr->person.gpa) ;

struct identity{ char FirstName[30]; char LastName[30]; unsigned age; struct personal person;};





Arrays of Structures

personage LastName FirstName

gpa id20 9140153 14 Jafarinezhad omid students[0]

20 2222222 90 Shekarestani Samad Students[1]

20 11111111 100 Shekarestani Khaje Nezam students[2]

students[3]

struct identity{ char FirstName[30]; char LastName[30]; unsigned age; struct personal person;} students[4];



struct identity sharifC40153[80] = {"omid", "Jafarinezhad", 14, 9140153, 20, "Samad", "Shekarestani", 90, 2222222, 20} ;

strcpy(sharifC40153[2].FirstName, "Khaje Nezam");strcpy(sharifC40153[2].LastName, "Shekarestani");sharifC40153[2]. age = 100;sharifC40153[2]. person.id = 11111111;sharifC40153[2]. person. gpa = 20;



bool check_birthday(struct Date today, struct Date myFriend){ if ((today.month == myFriend.month) && (today.day == myFriend.day)) return (true); return (false);}int main(){ struct Friend friends[NFRIENDS]; struct Date today = {2012, 3, 11}; // ... for (i = 0; i < NFRIENDS; i++) { if(check_birthday(today, friends[i].Birthday)) printf ("%s %s\n", friends[i].FirstName, oj.LastName) ; } // …

An Example#define NFRIENDS 10struct Date{ unsigned year; unsigned month; unsigned day;};struct Friend { char FirstName[30]; char LastName[30]; struct Date Birthday;};

typedef struct{ unsigned year; unsigned month; unsigned day;} Date;bool check_birthday(Date today, Date myFriend){ //…}



Pointers to StructuresDate create_date1(int month, int day, int year){ Date d;

d.month = month; d.day = day; d.year = year;

return (d);}

void create_date2(Date *d, int month, int day, int year){ d->month = month; d->day = day; d->year = year;}

Copies date

Pass-by-reference

Date today;

today = create_date1(9, 4, 2008);create_date2(&today, 9, 4, 2008);



Pointers to Structuresvoid create_date2(Date *d, int month, int day, int year){ d->month = month; d->day = day; d->year = year;}

void foo(void){ Date today; create_date2(&today, 9, 4, 2008);} today.month:

today.day:

today.year:

0x1000

0x1004

0x1008

month: 9

day: 4

year: 2008

0x30A0

0x30A4

0x30A8

d: 0x10000x3098

9

4

2008



Pointers to Structures

Date * create_date3(int month, int day, int year){ Date *d;

d->month = month; d->day = day; d->year = year;

return (d);}

What is d pointing to?!?!



Pointers to Structures

void changeByValue(Date date){ date.day ++;}void changeByRef(Date *date){ date->day++;}void printDate(const Date date){ printf("today(d/m/y) is : \n"); printf("%d/%d/%d\n", date.day, date.month, date.year);}

Date today = {2012, 3, 11};printDate(today);changeByValue(today);printDate(today);changeByRef(&today);printDate(today);

today(d/m/y) is :11/3/2012today(d/m/y) is :11/3/2012today(d/m/y) is :12/3/2012



Compression of Structures• Structures may not be compared using

operators == and !=, because structure members are not necessarily stored in consecutive bytes of memory

struct a { int a; // OK int b;}; struct a b, c; b.a = 10; b.b = 30; c = b; if(c == b) // Error



Enumeration• Enumeration is a user-defined data type. It is defined using the

keyword enum and the syntax is: enum tag_name {name_0, …, name_n} ;

• The tag_name is not used directly. The names in the braces are symbolic constants that take on integer values from zero through n. As an example, the statement: enum colors { red, yellow, green } ;

– creates three constants. red is assigned the value 0, yellow is assigned 1 and green is assigned 2



Enumeration• Values in an enum start with 0, unless specified otherwise, and are

incremented by 1

• The identifiers in an enumeration must be unique

• The value of each enumeration constant of an enumeration can be set explicitly in the definition by assigning a value to the identifier

• Multiple members of an enumeration can have the same constant value

• Assigning a value to an enumeration constant after it has been defined is a syntax error

• Use only uppercase letters enumeration constant names. This makes these constants stand out in a program and reminds you that enumeration constants are not variables



An Example/* This program uses enumerated data types to access the elements of an array */#include <stdio.h>int main( ){

int March[5][7]={{0,0,1,2,3,4,5}, {6,7,8,9,10,11,12},

{13,14,15,16,17,18,19},{20,21,22,23,24,25,26},{27,28,29,30,31,0,0}};

enum days {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}; enum week {week_one, week_two, week_three, week_four, week_five};

printf ("Monday the third week of March is March %d\n",March [week_three] [Monday] );

