Wednesday, November 12, 2008

Q: Write a program base on user inputs wich performs these three functions:

( 01 ) Check for Plindrome by using stack
( 02 ) convert singlty link list to a doubly link list
( 03 ) Insert elements in queue and then find and delete the target element

void STACK();
void doubly_link_list();
void Qeue();
#define size 100
int queue[size];
int rear=0;
void insert();
void del();
void Show();
const int SIZE=10000;
static char stack[SIZE];
int top=0;
void push(int c);
void show();
int w;
typedef struct item
int val;
int data;
void main()
int choice;
cout<<"ENTER YOUR CHOICE \n\t1->Plindrome by using stack";
cout<<"\n\t2->Doubly link list \n\t3->Queue\n\t4->exit\n";
cout<<"YOUR CHOICE---> ";
case 1:STACK();
case 2:doubly_link_list();
case 3:Qeue();
case 4:
cout<<"\nINVALID OPTION!!!!!!!!!!!\n\n"; } cout<<"ENTER YOUR CHOICE \n\t1->Plindrome by using Stack \n\t2->Doubly link list \n\t3->Queue\n\t4-
cout<<"\tYOUR CHOICE---> ";
void STACK()
char word[SIZE];
int i;
cout<<"NOW ENTER WORD OF "<<<" CHARACTERS: \n"; for(i=0;i<<"CHAR NO "<<<" "; cin>>word[i];
char temp;
for(i=0;i<<"YES IT IS A PALINDROM\n"; } else { cout<<"NO IT IS NOT A PALINDROM\n"; } } void push(int c) { stack[top]=c; top++; } void show() { if(top<=0) {cout<<"STACK is empty \n";} else cout<<"THE WORD IN STACK IS: "; for(int i=0;i<<<"\n\n"; } //////////////////////////////////////////////// /////////////////////////////////////////////// void doubly_link_list() { int i,n; item*head; item*curr; curr=new item; head=curr; head->next=NULL;
link=new item;
cout<<"Enter numbers of nodes: "; cin>>n;
if(n<=0) { cout<<"NOT POSSIBLE \n"; } else { for(i=1;i<=n;i++) { curr=new item; link=new item; cout<<"ENTER INTEGER: "; cin>>curr->val;
cout<<"\n NODE NUMBER "<<<" \n"; cout<<"BY USING SINGL LINK LIST ----->"<val<<" \n\n"; link->data=curr->val;
cout<<"BY USING DOUBLY LINK LIST----->"<data<<" \n\n"; link->next=link;
void Qeue()
int option;
cout<<"\nENTER 1->Insert 2->Delete 3->Show 4->Go back to main() ";
case 1:insert();
case 2:del();
case 3:Show();
case 4:main();
default:cout<<"Invalid option!!!"; } cout<<"\nENTER 1->Insert 2->Delete 3->Show 4->Go back to main() ";
void insert()
cout<<"Queue is full\n\n"; } cout<<"Enter element: "; cin>>queue[rear];
cout<<"\nQueue is full\n\n"; } else rear++; } void Show() { if(rear==0) { cout<<"queue is empty\n"; } else for(int i=0;i<<" "<<<"queue is empty\n"; } cout<<"Enter element to delete "; cin>>search;
for(int j=0;j
{ if(search==queue[j])
for(int k=0;k

Data Structures Using C


A string is a group of characters, usually letters of the alphabet. In

order to format your printout in such a way that it looks nice, has

meaningful titles and names, and is aesthetically pleasing to you and the

people using the output of your program, you need the ability to output

text data. We have used strings extensively already, without actually

defining them. A complete definition of a string is ‘a sequence of char

type data terminated by a NULL character,’.

When C is going to use a string of data in some way, either to compare

it with another, output it, copy it to another string, or whatever, the

functions are set up to do what they are called to do until a NULL

character (which is usually a character with a zero ASCII code number)

is detected. You should also recall (from Module 813: Fundamental

Programming Structures in C) that the char type is really a special form

of integer – one that stores the ASCII code numbers which represent

characters and symbols.

An array (as we shall discover shortly) is a series of homogeneous

pieces of data that are all identical in type. The data type can be quite

complex as we will see when we get to the section of this module

discussing structures. A string is simply a special case of an array, an

array of char type data. The best way to see these principles is by use of

an example [CHRSTRG.C].

#include "stdio.h"

void main( )


char name[5]; /* define a string of characters */

name[0] = 'D';

name[1] = 'a';

name[2] = 'v';

name[3] = 'e';

name[4] = 0; /* Null character - end of text */

printf("The name is %s\n",name);

printf("One letter is %c\n",name[2]);

printf("Part of the name is %s\n",&name[1]);


The data declaration for the string appears on line 4. We have used this

declaration in previous examples but have not indicated its meaning.

The data declaration defines a string called name which has, at most, 5

characters in it. Not only does it define the length of the string, but it

also states, through implication, that the characters of the string will be

numbered from 0 to 4. In the C language, all subscripts start at 0 and

increase by 1 each step up to the maximum which in this case is 4. We

have therefore named 5 char type variables: name[0], name[1],

name[2], name[3], and name[4]. You must keep in mind that in C the

subscripts actually go from 0 to one less than the number defined in the

definition statement. This is a property of the original definition of C

and the base limit of the string (or array), i.e. that it always starts at

zero, cannot be changed or redefined by the programmer.

Using strings

The variable name is therefore a string which can hold up to 5

characters, but since we need room for the NULL terminating

character, there are actually only four useful characters. To load

something useful into the string, we have 5 statements, each of which

assigns one alphabetical character to one of the string characters.

Finally, the last place in the string is filled with the numeral 0 as the end

indicator and the string is complete. (A #define statement which sets

NULL equal to zero would allow us to use NULL instead of an actual

zero, and this would add greatly to the clarity of the program. It would

be very obvious that this was a NULL and not simply a zero for some

other purpose.) Now that we have the string, we will simply print it out

with some other string data in the output statement.

You will, by now, be familiar with the %s is the output definition to

output a string and the system will output characters starting with the

first one in name until it comes to the NULL character; it will then quit.

Notice that in the printf statement only the variable name needs to be

given, with no subscript, since we are interested in starting at the

beginning. (There is actually another reason that only the variable name

is given without brackets. The discussion of that topic will be found in

the next section.)

Outputting part of a string

The next printf illustrates that we can output any single character of the

string by using the %c and naming the particular character of name we

want by including the subscript. The last printf illustrates how we can

output part of the string by stating the starting point by using a

subscript. The & specifies the address of name[1].

This example may make you feel that strings are rather cumbersome to

use since you have to set up each character one at a time. That is an

incorrect conclusion because strings are very easy to use as we will see

in the next example program.

Some string subroutines

The next example [STRINGS.C] illustrates a few of the more common

string handling functions. These functions are found in the C standard

library and are defined in the header file string.h.

Page 813-5

#include "stdio.h"

#include "string.h"

void main( )


char name1[12], name2[12], mixed[25];

char title[20];

strcpy(name1, "Rosalinda");

strcpy(name2, "Zeke");

strcpy(title,"This is the title.");

printf(" %s\n\n" ,title);

printf("Name 1 is %s\n", name1);

printf("Name 2 is %s\n", name2);

if(strcmp(name1,name2)>0) /* returns 1 if name1 > name2 */




printf("The biggest name alpabetically is %s\n" ,mixed);

strcpy(mixed, name1);

strcat(mixed," ");

strcat(mixed, name2);

printf("Both names are %s\n", mixed);


First, four stringsare defined. Next, a new function that is commonly

found in C programs, the strcpy function, or string copy function is

used. It copies from one string to another until it comes to the NULL

character. Remember that the NULL is actually a 0 and is added to the

character string by the system. It is easy to remember which one gets

copied to which if you think of the function as an assignment statement.

Thus if you were to say, for example, ‘x = 23;’, the data is copied from

the right entity to the left one. In the strcpy function the data are also

copied from the right entity to the left, so that after execution of the

first statement, name1 will contain the string Rosalinda, but without the

double quotes. The quotes define a literal string and are an indication

to the compiler that the programmer is defining a string.

