Department of Computer and Information Science,
School of Science, IUPUI
CSCI 230
Structures
Dale Roberts, Lecturer
Computer Science, IUPUI
E-mail: droberts@cs.iupui.edu
Dale Roberts
Introduction
Structures
A collection of one or more variables, possibly of different types,
grouped together under a single name for continent handling.
Commonly used to define records to be stored in files
Combined with pointers, can create linked lists, stacks, queues, and
trees
Example:
struct card {
char *face;
char *suit;
};
struct introduces the definition for structure card
card is the structure name and is used to declare variables of the
structure type
card contains two members of type char *
These members are face and suit
Dale Roberts
Structure Definitions
Example:
A date consists of several parts, such as the day, month, and year, and the day of the
year, and the month name
struct date {
int day;
int month;
int year;
int year_date;
char month_name[4];
};
date: the name of the structure, called structure tag.
day, month, …: the elements or variables mentioned in a structure
are called members.
struct information
A struct cannot contain an instance of itself
Can contain a member that is a pointer to the same structure type
A structure definition does not reserve space in memory
Instead creates a new data type used to declare structure variables
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Declaration of Variables of Structure
Declarations
method 1: declared like other variables: declare tag first, and then
declare variable.
struct card {
char *face;
char *suit;
};
struct card oneCard, deck[ 52 ],
*cPtr;
struct date {
.. .. ..
};
struct date d1, d2, d3, d4, d5;
method 2: A list of variables can be declared after the right brace and
use comma separated list:
struct card {
char *face;
char *suit;
} oneCard, deck[ 52 ], *cPtr;
method 3: Declare only variables.
struct {
char *face;
char *suit;
} oneCard, deck[ 52 ], *cPtr;
Dale Roberts
struct date {
.. .. ..
} d1, d2, d3;
struct date d4, d5;
struct {
.. .. ..
} d1, d2, d3, d4, d5;
Structure Definitions
Valid Operations
Assigning a structure to a structure of the same type
Taking the address (&) of a structure
Accessing the members of a structure
Using the sizeof operator to determine the size of a structure
Initialization of Structures
Initializer lists
Example:
struct card oneCard = { "Three", "Hearts" };
Example:
struct date d1 = {4, 7, 1776, 186, “Jul”};
struct date d2 = {4, 7, 1776, 186, {‘J’,’u’,’l’,’\0’}};
Assignment statements
Example:
card threeHearts = oneCard;
Dale Roberts
Accessing Members of Structures
Accessing structure members
Dot (.) is a member operator used with structure
variables
Syntax: structure_name. member
struct card myCard;
printf( "%s", myCard.suit );
One could also declare and initialize threeHearts as follows:
struct card threeHearts;
threeHearts.face = “Three”;
threeHearts.suit = “Hearts”;
Arrow operator (->) used with pointers to structure
variables
struct card *myCardPtr = &myCard;
printf( "%s", myCardPtr->suit );
myCardPtr->suit is equivalent to (*myCardPtr).suit
Dale Roberts
Structures
Structure can be nested
• Name Rule
• Members in different structure
can have the same name, since
they are at different position.
struct date {
int day;
int month;
int year;
int year_date;
char month_name[4];
};
struct person {
char name [NAME_LEN];
char address[ADDR_LEN};
long zipcode;
long ss__number;
double salary;
};
struct date birthday;
struct s1 {
.. .. .. ..
char name[10];
.. .. .. ..
} d1;
struct s2 {
.. .. .. ..
int name;
.. .. .. ..
} d2;
struct s3 {
.. .. .. ..
int name;
struct s3
t3; /* name */
.. .. .. ..
} d2;
struct person emp;
emp.birthday.month = 6;
emp.birthday.year = 1776;
Dale Roberts
float name;
Memory Layout
Example:
struct data1 {
int day1;
char month[9];
int year;
};
Word (2 bytes) alignment machine – begins (aligns)
at even address, such as PC, SUN workstation
day1
int
month
char array
(hole)
year
int
2
9
1
2
bytes
bytes
bytes
bytes
Quad (4 bytes) address alignment – begins (aligns)
at quad address, such as VAX 8200
day1
int
month
char array
(hole)
year
int
4
9
3
4
bytes
bytes
bytes
bytes
0
1
2 - 10
11
12
13
0-3
4 - 12
13 - 15
16-19
integer
9 character
(hole)
integer
integer
9 character
(hole)
integer
You must take care of hole, if you want to access data from very low level (i.e.
low-level I/O, byte operations, etc.)
