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Fundamental of Programming (C)

Fundamental of Programming (C). Lecture 8 Structures, Unions, Bit Manipulations and Enumerations. Outline. Structures Be able to use compound data structures in programs Unions Be able to share storage space of their members Bit fields Structures

aphrodite-harrington
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Fundamental of Programming (C)

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  1. Fundamental of Programming (C) Lecture 8 Structures, Unions, Bit Manipulations and Enumerations

  2. 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

  3. User Defined Data Types (typedef) • The C language provides a facility called typedeffor creating synonymsfor previously defined data type names. • For example, the declaration: typedefint Length; makes the name Lengtha 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]; Stringname; Array ages;

  4. Structures (struct) • Structures—sometimes referred to as aggregates—are collections of related variables under one name • Structuresmay contain variables of many different data types—in contrast to arraysthat contain only elements of the same data type • Structures are commonly used to define records to be stored in files • Pointers and structuresfacilitate the formation of more complex data structures such as linked lists, queues, stacksand trees • Structures are derived data types—they are constructed using objects of other types

  5. 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 structemployee { char firstName[ 20 ]; char lastName[ 20 ];int age; char gender; double hourlySalary; }; struct employee Ali, emp[10]; struct employee { char firstName[ 20 ]; char lastName[ 20 ];int age; char gender; double hourlySalary; } Ali, Sara,empDTS[20]; struct employee Reza, *emp; struct { char firstName[ 20 ]; char lastName[ 20 ];int age; char gender; double hourlySalary; } Ali;

  6. Declaring Structures (struct) • Often, typedefis used in combination with structto declare a synonym (or an alias) for a structure: typedefstruct { char firstName[ 20 ]; char lastName[ 20 ];int age; char gender; double hourlySalary; } employee;/* The "alias" employeeAli; /* Create a struct variable */ struct employee { char firstName[ 20 ]; char lastName[ 20 ];int age; char gender; double hourlySalary; } Ali, Sara,empDTS[20]; struct employee Reza, *emp;

  7. Declaring Structures (struct) • Membersof 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 ]; // OKint age; char gender; }; • struct Student Ce40153[80];

  8. 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 */};

  9. 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 { char firstName[ 20 ]; char lastName[ 20 ];int age; char gender; double hourlySalary; } Ali;

  10. 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));

  11. Memory layout struct COST { int amount; char currency_type[2]; } struct PART { char id[2]; struct COST cost; intnum_avail; } • Here, the system uses 4-byte alignment of integers, so amount and num_avail must be aligned Four bytes wasted for each structure! currency_type id amount num_avail cost

  12. Memory layout struct COST { int amount; char currency_type[2]; } struct PART { struct COST cost; char id[2]; intnum_avail; } • Implementation dependent!!! currency_type amount id num_avail cost

  13. Accessing Struct Members • Individual membersof 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;

  14. An Example - Initialization //Create a struct but don’t reserve space struct personal { long id; // student ID float gpa; // grade point average }; struct identity { char FirstName[30]; char LastName[30]; unsigned age; struct personal person; }; 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) ; js.personal.id Error js = {"Joe", "Smith", 25, 9, 10} strcpy(js.FirstName, "Joe");

  15. An Example - Assignment //Create a struct but don’t reserve space struct personal { long id; // student ID float gpa; // grade point average }; struct identity { char FirstName[30]; char LastName[30]; unsigned age; struct personal person; }; 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) ;

  16. Arrays of Structures struct identity { char FirstName[30]; char LastName[30]; unsigned age; struct personal person; } students[4]; //Create a struct but don’t reserve space struct personal { long id; // student ID float gpa; // grade point average }; struct identity sharifC40153[80] = {"omid", "Jafarinezhad", 14, 9140153, 20, "Samad", "Shekarestani", 90, 2222222, 20} ; strcpy(sharifC40153[2].FirstName, "KhajeNezam"); strcpy(sharifC40153[2].LastName, "Shekarestani"); sharifC40153[2]. age = 100; sharifC40153[2]. person.id = 11111111; sharifC40153[2]. person. gpa = 20;

  17. An Example boolcheck_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) ; } // … #define NFRIENDS 10 struct Date { unsigned year; unsigned month; unsigned day; }; struct Friend { char FirstName[30]; char LastName[30]; struct Date Birthday; }; typedefstruct { unsigned year; unsigned month; unsigned day; } Date; boolcheck_birthday(Date today, Date myFriend) { //… }

  18. Pointers to Structures Date create_date1(intmonth, 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; } Pass-by-reference Date today; today = create_date1(9, 4, 2008); create_date2(&today, 9, 4, 2008); Copies date

  19. year: 2008 0x30A8 day: 4 0x30A4 month: 9 0x30A0 d: 0x1000 0x3098 today.year: 0x1008 today.day: 0x1004 today.month: 0x1000 Pointers to Structures void 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); } 2008 4 9

  20. Pointers to Structures Date * create_date3(intmonth, int day, int year) { Date *d; d->month = month; d->day = day; d->year = year; return (d); } What is d pointing to?!?!

  21. 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 : 11/3/2012 today(d/m/y) is : 11/3/2012 today(d/m/y) is : 12/3/2012

  22. 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

  23. Enumeration • Enumeration is a user-defined data type. It is defined using the keyword enumand the syntax is: enumtag_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

  24. Enumeration • Values in an enumstart with 0, unless specified otherwise, and are incremented by 1 • The identifiersin 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

  25. 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] ); }

  26. 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] ); }

  27. 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

  28. Unions representation c union myDataUnion{ inti; char c; float f; } u1, u2; union myDataUnion u3; u1.i = 4; u1.c = ’a’; u2.i = 0xDEADBEEF; i f

  29. Unions • The operations that can be performed on a union are the following: • assigninga 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

  30. 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);

  31. ? How should your program keep track whether elt1, elt2 hold an int or a char? ? Unions • A union value doesn’t "know" which case it contains union AnElt { inti; char c; } elt1, elt2; elt1.i = 4; elt2.c = ’a’; elt2.i = 0xDEADBEEF; if (elt1 currently has a char) … Basic answer: Another variable holds that info

  32. Enum must be external to struct, so constants are globally visible Struct field must be named Tagged Unions • Tag every value with its case enum Union_Tag {IS_INT, IS_CHAR}; struct TaggedUnion { enum Union_Tag tag; union { int i; char c; } data; };

  33. 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)

  34. f1 f2 f3 1 1 0 1 1 0 … … 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;

  35. Unnamed Bit-field struct 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

  36. 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

  37. An Example - disk drive controller struct 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) ...... }

  38. 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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