Similarly, Zeke is copied into name2 by the second statement, then the

title is copied. Finally, the title and both names are printed out. Note

that it is not necessary for the defined string to be exactly the same size

as the string it will be called upon to store, only that it is at least as

long as the string plus one more character for the NULL.

Alphabetical sorting of strings

The next function to be considered is the strcmp or the string compare

function. It will return a 1 if the first string is lexicographically larger

than the second, zero if they are the two strings are identical, and -1 if

the first string is lexicographically smaller than the second. A

lexicographical comparison uses the ASCII codes of the characters as

the basis for comparison. Therefore, ‘A’ will be “smaller” than ‘Z’

because the ASCII code for ‘A’ is 65 whilst the code for ‘Z’ is 90.

Sometimes however, strange results occur. A string that begins with the

letter ‘a’ is larger than a string that begins with the letter ‘Z’ since the

ASCII code for ‘a’ is 97 whilst the code for ‘Z’ is 90.

Page 813-6

One of the strings, depending on the result of the compare, is copied

into the variable mixed, and therefore the largest name alphabetically is

printed out. It should come as no surprise to you that Zeke wins

because it is alphabetically larger: length doesn’t matter, only the ASCII

code values of the characters.

Combining strings

The last four statements in the [STRINGS.C] example have another

new feature, the strcat, or string concatenation function. This function

simply adds the characters from one string onto the end of another

string taking care to adjust the NULL so a single string is produced. In

this case, name1 is copied into mixed, then two blanks are concatenated

to mixed, and finally name2 is concatenated to the combination. The

result is printed out which shows both names stored in the one variable

called mixed.

Exercise 1

Write a program with three short strings, about 6 characters each, and

use strcpy to copy one, two, and three into them. Concatenate the three

strings into one string and print the result out 10 times.

Write a program thatwill output the characters of a string backwards.

For example, given the string “computer”, the program will produce

Page 813-7

Objective 2 After working through this module you should be able to declare and

manipulate single and multi-dimensional arrays of the C data types.


The last objective discussed the structure of strings which are really

special cases of an array. Arrays are a data type that are used to

represent a large number of homogeneous values, that is values that are

all of the one data type. The data type could be of type char, in which

case we have a string. The data type could just as easily be of type int,

float or even another array.

An array of integers

The next program [INTARRAY.C] is a short example of using an array

of integers.

#include "stdio.h"

void main( )


int values[12];

int index;

for (index = 0;index <>

values[index] = 2 * (index + 4);

for (index = 0;index <>

printf("The value at index = %2d is %3d\n", index, values[index]);


Notice that the array is defined in much the same way we defined an

array of char in order to do the string manipulations in the last section.

We have 12 integer variables to work with not counting the one named

index. The names of the variables are values[0], values[1], ... , and

values[11]. Next we have a loop to assign nonsense, but well defined,

data to each of the 12 variables, then print all 12 out. Note carefully

that each element of the array is simply an int type variable capable of

storing an integer. The only difference between the variables index and

values[2], for example, is in the way that you address them. You should

have no trouble following this program, but be sure you understand it.

Compile and run it to see if it does what you expect it to do.

An array of floating point data

Now for an example of a program [BIGARRAY.C] with an array of

float type data. This program also has an extra feature to illustrate how

strings can be initialised.

#include "stdio.h"

#include "string.h"

char name1[ ] = "First Program Title";

void main( )


int index;

int stuff[12];

Page 813-8

float weird[12];

static char name2[] = "Second Program Title";

for (index = 0; index <>

stuff[index] = index + 10;

weird[index] = 12.0 * (index + 7);


printf("%s\n", name1);

printf("%s\n\n", name2);

for (index = 0; index <>

printf("%5d %5d %10.3f\n", index, stuff[index], weird[index]);


The first line of the program illustrates how to initialise a string of

characters. Notice that the square brackets are empty, leaving it up to

the compiler to count the characters and allocate enough space for our

string including the terminating NULL. Another string is initialised in

the body of the program but it must be declared static here. This

prevents it from being allocated as an automatic variable and allows it

to retain the string once the program is started. You can think of a staic

declaration as a local constant.

There is nothing else new here, the variables are assigned nonsense data

and the results of all the nonsense are printed out along with a header.

This program should also be easy for you to follow, so study it until

you are sure of what it is doing before going on to the next topic.

Getting data back from a function

The next program [PASSBACK.C] illustrates how a function can

manipulate an array as a variable parameter.

#include "stdio.h"

void main( )


int index;

int matrix[20];

/* generate data */

for (index = 0 ;index <>

matrix[index] = index + 1;

/* print original data */

for (index = 0; index <>

printf("Start matrix[%d] = %d\n", index, matrix[index]);

/* go to a function & modify matrix */


/* print modified matrix */

for (index = 0; index <>

printf("Back matrix[%d] = %d\n", index, matrix[index]);


dosome(list) /* This will illustrate returning data */

int list[ ];


int i;

/* print original matrix */

for (i = 0;i <>

printf("Before matrix[%d] = %d\n", i list[i]);

/* add 10 to all values */

for (i = 0; i <>

list[i] += 10;

/* print modified matrix */

for (i = 0; i <>

printf("After matrix[%d] = %d\n", i, list[i]);

Page 813-9


An array of 20 variables named matrix is defined, some nonsense data is

assigned to the variables, and the first five values are printed out. Then

we call the function dosome taking along the entire array as a


The function dosome has a name in its parentheses also but it uses the

local name ‘list’ for the array. The function needs to be told that it is

really getting an array passed to it and that the array is of type int. Line

20 does that by defining list as an integer type variable and including the

square brackets to indicate an array. It is not necessary to tell the

function how many elements are in the array, but you could if you so

desired. Generally a function works with an array until some end-ofdata

marker is found, such as a NULL for a string, or some other

previously defined data or pattern. Many times, another piece of data is

passed to the function with a count of how many elements to work

with. In our present illustration, we will use a fixed number of elements

to keep it simple.

So far nothing is different from the previous functions we have called

except that we have passed more data points to the function this time

than we ever have before, having passed 20 integer values. We print out

the first 5 again to see if they did indeed get passed here. Then we add

ten to each of the elements and print out the new values. Finally we

return to the main program and print out the same 5 data points. We

find that we have indeed modified the data in the function, and when we

returned to the main program, we brought the changes back. Compile

and run this program to verify this conclusion.

Arrays pass data both ways

It was stated during our study of functions that when data was passed

to a function, the system made a copy to use in the function which was

thrown away when we returned. This is not the case with arrays. The

actual array is passed to the function and the function can modify it any

way it wishes to. The result of the modifications will be available back

in the calling program. This may seem strange (that arrays are handled

differently from single point data) but it is correct. The reason is that

when an array is passed to a function, the address of (or a pointer to)

the function is the piece of data that is used. The is analagous to the

Pascal construct of placing a VAR in front of a parameter in a

procedure definition.

Multiple-dimensional arrays

Arrays need not be one-dimensional as the last three examples have

shown, but can have any number of dimensions. Most arrays usually

have either one or two dimensions. Higher dimesions have applications

Page 813-10

in mathematics but are rarely used in data processing environments. The

use of doubly-dimensioned arrays is illustrated in the next program


#include "stdio.h"

void main( )


int i, j;

int big[8][8], large[25][12];

/* Create big as a a multiplication table */

for (i = 0; i <>

for (j = 0; j <>

big[i][j] = i * j;

/* Create large as an addition table */

for (i = 0; i <>

for (j = 0; j <>

large[i][j] = i + j;

big[2][6] = large[24][10] * 22;

big[2][2] = 5;

big[big[2][2]][big[2][2]] = 177;

/* this is big[5][5] = 177; */

for (i = 0; i <>

for (j = 0; j <>

printf("%5d ", big[i][j]);

printf("\n"); /* newline for each increase in i */



The variable big is an 8 by 8 array that contains 8 times 8 or 64

elements in total. The first element is big[0][0], and the last is big[7][7].

Another array named large is also defined which is not square to

illustrate that the array need not be square! Both are filled up with data,

one representing a multiplication table and the other being formed into

an addition table.