Dale Roberts
sizeof Operator
sizeof(struct tag)
struct test {
char name[5];
int i;
char s;
} t1, t2;
/* assume int is 2 bytes */
main()
{
printf(“sizeof(struct test) = %d\n”, sizeof (struct test));
printf(“address of t1 = %d\n”, &t1);
printf(“address of t2 = %d\n”, &t2);
printf(“address of t1.name = %d\n”, t1.name); t1 992
5 bytes
printf(“address of t1.i = %d\n”, &t1.i);
1 byte (hole)
997
printf(“address of t1.s = %d\n”, &t1.s);
2 bytes
}
998
output:
t2
sizeof(struct test) = 10
address of t1 = 992
address of t2 = 1002
address of t1.name = 992
address of t1.i = 998
address of t1.s = 1000
1000
1001
1002
1 byte
1 byte (hole)
5 bytes
1 byte (hole)
2 bytes
1 byte
1 byte (hole)
Dale Roberts
Using Structures With Functions
Passing structures to functions
Pass entire structure or pass individual members
Both pass call by value
It is not a good idea to pass a structure to or return from function.
The better way is passing a pointer to the structure to the functions
and returning a pointer from function.
To pass structures call-by-reference
Pass its address
Pass reference to it
To pass arrays call-by-value
Create a structure with the array as a member
Pass the structure
Dale Roberts
Using Structures With Functions (cont.)
Example:
day_of_year(struct date *pd)
{
int i, day, leap;
day = pd -> day;
leap = pd->year%4 ==0 && pd->year %100 ==0 || pd->year%400 ==0;
for (i=1; i<pd -> month; i++)
day += day_tab[leap][i];
return (day);
}
The declaration
struct date *pd;
says that pd is a pointer to a structure of the type date
If p is a pointer to a structure, then p-> member_of_structure refers to the
particular members, like pd -> year
p-> member_of_structure is equivalent to (*p).member_of_structure
Notice: ‘.’ has higher precedence than ‘*’; *pd.year is wrong, since pd.year
is not a pointer.
Both -> and . associate from left to right. So p -> q -> member
are (p->q)->member.
Example: emp.birthday.month are (emp.birthday).month
Dale Roberts
Using Structures With Functions (cont.)
-> and . both are at the highest precedence (together with () for
function and [] for array subscripts)
Example:
++p->x;
(++p)->x;
*p->y;
*p->y++;
(*p->y)++;
*p++->y;
struct {
int *x;
int *y;
} *p;
is equivalent to ++(p->x) /* increment x, not p */
/* increment p before access x */
/* fetch whatever y points to */
/* increments y after accessing whatever y point to */
/* increments whatever y point to, just like *p->y++ */
/* increments p after accessing whatever y point to */
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typedef
typedef
Creates synonyms (aliases) for previously defined data types
Use typedef to create shorter type names
Example:
typedef struct card *CardPtr;
Defines a new type name CardPtr as a synonym for type struct card *
typedef does not create a new data type while it only creates an alias
Example:
struct card {
const char *face;
const char *suit;
};
typedef struct card Card;
void fillDeck( Card * const, const char *[], const char *[] );
int main()
{
Card deck[ 52 ];
const char *face[] = {"Ace", "Deuce", "Three", "Four", "Five", "Six",
Seven", "Eight", “Nine", "Ten", "Jack", "Queen", "King"};
const char *suit[] = { "Hearts", "Diamonds", "Clubs", "Spades"};
.. ..
fillDeck( deck, face, suit );
.. ..
}
void fillDeck(Card * const wDeck, const char * wFace[], const char * wSuit[])
{
.. ..
}
Dale Roberts
Unions
union
Memory that contains a variety of objects over time
Only contains one data member at a time
Members of a union share space
Conserves storage
Only the last data member defined can be accessed
union declarations
Same as struct
union Number {
int x;
float y;
};
union Number value;
Valid union operations
Assignment to union of same type: =
Taking address: &
Accessing union members: .
Accessing members using pointers: ->
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/* Fig. 10.5: fig10_05.c
An example of a union */
#include <stdio.h>
union number {
int x;
double y;
};
Define union
int main()
{
union number value;
Initialize variables
value.x = 100;
printf( "%s\n%s\n%s%d\n%s%f\n\n",
"Put a value in the integer member",
"and print both members.",
"int:
", value.x,
"double:\n", value.y );
value.y = 100.0;
printf( "%s\n%s\n%s%d\n%s%f\n",
"Put a value in the floating member",
"and print both members.",
"int:
", value.x,
"double:\n", value.y );
return 0;
}
Set variables
Print
Program Output
Put a value in the integer member
and print both members.
int:
100
double:
9255959211743313600000000000000000000000000
0000000000000000000.00000
Put a value in the floating member
and print both members.
int:
0
double:
100.000000
Dale Roberts
Array of Structures
struct person_data{
char name[NAMESIZE];
int tscore;
int math;
int english;
} person[PERSON];
Example: (before)
char name[PERSON][NAMESIZE];
int tscore[PERSON]
int math[PERSON]
int english[PERSON]
(now)

Initialization of structure array
struct person_data{
.. .. .. ..
} person[]={
{“Jane”,180,89,91},
{“John”,190,90,100},
.. .. .. ..
}; /* similar to 2D array */
the inner brace is not necessary

“Jane”,180,89,91,
“John”,190,90,100,
.. .. .. ..