}



An Example/* enumeration constants represent months of the year */ enum months {JAN = 1, FEB, MAR, APR, MAY, JUN, JUL, AUG, SEP, OCT, NOV, DEC };

enum months month;

/* initialize array of pointers */const char *monthName[] = { "", "January", "February", "March", "April", "May", "June", "July", "August", "September", "October",

/* loop through months */for (month = JAN; month <= DEC; month++ ) { printf( "%2d%11s\n", month, monthName[month] );}



Unions• A union is a derived data type—like a structure—with members that

share the same storage space

• For different situations in a program, some variables may not be relevant, but other variables are—so a union shares the space instead of wasting storage on variables that are not being used

• The members of a union can be of any data type

• The number of bytes used to store a union must be at least enough to hold the largest member

• Only one member, and thus one data type, can be referenced at a time



Unions representation

union myDataUnion { int i; char c; float f;} u1, u2;union myDataUnion u3;

u1.i = 4;u1.c = ’a’;u2.i = 0xDEADBEEF;

c

i

f



Unions• The operations that can be performed on a union

are the following: – assigning a union to another union of the same type – taking the address (&) of a union variable– accessing union members using the structure member

operator and the structure pointer operator

• Unions may not be compared using operators == and != for the same reasons that structures cannot be compared



Unions• In a declaration, a union may be initialized

with a value of the same type as the first union member

union a { int a; // OK char b[4];};

union a b = {10};printf("%d", b.a);



Unions• A union value doesn’t "know" which case it contains

union AnElt { int i; char c;} elt1, elt2;

elt1.i = 4;elt2.c = ’a’;elt2.i = 0xDEADBEEF;

if (elt1 currently has a char) …

How should your program keep track whether elt1, elt2 hold an int or a char?

?

?Basic answer: Another variable holds that info



enum Union_Tag {IS_INT, IS_CHAR};struct TaggedUnion { enum Union_Tag tag; union { int i; char c; } data;};

Tagged Unions• Tag every value with its case

Enum must be external to struct,so constants are globally visible

Struct field must be named



Bit-field Structures• C enables you to specify the number of bits in which an unsigned or int

member of a structure or union is stored

• This is referred to as a bit field

• Bit fields enable better memory utilization by storing data in the minimum number of bits required

• Bit field members must be declared as int or unsigned

• A bit field is declared by following an unsigned or int member name with a colon (:) and an integer constant representing the width of the field (i.e., the number of bits in which the member is stored)



…8 bit … …8 bit … …8 bit …

Bit-field Structures• Notice that bit field members of structures are

accessed exactly as any other structure member

• Padded to be an integral number of words– Placement is compiler-specific

struct Flags { int f1:3; unsigned int f2:1; unsigned int f3:2;} foo;foo.f1 = -2;foo.f2 = 1;foo.f3 = 2;

1 1 0 1 1 0 … …

f1 f2 f3



Unnamed Bit-fieldstruct example { unsigned a : 13;

unsigned : 19; unsigned b : 4;};

• uses an unnamed 19-bit field as padding—nothing can be stored in those 19 bits• An unnamed bit field with a zero width is used to align the next bit field on a

new storage-unit boundary• For example, the structure definition

struct example { unsigned a : 13;

unsigned : 0; unsigned b : 4;

};

uses an unnamed 0-bit field to skip the remaining bits (as many as there are) of the storage unit in which a is stored and to align b on the next storage-unit boundary



An Example - disk drive controller• Frequently device controllers (e.g. disk drives)

and the operating system need to communicate at a low level. Device controllers contain several registers which may be packed together in one integer



An Example - disk drive controllerstruct DISK_REGISTER { unsigned ready:1; unsigned error_occured:1; unsigned disk_spinning:1; unsigned write_protect:1; unsigned head_loaded:1; unsigned error_code:8; unsigned track:9; unsigned sector:5; unsigned command:5;};struct DISK_REGISTER *disk_reg = (struct DISK_REGISTER *) DISK_REGISTER_MEMORY;/* Define sector and track to start read */disk_reg->sector = new_sector;disk_reg->track = new_track;disk_reg->command = READ;/* wait until operation done, ready will be true */while ( ! disk_reg->ready ) ;/* check for errors */if (disk_reg->error_occured) { /* interrogate disk_reg->error_code for error type */ switch (disk_reg->error_code) ...... }



Notes of caution• Bit-field manipulations are machine dependent

• Attempting to access individual bits of a bit field as if they were elements of an array is a syntax error. Bit fields are not "arrays of bits"

• Attempting to take the address of a bit field (the & operator may not be used with bit fields because they do not have addresses)

• Although bit fields save space, using them can cause the compiler to generate slower-executing machine-language code. This occurs because it takes extra machine language operations to access only portions of an addressable storage unit. This is one of many examples of the kinds of space–time trade-offs that occur in computer science

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Fundamental of Programming (C)

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