To illustrate that individual elements can be modified at will, one of the

elements of big is assigned the value from one of the elements of large

after being multiplied by 22. Next, big[2][2] is assigned the arbitrary

value of 5, and this value is used for the subscripts of the next

assignment statement. The third assignment statement is in reality

big[5][5] = 177 because each of the subscripts contain the value 5. This

is only done to demonstrate that any valid expression can be used for a

subscript. It must only meet two conditions: it must be an integer

(although a char will work just as well), and it must be within the range

of the subscript for which it is being used.

The entire matrix variable big is printed out in a square form so you can

check the values to see if they were set the way you expected them to

be. You will find many opportunities to use arrays, so do not

underestimate the importance of the material in this section.

Page 813-11

Exercise 2

Define two integer arrays, each 10 elements long, called array1 and

array2. Using a loop, put some kind of nonsense data in each and add

them term for term into another 10 element array named arrays. Finally,

print all results in a table with an index number, for example

1 2 + 10 = 12

2 4 + 20 = 24

3 6 + 30 = 36 etc.

Hint: the print statement will be similar to this:

printf("%4d %4d + %4d = %4d\n", index, array1[index],

array2[index], arrays[index]);

Page 813-12

Objective 3 After working through this module you should be able to create

manipulate and manage C pointers to data elements.


Simply stated, a pointer is an address. Instead of being a variable, it is a

pointer to a variable stored somewhere in the address space of the

program. It is always best to use an example, so examine the next

program [POINTER.C] which has some pointers in it.

#include "stdio.h"

/* illustration of pointer use */

void main( )


int index, *pt1, *pt2;

index = 39; /* any numerical value */

pt1 = &index; /* the address of index */

pt2 = pt1;

printf("The value is %d %d %d\n", index, *pt1, *pt2);

*pt1 = 13; /* this changes the value of index */

printf("The value is %d %d %d\n", index, *pt1, *pt2);


Ignore for the moment the data declaration statement where we define

index and two other fields beginning with an star. (It is properly called

an asterisk, but for reasons we will see later, let’s agree to call it a star.)

If you observe the first statement, it should be clear that we assign the

value of 39 to the variable index; this is no surprise. The next statement,

however, says to assign to pt1 a strange looking value, namely the

variable index with an ampersand (&) in front of it. In this example, pt1

and pt2 are pointers, and the variable index is a simple variable.

Now we have a problem. We need to learn how to use pointers in a

program, but to do so requires that first we define the means of using

the pointers in the program!

Two very important rules

The following two rules are very important when using pointers and

must be thoroughly understood. They may be somewhat confusing to

you at first but we need to state the definitions before we can use them.

Take your time, and the whole thing will clear up very quickly.

1. A variable name with an ampersand (&) in front of it defines the

address of the variable and therefore points to the variable. You

can therefore read line 7 as ‘pt1 is assigned the value of the address

2. A pointer with a star in front of it refers to the value of the variable

pointed to by the pointer. Line 10 of the program can be read as

‘The stored (starred) value to which the pointer pt1 points is

Page 813-13

assigned the value 13’. Now you can see why it is perhaps

convenient to think of the asterisk as a star,: it sort of sounds like

the word ‘store’!

Memory aids

1. Think of & as an address.

2. Think of * as a star referring to stored.

Assume for the moment that pt1 and pt2 are pointers (we will see how

to define them shortly). As pointers they do not contain a variable value

but an address of a variable and can be used to point to a variable. Line

7 of the program assigns the pointer pt1 to point to the variable we

have already defined as index because we have assigned the address of

index to pt1. Since we have a pointer to index, we can manipulate the

value of index by using either the variable name itself, or the pointer.

Line 10 modifies the value by using the pointer. Since the pointer pt1

points to the variable index, then putting a star in front of the pointer

name refers to the memory location to which it is pointing. Line 10

therefore assigns to index the value of 13. Anywhere in the program

where it is permissible to use the variable name index, it is also

permissible to use the name *pt1 since they are identical in meaning

until the pointer is reassigned to some other variable.

Just to add a little intrigue to the system, we have another pointer

defined in this program, pt2. Since pt2 has not been assigned a value

prior to line 8, it doesn’t point to anything, it contains garbage. Of

course, that is also true of any variable until a value is assigned to it.

Line 8 assigns pt2 the same address as pt1, so that now pt2 also points

to the variable index. So to continue the definition from the last

paragraph, anywhere in the program where it is permissible to use the

variable index, it is also permissible to use the name *pt2 because they

are identical in meaning. This fact is illustrated in the first printf

statement since this statement uses the three means of identifying the

same variable to print out the same variable three times.

Note carefully that, even though it appears that there are three

variables, there is really only one variable. The two pointers point to the

single variable. This is illustrated in the next statement which assigns the

value of 13 to the variable index, because that is where the pointer pt1

is pointing. The next printf statement causes the new value of 13 to be

printed out three times. Keep in mind that there is really only one

variable to be changed, not three.

This is admittedly a difficult concept, but since it is used extensively in

all but the most trivial C programs, it is well worth your time to stay

with this material until you understand it thoroughly.

Page 813-14

Declaring a pointer

Refer to the fifth line of the program and you will see the familiar way

of defining the variable index, followed by two more definitions. The

second definition can be read as the storage location to which pt1

points will be an int type variable. Therefore, pt1 is a pointer to an int

type variable. Similarly, pt2 is another pointer to an int type variable.

A pointer must be defined to point to some type of variable. Following

a proper definition, it cannot be used to point to any other type of

variable or it will result in a type incompatibility error. In the same

manner that a float type of variable cannot be added to an int type

variable, a pointer to a float variable cannot be used to point to an

integer variable.

Compile and run this program and observe that there is only one

variable and the single statement in line 10 changes the one variable

which is displayed three times.

A second pointer program

Consider the next program [POINTER2.C], which illustrates some of

the more complex operations that are used in C programs:

#include "stdio.h"

void main( )


char strg[40],*there,one,two;

int *pt,list[100],index;

strcpy(strg,"This is a character string.");

one = strg[0]; /* one and two are identical */

two = *strg;

printf("The first output is %c %c\n", one, two);

one = strg[8]; /* one and two are indentical */

two = *(strg+8);

printf("the second output is %c %c\n", one, two);

there = strg+10; /* strg+10 is identical to strg[10] */

printf("The third output is %c\n", strg[10]);

printf("The fourth output is %c\n", *there);

for (index = 0; index <>

list[index] = index + 100;

pt = list + 27;

printf("The fifth output is %d\n", list[27]);

printf("The sixth output is %d\n", *pt);


In this program we have defined several variables and two pointers. The

first pointer named there is a pointer to a char type variable and the

second named pt points to an int type variable. Notice also that we have

defined two array variables named strg and list. We will use them to

show the correspondence between pointers and array names.

Page 813-15

String variables as pointers

In C a string variable is defined to be simply a pointer to the beginning

of a string – this will take some explaining. You will notice that first we

assign a string constant to the string variable named strg so we will

have some data to work with. Next, we assign the value of the first

element to the variable one, a simple char variable. Next, since the

string name is a pointer by definition, we can assign the same value to

two by using the star and the string name. The result of the two

assignments are such that one now has the same value as two, and both

contain the character ‘T’, the first character in the string. Note that it

would be incorrect to write the ninth line as two = *strg[0]; because the

star takes the place of the square brackets.

For all practical purposes, strg is a pointer. It does, however, have one

restriction that a true pointer does not have. It cannot be changed like a

variable, but must always contain the initial value and therefore always

points to its string. It could be thought of as a pointer constant, and in

some applications you may desire a pointer that cannot be corrupted in

any way. Even though it cannot be changed, it can be used to refer to

other values than the one it is defined to point to, as we will see in the

next section of the program.

Moving ahead to line 9, the variable one is assigned the value of the

ninth variable (since the indexing starts at zero) and two is assigned the

same value because we are allowed to index a pointer to get to values

farther ahead in the string. Both variables now contain the character ‘a’.