Example: using separated arrays
Example: using pointer to structure
average (int tscore, int math, int
eng, int n)
{
int i, total=0,mathtotal = 0,
engtotal=0;
for (i=0; i<n, i++) {
total += *tscore++;
mathtotal += *math++;
engtotal += *eng++;
}
average (struct person_data
*person, int n)
{
int i, total=0,mathtotal = 0,
engtotal=0;
for (i=0; i<n, i++) {
total += person->tscore;
mathtotal += person->math;
engtotal += person->eng;
person++;
}
Dale Roberts
Self-Reference Structure
Dynamic Data Structure
Data Structure: link list, tree, graph, …
Example: Binary Tree
Struct treeNode {
char word[SIZE];
int count;
struct treeNode *left;
struct treeNode *right;
/* or struct treeNode *left, *right;
};
/* Inorder traversal */
tree_print(struct treeNode *p)
{
if (p!=NULL){
tree_print(p->left);
printf(“.. ..”, p->word, .. ..);
tree_print(p-> right);
}
}
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*/
Link list
Example:
Struct listNode {
char word[SIZE];
int count;
struct listNode *next;
};
/* Inorder traversal */
list_print(struct listNode *p)
{
while (p!=NULL)
{
printf(“.. ..”, p->word, .. ..);
p = p->next;
}
}
Dale Roberts
Bit Fields
Bit field
Member of a structure whose size (in bits) has been specified
Enable better memory utilization
Must be declared as int or unsigned
Cannot access individual bits
Declaring bit fields
Follow unsigned or int member with a colon (:) and an integer constant representing
the width of the field
Example:
struct BitCard {
unsigned face : 4;
unsigned suit : 2;
unsigned color : 1;
};
Unnamed bit field
struct Example {
unsigned a : 13;
unsigned
: 3;
unsigned b : 4;
}
Field used as padding in the structure
Nothing may be stored in the bits
Unnamed bit field with zero width aligns next bit field to a new storage unit boundary
Dale Roberts
Enumeration Constants
Enumeration
Set of integer constants represented by identifiers
Enumeration constants are like symbolic constants whose values are
automatically set
Values start at 0 and are incremented by 1
Values can be set explicitly with =
Need unique constant names
Example:
enum Months { JAN = 1, FEB, MAR, APR, MAY, JUN, JUL,
AUG, SEP, OCT, NOV, DEC};
Creates a new type enum Months in which the identifiers are set to the
integers 1 to 12
Enumeration variables can only assume their enumeration constant
values (not the integer representations)
Dale Roberts
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/* Fig. 10.18: fig10_18.c
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Using an enumeration type */
#include <stdio.h>
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enum months { JAN = 1, FEB, MAR, APR, MAY, JUN,
6
JUL, AUG, SEP, OCT, NOV, DEC };
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int main()
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{
10
enum months month;
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const char *monthName[] = { "", "January", "February",
12
"March", "April", "May",
13
"June", "July", "August",
14
"September", "October",
15
"November", "December" };
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for ( month = JAN; month <= DEC; month++ )
printf( "%2d%11s\n", month, monthName[ month ] );
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return 0;
21 }
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January
February
March
April
May
June
July
August
September
October
November
December
Storage Management
C supports 4 functions, malloc(), calloc(),free(),
and cfree() for storage management
malloc(n):
allocate a node while its content is still ‘garbage’
n is an integer, indicating the size of memory in byte which you would like
to allocate
malloc() return a character pointer to that memory
So, you have to use cast operator (type), to change the type of the
pointer.
Example:
int *ip;
ip = (int*) malloc(sizeof(int));
struct treeNode *tp;
tp = (struct tnode *) malloc(sizeof(struct tnode));
Dale Roberts
Storage Management (cont.)
free(p):
free() will release the memory allocated by malloc().
p is the pointer containing the address returning from malloc().
Example:
Example:
int *ip;
ip = (int*) malloc(sizeof(int));
... .. ..
free(ip);
/* Question: can you free(ip) after ip++ ? */
struct treeNode *tp;
tp=(struct treeNode *)malloc(sizeof(struct treeNode ));
... .. ..
free(tp);
When there is no further memory, malloc() will return NULL pointer. It is a
good idea to check the returning value of malloc().
if ((ip=(int *)malloc(sizeof(int))) == NULL){
printf(“\nMemory is FULL\n”);
exit(1);
}
When you free the memory, you must be sure that you pass the original
address returning from malloc() to function free(). Otherwise, system
exception may be happened
Dale Roberts
Storage Management (cont.)
calloc(n,size):
calloc() allow you to allocate an n elements array of same data type.
Because n can be an integer variable, you can use calloc() to allocate a
dynamic size array.
n is the element number of array that you want to allocate.
size is the number of byte of each element.
Unlike malloc(), calloc() guarantees that memory contents are all zero
Example: allocate an array of 10 elements
int *ip;
ip = (int*) calloc(10, sizeof(int));
*(ip+1) refer to the 2nd element, the same as ip[1]
*(ip+i) refer to the i+1th element, the same as ip[i]
Like malloc(), calloc() will return NULL, if no further memory is available.
cfree(p):
cfree() releases the memory allocated by calloc().
Example:
cfree(ip);
Dale Roberts