The C compiler takes care of indexing for us by automatically adjusting

the indexing for the type of variable the pointer is pointing to.(This is

why the data type of a variable must be declared before the variable is

used.) In this case, the index of 8 is simply added to the pointer value

before looking up the desired result because a char type variable is one

byte long. If we were using a pointer to an int type variable, the index

would be doubled and added to the pointer before looking up the value

because an int type variable uses two bytes per value stored. When we

get to the section on structures, we will see that a variable can have

many, even into the hundreds or thousands, of bytes per variable, but

the indexing will be handled automatically for us by the system.

Since there is already a pointer, it can be assigned the address of the

eleventh element of strg by the statement in line 12 of the program.

Remember that since there is a true pointer, it can be assigned any value

as long as that value represents a char type of address. It should be

clear that the pointers must be typed in order to allow the pointer

arithmetic described in the last paragraph to be done properly. The third

and fourth outputs will be the same, namely the letter ‘c’.

Page 813-16

Pointer arithmetic

Not all forms of arithmetic are permissible on a pointer: only those

things that make sense. Considering that a pointer is an address

somewhere in the computer, it would make sense to add a constant to

an address, thereby moving it ahead in memory that number of places.

Similarly, subtraction is permissible, moving it back some number of

locations. Adding two pointers together would not make sense because

absolute memory addresses are not additive. Pointer multiplication is

also not allowed, as that would be a ‘funny’ number. If you think about

what you are actually doing, it will make sense to you what is allowed,

and what is not.

Integer pointers

The array named list is assigned a series of values from 100 to 199 in

order to have some data to work with. Next we assign the pointer pt

the address of the 28th element of the list and print out the same value

both ways to illustrate that the system truly will adjust the index for the

int type variable. You should spend some time in this program until you

feel you fairly well understand these lessons on pointers.

Function data return with a pointer

You may recall that back in the objective dealing with functions it was

mentioned that a function could use variable data if the parameter was

declared as an array. This works because an array is really a pointer to

the array elements. Functons can manipulate variable data if that data is

passed to the function as a pointer. The following program

[TWOWAY.C] illustrates the general approach used to manipulate

variable data in a function.

#include "stdio.h"

void fixup(nuts,fruit); /* prototype the function */

void main( )


int pecans,apples;

pecans = 100;

apples = 101;

printf("The starting values are %d %d\n",pecans,apples);


/* when we call "fixup" we send the value */

/* of pecans and the address of apples */

printf("The ending values are %d %d\n", pecans, apples);


fixup(nuts,fruit) /* nuts is an integer value */

int nuts,*fruit; /* fruit points to an integer */


printf("The values are %d %d\n", nuts, *fruit);

nuts = 135;

*fruit = 172;

printf("The values are %d %d\n", nuts ,*fruit);


Page 813-17

There are two variables defined in the main program: pecans and

apples; notice that neither of these is defined as a pointer. We assign

values to both of these and print them out, then call the function fixup

taking with us both of these values. The variable pecans is simply sent

to the function, but the address of the variable apples is sent to the

function. Now we have a problem. The two arguments are not the

same, the second is a pointer to a variable. We must somehow alert the

function to the fact that it is supposed to receive an integer variable and

a pointer to an integer variable. This turns out to be very simple. Notice

that the parameter definitions in the function define nuts as an integer,

and fruit as a pointer to an integer. The call in the main program

therefore is now in agreement with the function heading and the

program interface will work just fine.

In the body of the function, we print the two values sent to the function,

then modify them and print the new values out. The surprise occurs

when we return to the main program and print out the two values again.

We will find that the value of pecans will be restored to its value before

the function call because the C language made a copy of the variable

pecans and takes the copy to the called function, leaving the original

intact. In the case of the variable apples, we made a copy of a pointer

to the variable and took the copy of the pointer to the function. Since

we had a pointer to the original variable, even though the pointer was a

copy, we had access to the original variable and could change it in the

function. When we returned to the main program, we found a changed

value in apples when we printed it out.

By using a pointer in a function call, we can have access to the data in

the function and change it in such a way that when we return to the

calling program, we have a changed value for the data. It must be

pointed out however, that if you modify the value of the pointer itself in

the function, you will have a restored pointer when you return because

the pointer you use in the function is a copy of the original. In this

example, there was no pointer in the main program because we simply

sent the address to the function, but in many programs you will use

pointers in function calls.

Compile and run the program and observe the output.

Programming Exercises

Define a character array and use strcpy to copy a string into it. Print the

string out by using a loop with a pointer to print out one character at a

time. Initialise the pointer to the first element and use the double plus

sign to increment the pointer. Use a separate integer variable to count

the characters to print.

Page 813-18

Modify the program to print out the string backwards by pointing to the

end and using a decrementing pointer.

Page 813-19

Objective 4 After working through this module you should be able to create and

manage complex data types in C.


A structure is a user-defined data type. You have the ability to define a

new type of data considerably more complex than the types we have

been using. A structure is a collection of one or more variables, possibly

of different types, grouped together under a single name for convenient

handling. Structures are called “records” in some languages, notably

Pascal. Structures help organise complicated data, particularly in large

programs, because they permit a group of related variables to be treated

as a unit instead of as separate entities.

The best way to understand a structure is to look at an example


#include "stdio.h"

void main( )


struct {

char initial; /* last name initial */

int age; /* childs age */

int grade; /* childs grade in school */

} boy, girl;

boy.initial = 'R';

boy.age = 15;

boy.grade = 75;

girl.age = boy.age - 1; /* she is one year younger */

girl.grade = 82;

girl.initial = 'H';

printf("%c is %d years old and got a grade of %d\n",

girl.initial, girl.age, girl.grade);

printf("%c is %d years old and got a grade of %d\n",

boy.initial, boy.age, boy.grade);


The program begins with a structure definition. The key word struct is

followed by some simple variables between the braces, which are the

components of the structure. After the closing brace, you will find two

variables listed, namely boy, and girl. According to the definition of a

structure, boy is now a variable composed of three elements: initial,

age, and grade. Each of the three fields are associated with boy, and

each can store a variable of its respective type. The variable girl is also

a variable containing three fields with the same names as those of boy

but are actually different variables. We have therefore defined 6 simple


A single compound variable

Let’s examine the variable boy more closely. As stated above, each of

the three elements of boy are simple variables and can be used anywhere

in a C program where a variable of their type can be used. For example,

Page 813-20

the age element is an integer variable and can therefore be used

anywhere in a C program where it is legal to use an integer variable: in

calculations, as a counter, in I/O operations, etc. The only problem we

have is defining how to use the simple variable age which is a part of

the compound variable boy. We use both names with a decimal point

between them with the major name first. Thus boy.age is the complete

variable name for the age field of boy. This construct can be used

anywhere in a C program that it is desired to refer to this field. In fact,

it is illegal to use the name boy or age alone because they are only

partial definitions of the complete field. Alone, the names refer to


Assigning values to the variables

Using the above definition, we can assign a value to each of the three

fields of boy and each of the three fields of girl. Note carefully that

boy.initial is actually a char type variable, because it was assigned that

in the structure, so it must be assigned a character of data. Notice that

boy.initial is assigned the character R in agreement with the above

rules. The remaining two fields of boy are assigned values in accordance

with their respective types. Finally the three fields of girl are assigned

values but in a different order to illustrate that the order of assignment

is not critical.

Using structure data

Now that we have assigned values to the six simple variables, we can

do anything we desire with them. In order to keep this first example

simple, we will simply print out the values to see if they really do exist

as assigned. If you carefully inspect the printf statements, you will see

that there is nothing special about them. The compound name of each

variable is specified because that is the only valid name by which we can

refer to these variables.

Structures are a very useful method of grouping data together in order

to make a program easier to write and understand. This first example is

too simple to give you even a hint of the value of using structures, but

continue on through these lessons and eventually you will see the value

of using structures.

An array of structures

The next program [STRUCT2.C] contains the same structure definition

as before but this time we define an array of 12 variables named kids.

This program therefore contains 12 times 3 = 36 simple variables, each

of which can store one item of data provided that it is of the correct

type. We also define a simple variable named index for use in the for


Page 813-21

#include "stdio.h"

void main( )


struct {

char initial;

int age;

int grade;

} kids[12];

int index;

for (index = 0; index <>

kids[index].initial = 'A' + index;

kids[index].age = 16;

kids[index].grade = 84;


kids[3].age = kids[5].age = 17;

kids[2].grade = kids[6].grade = 92;

kids[4].grade = 57;

kids[10] = kids[4]; /* Structure assignment */

for (index = 0; index <>

printf("%c is %d years old and got a grade of %d\n",

kids[index].initial, kids[index].age,



To assign each of the fields a value we use a for loop and each pass

through the loop results in assigning a value to three of the fields. One

pass through the loop assigns all of the values for one of the kids. This

would not be a very useful way to assign data in a real situation, but a

loop could read the data in from a file and store it in the correct fields.

You might consider this the crude beginning of a data base – which, of

course, it is.

In the next few instructions of the program we assign new values to

some of the fields to illustrate the method used to accomplish this. It

should be self explanatory, so no additional comments will be given.

Copying structures

C allows you to copy an entire structure with one statement. Line 17 is

an example of using a structure assignment. In this statement, all 3

fields of kids[4] are copied into their respective fields of kids[10].

The last few statements contain a for loop in which all of the generated

values are displayed in a formatted list. Compile and run the program to

see if it does what you expect it to do.

Using pointers and structures together

The next program [STRUCT3.C] is an example of using pointers with

structures; it is identical to the last program except that it uses pointers

for some of the operations.

#include "stdio.h"

void main( )


struct {

char initial;

int age;

Page 813-22

int grade;

} kids[12], *point, extra;

int index;

for (index = 0; index <>

point = kids + index;

point->initial = 'A' + index;

point->age = 16;

point->grade = 84;


kids[3].age = kids[5].age = 17;

kids[2].grade = kids[6].grade = 92;

kids[4].grade = 57;

for (index = 0; index <>

point = kids + index;

printf("%c is %d years old and got a grade of %d\n",

(*point).initial, kids[index].age, point->grade);


extra = kids[2]; /* Structure assignment */

extra = *point; /* Structure assignment */


The first difference shows up in the definition of variables following the

structure definition. In this program we define a pointer named point

which is defined as a pointer that points to the structure. It would be

illegal to try to use this pointer to point to any other variable type.

There is a very definite reason for this restriction in C as we have

alluded to earlier and will review in the next few paragraphs.

The next difference is in the for loop where we use the pointer for

accessing the data fields. Since kids is a pointer variable that points to

the structure, we can define point in terms of kids. The variable kids is a

constant so it cannot be changed in value, but point is a pointer variable

and can be assigned any value consistent with its being required to point

to the structure. If we assign the value of kids to point then it should be

clear that it will point to the first element of the array, a structure

containing three fields.

Pointer arithmetic

Adding 1 to point will now cause it to point to the second field of the

array because of the way pointers are handled in C. The system knows

that the structure contains three variables and it knows how many

memory elements are required to store the complete structure.

Therefore if we tell it to add one to the pointer, it will actually add the

number of memory elements required to get to the next element of the

array. If, for example, we were to add 4 to the pointer, it would

advance the value of the pointer 4 times the size of the structure,

resulting in it pointing 4 elements farther along the array. This is the

reason a pointer cannot be used to point to any data type other than the

one for which it was defined.

Now return to the program. It should be clear from the previous

discussion that as we go through the loop, the pointer will point to the

beginning of one of the array elements each time. We can therefore use

Page 813-23

the pointer to reference the various elements of the structure. Referring

to the elements of a structure with a pointer occurs so often in C that a

special method of doing that was devised. Using point->initial is the

same as using (*point).initial which is really the way we did it in the

last two programs. Remember that *point is the stored data to which

the pointer points and the construct should be clear. The -> is made up

of the minus sign and the greater than sign.

Since the pointer points to the structure, we must once again define

which of the elements we wish to refer to each time we use one of the

elements of the structure. There are, as we have seen, several different

methods of referring to the members of the structure, and in the for

loop used for output at the end of the program, we use three different

methods. This would be considered very poor programming practice,

but is done this way here to illustrate to you that they all lead to the

same result. This program will probably require some study on your

part to fully understand, but it will be worth your time and effort to

grasp these principles.

Nested and named structures

The structures we have seen so far have been useful, but very simple. It

is possible to define structures containing dozens and even hundreds or

thousands of elements but it would be to the programmer’s advantage

not to define all of the elements at one pass but rather to use a

hierarchical structure of definition. This will be illustrated with the next

program [NESTED.C], which shows a nested structure.

#include "stdio.h"

void main( )


struct person {

char name[25];

int age;

char status; /* M = married, S = single */

} ;

struct alldat {

int grade;

struct person descrip;

char lunch[25];

} student[53];

struct alldat teacher,sub;

teacher.grade = 94;

teacher.descrip.age = 34;

teacher.descrip.status = 'M';

strcpy(,"Mary Smith");

strcpy(teacher.lunch,"Baloney sandwich");

sub.descrip.age = 87;

sub.descrip.status = 'M';

strcpy(,"Old Lady Brown");

sub.grade = 73;

strcpy(sub.lunch,"Yogurt and toast");

student[1].descrip.age = 15;

student[1].descrip.status = 'S';

strcpy(student[1],"Billy Boston");

strcpy(student[1].lunch,"Peanut Butter");

student[1].grade = 77;

Page 813-24

student[7].descrip.age = 14;

student[12].grade = 87;


The first structure contains three elements but is followed by no

variable name. We therefore have not defined any variables only a

structure, but since we have included a name at the beginning of the

structure, the structure is named person. The name person can be used

to refer to the structure but not to any variable of this structure type. It

is therefore a new type that we have defined, and we can use the new

type in nearly the same way we use int, char, or any other types that

exist in C. The only restriction is that this new name must always be

associated with the reserved word struct.

The next structure definition contains three fields with the middle field

being the previously defined structure which we named person. The

variable which has the type of person is named descrip. So the new

structure contains two simple variables, grade and a string named

lunch[25], and the structure named descrip. Since descrip contains

three variables, the new structure actually contains 5 variables. This

structure is also given a name, alldat, which is another type definition.

Finally we define an array of 53 variables each with the structure

defined by alldat, and each with the name student. If that is clear, you

will see that we have defined a total of 53 times 5 variables, each of

which is capable of storing a value.

Since we have a new type definition we can use it to define two more

variables. The variables teacher and sub are defined in line 13 to be

variables of the type alldat, so that each of these two variables contain

5 fields which can store data.

Using fields

In the next five lines of the program we assign values to each of the

fields of teacher. The first field is the grade field and is handled just like

the other structures we have studied because it is not part of the nested

structure. Next we wish to assign a value to her age which is part of the

nested structure. To address this field we start with the variable name

teacher to which we append the name of the group descrip, and then

we must define which field of the nested structure we are interested in,

so we append the name age. The teacher’s status field is handled in

exactly the same manner as her age, but the last two fields are assigned

strings using the string copy function strcpy, which must be used for

string assignment. Notice that the variable names in the strcpy function

are still variable names even though they are each made up of several


The variable sub is assigned nonsense values in much the same way, but

in a different order since they do not have to occur in any required

Page 813-25

order. Finally, a few of the student variables are assigned values for

illustrative purposes and the program ends. None of the values are

printed for illustration since several were printed in the last examples.

It is possible to continue nesting structures until you get totally

confused. If you define them properly, the computer will not get

confused because there is no stated limit as to how many levels of

nesting are allowed. There is probably a practical limit of three beyond

which you will get confused, but the language has no limit. In addition

to nesting, you can include as many structures as you desire in any level

of structures, such as defining another structure prior to alldat and

using it in alldat in addition to using person. The structure named

person could be included in alldat two or more times if desired.

Structures can contain arrays of other structures which in turn can

contain arrays of simple types or other structures. It can go on and on

until you lose all reason to continue. Be conservative at first, and get

bolder as you gain experience.

Exercise 4

Define a named structure containing a string field for a name, an integer

for feet, and another for arms. Use the new type to define an array of

about 6 items. Fill the fields with data and print them out as follows:

A human being has 2 legs and 2 arms.

A dog has 4 legs and 0 arms.

A television set has 4 legs and 0 arms.

A chair has 4 legs and 2 arms.


2. Rewrite the previous exercise using a pointer to print the data out.

Page 813-26

Objective 5 After working through this module you should be able to use unions to

define alternate data sets for use in C programs.


A union is a variable that may hold (at different times) data of different

types and sizes. For example, a programmer may wish to define a

structure that will record the citation details about a book, and another

that records details about a journal. Since a library item can be neither a

book or a journal simultaneously, the programmer would declare a

library item to be a union of book and journal structures. Thus, on one

occasion item might be used to manipulate book details, and on another

occassion, item might be used to manipulate journal details. It is up to

the programmer, of course, to remember the type of item with which

they are dealing.

Examine the next program [UNION1.C] for examples.

void main( )


union {

int value; /* This is the first part of the union


struct {

char first; /* These two values are the second part


char second;

} half;

} number;

long index;

for (index = 12; index <>

number.value = index;

printf("%8x %6x %6x\n", number.value, number.half.first,




In this example we have two elements to the union, the first part being

the integer named value, which is stored as a two byte variable

somewhere in the computer’s memory. The second element is made up

of two character variables named first and second. These two variables

are stored in the same storage locations that value is stored in, because

that is what a union does. A union allows you to store different types of

data in the same physical storage locations. In this case, you could put

an integer number in value, then retrieve it in its two halves by getting

each half using the two names first and second. This technique is often

used to pack data bytes together when you are, for example, combining

bytes to be used in the registers of the microprocessor.

Accessing the fields of the union is very similar to accessing the fields of

a structure and will be left to you to determine by studying the example.

One additional note must be given here about the program. When it is

run using the C compiler the data will be displayed with two leading f’s,

Page 813-27

due to the hexadecimal output promoting the char type variables to int

and extending the sign bit to the left. Converting the char type data

fields to int type fields prior to display should remove the leading f’s

from your display. This will involve defining two new int type variables

and assigning the char type variables to them. This will be left as an

exercise for you. Note that the same problem will also come up in a few

of the later examples.

Compile and run this program and observe that the data is read out as

an int and as two char variables. The char variables are reversed in

order because of the way an int variable is stored internally in your

computer. Don’t worry about this. It is not a problem but it can be a

very interesting area of study if you are so inclined.

The next program [UNION2.C] contains another example of a union,

one which is much more common. Suppose you wished to build a large

database including information on many types of vehicles. It would be

silly to include the number of propellers on a car, or the number of tires

on a boat. In order to keep all pertinent data, however, you would need

those data points for their proper types of vehicles. In order to build an

efficient data base, you would need several different types of data for

each vehicle, some of which would be common, and some of which

would be different. That is exactly what we are doing in the example

program, in which we define a complete structure, then decide which of

the various types can go into it.

#include "stdio.h"

#define AUTO 1

#define BOAT 2

#define PLANE 3

#define SHIP 4

void main( )


struct automobile { /* structure for an automobile */

int tires;

int fenders;

int doors;

} ;

typedef struct { /* structure for a boat or ship */

int displacement;

char length;


struct {

char vehicle; /* what type of vehicle? */

int weight; /* gross weight of vehicle */

union { /* type-dependent data */

struct automobile car; /* part 1 of the union */

BOATDEF boat; /* part 2 of the union */

struct {

char engines;

int wingspan;

} airplane; /* part 3 of the union */

BOATDEF ship; /* part 4 of the union */

} vehicle_type;

int value; /* value of vehicle in dollars */

char owner[32]; /* owners name */

} ford, sun_fish, piper_cub; /* three variable structures */

Page 813-28

/* define a few of the fields as an illustration */

ford.vehicle = AUTO;

ford.weight = 2742; /* with a full petrol tank */ = 5; /* including the spare */ = 2;

sun_fish.value = 3742; /* trailer not included */

sun_fish.vehicle_type.boat.length = 20;

piper_cub.vehicle = PLANE;

piper_cub.vehicle_type.airplane.wingspan = 27;

if (ford.vehicle == AUTO) /* which it is in this case */

printf("The ford has %d tires.\n",;

if (piper_cub.vehicle == AUTO) /* which it is not */

printf("The plane has %d tires.\n",;


First, we define a few constants with the #defines, and begin the

program itself. We define a structure named automobile containing

several fields which you should have no trouble recognising, but we

define no variables at this time.

The typedef

Next we define a new type of data with a typedef. This defines a

complete new type that can be used in the same way that int or char can

be used. Notice that the structure has no name, but at the end where

there would normally be a variable name there is the name BOATDEF.

We now have a new type, BOATDEF, that can be used to define a

structure anywhere we would like to. Notice that this does not define

any variables, only a new type definition. Capitalising the name is a

common preference used by programmers and is not a C standard; it

makes the typedef look different from a variable name.

We finally come to the big structure that defines our data using the

building blocks already defined above. The structure is composed of 5

parts, two simple variables named vehicle and weight, followed by the

union, and finally the last two simple variables named value and owner.

Of course the union is what we need to look at carefully here, so focus

on it for the moment. You will notice that it is composed of four parts,

the first part being the variable car which is a structure that we defined

previously. The second part is a variable named boat which is a

structure of the type BOATDEF previously defined. The third part of

the union is the variable airplane which is a structure defined in place in

the union. Finally we come to the last part of the union, the variable

named ship which is another structure of the type BOATDEF.

It should be stated that all four could have been defined in any of the

three ways shown, but the three different methods were used to show

you that any could be used. In practice, the clearest definition would

probably have occurred by using the typedef for each of the parts.

Page 813-29

We now have a structure that can be used to store any of four different

kinds of data structures. The size of every record will be the size of that

record containing the largest union. In this case part 1 is the largest

union because it is composed of three integers, the others being

composed of an integer and a character each. The first member of this

union would therefore determine the size of all structures of this type.

The resulting structure can be used to store any of the four types of

data, but it is up to the programmer to keep track of what is stored in

each variable of this type. The variable vehicle was designed into this

structure to keep track of the type of vehicle stored here. The four

defines at the top of the page were designed to be used as indicators to

be stored in the variable vehicle. A few examples of how to use the

resulting structure are given in the next few lines of the program. Some

of the variables are defined and a few of them are printed out for

illustrative purposes.

The union is not used frequently, and almost never by novice

programmers. You will encounter it occasionally so it is worth your

effort to at least know what it is.


The bitfield is a relatively new addition to the C programming language.

In the next program [BITFIELD.C] we have a union made up of a

single int type variable in line 5 and the structure defined in lines 6 to

10. The structure is composed of three bitfields named x, y, and z. The

variable named x is only one bit wide, the variable y is two bits wide and

adjacent to the variable x, and the variable z is two bits wide and

adjacent to y. Moreover, because the union causes the bits to be stored

in the same memory location as the variable index, the variable x is the

least significant bit of the variable index, y forms the next two bits, and

z forms the two high-order bits.

#include "stdio.h"

void main( )


union {

int index;

struct {

unsigned int x : 1;

unsigned int y : 2;

unsigned int z : 2;

} bits;

} number;

for (number.index = 0; number.index <>

printf("index = %3d, bits = %3d%3d%3d\n", number.index,

number.bits.z, number.bits.y, number.bits.x);



Compile and run the program and you will see that as the variable index

is incremented by 1 each time through the loop, you will see the

bitfields of the union counting due to their respective locations within

the int definition. One thing must be pointed out: the bitfields must be

Page 813-30

defined as parts of an unsigned int or your compiler will issue an error


The bitfield is very useful if you have a lot of data to separate into

separate bits or groups of bits. Many systems use some sort of a packed

format to get lots of data stored in a few bytes. Your imagination is

your only limitation to effective use of this feature of C.

Exercise 5

Page 813-31

Objective 6 After working through this module you should be able to allocate

memory to variables dynamically.


Dynamic allocation is very intimidating the first time you come across

it, but that need not be. All of the programs up to this point have used

static variables as far as we are concerned. (Actually, some of them

have been automatic and were dynamically allocated for you by the

system, but it was transparent to you.) This section discusses how C

uses dynamically allocated variables. They are variables that do not

exist when the program is loaded, but are created dynamically as they

are needed. It is possible, using these techniques, to create as many

variables as needed, use them, and deallocate their space so that it can

be used by other dynamic variables. As usual, the concept is best

presented by an example [DYNLIST.C].

#include "stdio.h"

#include "stdlib.h"

void main( )


struct animal {

char name[25];

char breed[25];

int age;

} *pet1, *pet2, *pet3;

pet1 = (struct animal *) malloc (sizeof(struct animal));


strcpy(pet1->breed,"Mixed Breed");

pet1->age = 1;

pet2 = pet1;

/* pet2 now points to the above data structure */

pet1 = (struct animal *) malloc (sizeof(struct animal));


strcpy(pet1->breed,"Labrador Retriever");

pet1->age = 3;

pet3 = (struct animal *) malloc (sizeof(struct animal));


strcpy(pet3->breed,"German Shepherd");

pet3->age = 4;

/* now print out the data described above */

printf("%s is a %s, and is %d years old.\n", pet1->name,

pet1->breed, pet1->age);

printf("%s is a %s, and is %d years old.\n", pet2->name,

pet2->breed, pet2->age);

printf("%s is a %s, and is %d years old.\n", pet3->name,

pet3->breed, pet3->age);

pet1 = pet3;

/* pet1 now points to the same structure that pet3 points to */

free(pet3); /* this frees up one structure */

free(pet2); /* this frees up one more structure */

/* free(pet1); this cannot be done, see explanation in text */


We begin by defining a named structure – animal – with a few fields

pertaining to dogs. We do not define any variables of this type, only

three pointers. If you search through the remainder of the program, you

will find no variables defined so we have nothing to store data in. All

Page 813-32

we have to work with are three pointers, each of which point to the

defined structure. In order to do anything we need some variables, so

we will create some dynamically.

Dynamic variable creation

The first program statement is line 11, which assigns something to the

pointer pet1; it will create a dynamic structure containing three

variables. The heart of the statement is the malloc function buried in the

middle. This is a memory allocate function that needs the other things

to completely define it. The malloc function, by default, will allocate a

piece of memory on a heap that is n characters in length and will be of

type character. The n must be specified as the only argument to the

function. We will discuss n shortly, but first we need to define a heap.

The heap

Every compiler has a set of limitations on it that define how big the

executable file can be, how many variables can be used, how long the

source file can be, etc. One limitation placed on users by the C compiler

is a limit of 64K for the executable code if you happen to be in the small

memory model. This is because the IBM-PC uses a microprocessor

with a 64K segment size, and it requires special calls to use data outside

of a single segment. In order to keep the program small and efficient,

these calls are not used, and the memory space is limited but still

adequate for most programs.

In this model C defines two heaps, the first being called a heap, and the

second being called the far heap. The heap is an area within the 64K

boundary that can store dynamically allocated data and the far heap is

an area outside of this 64K boundary which can be accessed by the

program to store data and variables.

The data and variables are put on the heap by the system as calls to

malloc are made. The system keeps track of where the data is stored.

Data and variables can be deallocated as desired leading to holes in the

heap. The system knows where the holes are and will use them for

additional data storage as more malloc calls are made. The structure of

the heap is therefore a very dynamic entity, changing constantly.

The data and variables are put on the far heap by utilising calls to

farmalloc, farcalloc, etc. and removed through use of the function

farfree. Study your C compiler’s Reference Guide for details of how to

use these features.

Page 813-33


C compilers give the user a choice of memory models to use. The user

has a choice of using a model with a 64K limitation for either program

or data leading to a small fast program or selecting a 640K limitation

and requiring longer address calls leading to less efficient addressing.

Using the larger address space requires inter-segment addressing,

resulting in the slightly slower running time. The time is probably

insignificant in most programs, but there are other considerations.

If a program uses no more than 64K bytes for the total of its code and

memory and if it doesn’t use a stack, it can be made into a .COM file.

With C this is only possible by using the tiny memory model. Since a

.COM file is already in a memory image format, it can be loaded very

quickly whereas a file in an .EXE format must have its addresses

relocated as it is loaded. Therefore a tiny memory model can generate a

program that loads faster than one generated with a larger memory

model. Don’t let this worry you, it is a fine point that few programmers

worry about.

Even more important than the need to stay within the small memory

model is the need to stay within the computer. If you had a program

that used several large data storage areas, but not at the same time, you

could load one block storing it dynamically, then get rid of it and reuse

the space for the next large block of data. Dynamically storing each

block of data in succession, and using the same storage for each block

may allow you to run your entire program in the computer without

breaking it up into smaller programs.

The malloc function

Hopefully the above description of the heap and the overall plan for

dynamic allocation helped you to understand what we are doing with

the malloc function. The malloc function forms part of the standard

library. Its prototype is defined in the stdlib.h file, hence this file is

included using the statement on line 2. malloc simply asks the system

for a block of memory of the size specified, and gets the block with the

pointer pointing to the first element of the block. The only argument in

the parentheses is the size of the block desired and in our present case,

we desire a block that will hold one of the structures we defined at the

beginning of the program. The sizeof is a new function that returns the

size in bytes of the argument within its parentheses. It therefore returns

the size of the structure named animal, in bytes, and that number is sent

to the system with the malloc call. At the completion of that call we

have a block on the heap allocated to us, with pet1 pointing to the

block of data.

Page 813-34


We still have a funny looking construct at the beginning of the malloc

function call that is called a cast. The malloc function returns a block

with the pointer pointing to it being a pointer of type char by default.

Many (perhaps most) times you do not want a pointer to a char type

variable, but to some other type. You can define the pointer type with

the construct given on the example line. In this case we want the

pointer to point to a structure of type animal, so we tell the compiler

with this strange looking construct. Even if you omit the cast, most

compilers will return a pointer correctly, give you a warning, and go on

to produce a working program. It is better programming practice to

provide the compiler with the cast to prevent getting the warning


Using the dynamically allocated memory block

If you remember our studies of structures and pointers, you will recall

that if we have a structure with a pointer pointing to it, we can access

any of the variables within the structure. In the next three lines of the

program we assign some silly data to the structure for illustration. It

should come as no surprise to you that these assignment statements

look just like assignments to statically defined variables.

In the next statement, we assign the value of pet1 to pet2 also. This

creates no new data; we simply have two pointers to the same object.

Since pet2 is pointing to the structure we created above, pet1 can be

reused to get another dynamically allocated structure which is just what

we do next. Keep in mind that pet2 could have just as easily been used

for the new allocation. The new structure is filled with silly data for


Finally, we allocate another block on the heap using the pointer pet3,

and fill its block with illustrative data. Printing the data out should pose

no problem to you since there is nothing new in the three print

statements. It is left for you to study.

Even though it is not illustrated in this example, you can dynamically

allocate and use simple variables such as a single char type variable.

This should be used wisely however, since it sometimes is very

inefficient. It is only mentioned to point out that there is nothing magic

about a data structure that would allow it to be dynamically allocated

while simple types could not.

Getting rid of dynamically allocated data

Another new function is used to get rid of the data and free up the

space on the heap for reuse. This function is called free. To use it, you

Page 813-35

simply call it with the pointer to the block as the only argument, and the

block is deallocated.

In order to illustrate another aspect of the dynamic allocation and

deallocation of data, an additional step is included in the program on

your monitor. The pointer pet1 is assigned the value of pet3 in line 31.

In doing this, the block that pet1 was pointing to is effectively lost since

there is no pointer that is now pointing to that block. It can therefore

never again be referred to, changed, or disposed of. That memory,

which is a block on the heap, is wasted from this point on. This is not

something that you would ever purposely do in a program. It is only

done here for illustration.

The first free function call removes the block of data that pet1 and pet3

were pointing to, and the second free call removes the block of data

that pet2 was pointing to. We therefore have lost access to all of our

data generated earlier. There is still one block of data that is on the heap

but there is no pointer to it since we lost the address to it. Trying to free

the data pointed to by pet1 would result in an error because it has

already been freed by the use of pet3. There is no need to worry, when

we return to DOS, the entire heap will be disposed of with no regard to

what we have put on it. The point does need to made that, if you lose a

pointer to a block on the heap, it forever removes that block of data

storage from our use and we may need that storage later. Compile and

run the program to see if it does what you think it should do based on

this discussion.

Our discussion of the last program has taken a lot of time – but it was

time well spent. It should be somewhat exciting to you to know that

there is nothing else to learn about dynamic allocation, the last few

pages covered it all. Of course, there is a lot to learn about the

technique of using dynamic allocation, and for that reason, there are

two more files to study. But the fact remains, there is nothing more to

learn about dynamic allocation than what was given so far in this


An array of pointers

Our next example [BIGDYNL.C] is very similar to the last, since we

use the same structure, but this time we define an array of pointers to

illustrate the means by which you could build a large database using an

array of pointers rather than a single pointer to each element. To keep it

simple we define 12 elements in the array and another working pointer

named point.

#include "stdio.h"

#include "stdlib.h"

void main( )


Page 813-36

struct animal {

char name[25];

char breed[25];

int age;

} *pet[12], *point; /* this defines 13 pointers, no variables


int index;

/* first, fill the dynamic structures with nonsense */

for (index = 0; index <>

pet[index] = (struct animal *) malloc (sizeof(struct animal));

strcpy(pet[index]->name, "General");

strcpy(pet[index]->breed, "Mixed Breed");

pet[index]->age = 4;


pet[4]->age = 12; /* these lines are simply to


pet[5]->age = 15; /* put some nonsense data into


pet[6]->age = 10; /* a few of the fields.


/* now print out the data described above */

for (index = 0; index <12>

point = pet[index];

printf("%s is a %s, and is %d years old.\n", point->name,

point->breed, point->age);


/* good programming practice dictates that we free up the


/* dynamically allocated space before we quit.


for (index = 0; index <>



The *pet[12] might seem strange to you so a few words of explanation

are in order. What we have defined is an array of 12 pointers, the first

being pet[0], and the last pet[11]. Actually, since an array is itself a

pointer, the name pet by itself is a pointer to a pointer. This is valid in

C, and in fact you can go farther if needed but you will get quickly

confused. A definition such as int ****pt is legal as a pointer to a

pointer to a pointer to a pointer to an integer type variable.

Twelve pointers have now been defined which can be used like any

other pointer, it is a simple matter to write a loop to allocate a data

block dynamically for each and to fill the respective fields with any data

desirable. In this case, the fields are filled with simple data for

illustrative purposes, but we could be reading in a database, reading

from some test equipment, or any other source of data.

A few fields are randomly picked to receive other data to illustrate that

simple assignments can be used, and the data is printed out to the

monitor. The pointer point is used in the printout loop only to serve as

an illustration, the data could have been easily printed using the pet[n]

notation. Finally, all 12 blocks of data are release with the free function

before terminating the program.

Compile and run this program to aid in understanding this technique. As

stated earlier, there was nothing new here about dynamic allocation,

only about an array of pointers.

Page 813-37

A linked list

The final example is what some programmers find the most intimidating

of all techniques: the dynamically allocated linked list. Examine the next

program [DYNLINK.C] to start our examination of lists.

#include "stdio.h"

#include "stdlib.h"

#define RECORDS 6

void main( )


struct animal {

char name[25]; /* The animal's name */

char breed[25]; /* The type of animal */

int age; /* The animal's age */

struct animal *next;

/* a pointer to another record of this type */

} *point, *start, *prior;

/* this defines 3 pointers, no variables */

int index;

/* the first record is always a special case */

start = (struct animal *) malloc (sizeof(struct animal));

strcpy(start->name, "General");

strcpy(start->breed, "Mixed Breed");

start->age = 4;

start->next = NULL;

prior = start;

/* a loop can be used to fill in the rest once it is started */

for (index = 0; index <>

point = (struct animal *) malloc (sizeof(struct animal));

strcpy(point->name, "Frank");

strcpy(point->breed, "Labrador Retriever");

point->age = 3;

prior->next = point;

/* point last "next" to this record */

point->next = NULL;

/* point this "next" to NULL */

prior = point;

/* this is now the prior record */


/* now print out the data described above */

point = start;

do {

prior = point->next;

printf("%s is a %s, and is %d years old.\n", point->name,

point->breed, point->age);

point = point->next;

} while (prior != NULL);

/* good programming practice dictates that we free up the */

/* dynamically allocated space before we quit. */

point = start; /* first block of group */

do {

prior = point->next; /* next block of data */

free(point); /* free present block */

point = prior; /* point to next */

} while (prior != NULL); /* quit when next is NULL



The program starts in a similar manner to the previous two, with the

addition of the definition of a constant to be used later. The structure is

nearly the same as that used in the last two programs except for the

addition of another field within the structure in line 10, the pointer. This

pointer is a pointer to another structure of this same type and will be

Page 813-38

used to point to the next structure in order. To continue the above

analogy, this pointer will point to the next note, which in turn will

contain a pointer to the next note after that.

We define three pointers to this structure for use in the program, and

one integer to be used as a counter, and we are ready to begin using the

defined structure for whatever purpose we desire. In this case, we will

once again generate nonsense data for illustrative purposes.

Using the malloc function, we request a block of storage on the heap

and fill it with data. The additional field in this example, the pointer, is

assigned the value of NULL, which is only used to indicate that this is

the end of the list. We will leave the pointer start at this structure, so

that it will always point to the first structure of the list. We also assign

prior the value of start for reasons we will see soon. Keep in mind that

the end points of a linked list will always have to be handled differently

than those in the middle of a list. We have a single element of our list

now and it is filled with representative data.

Filling additional structures

The next group of assignments and control statements are included

within a for loop so we can build our list fast once it is defined. We will

go through the loop a number of times equal to the constant

RECORDS defined at the beginning of our program. Each time

through, we allocate memory, fill the first three fields with nonsense,

and fill the pointers. The pointer in the last record is given the address

of this new record because the prior pointer is pointing to the prior

record. Thus prior->next is given the address of the new record we

have just filled. The pointer in the new record is assigned the value

NULL, and the pointer prior is given the address of this new record

because the next time we create a record, this one will be the prior one

at that time. That may sound confusing but it really does make sense if

you spend some time studying it.

When we have gone through the for loop 6 times, we will have a list of

7 structures including the one we generated prior to the loop. The list

will have the following characteristics:

1. start points to the first structure in the list.

2. Each structure contains a pointer to the next structure.

3. The last structure has a pointer that points to NULL and can be

used to detect the end of the list.

The following diagram may help you to understand the structure of the

data at this point.

Page 813-39

start® struct1




point® struct2




point® struct3




point® … … struct7




point® NULL

It should be clear (if you understand the above structure) that it is not

possible to simply jump into the middle of the structure and change a

few values. The only way to get to the third structure is by starting at

the beginning and working your way down through the structure one

record at a time. Although this may seem like a large price to pay for

the convenience of putting so much data outside of the program area, it

is actually a very good way to store some kinds of data.

A word processor would be a good application for this type of data

structure because you would never need to have random access to the

data. In actual practice, this is the basic type of storage used for the text

in a word processor with one line of text per record. Actually, a

program with any degree of sophistication would use a doubly linked

list. This would be a list with two pointers per record, one pointing

down to the next record, and the other pointing up to the record just

prior to the one in question. Using this kind of a record structure would

allow traversing the data in either direction.

Printing the data out

A method similar to the data generation process is used to print the data

out. The pointers are initialised and are then used to go from record to

record reading and displaying each record one at a time. Printing is

terminated when the NULL on the last record is found, so the program

doesn’t even need to know how many records are in the list. Finally, the

entire list is deleted to make room in memory for any additional data

that may be needed, in this case, none. Care must be taken to ensure

that the last record is not deleted before the NULL is checked; once the

data are gone, it is impossible to know if you are finished yet.

Page 813-40

It is not difficult, but it is not trivial, to add elements into the middle of

a linked lists. It is necessary to create the new record, fill it with data,

and point its pointer to the record it is desired to precede. If the new

record is to be installed between the 3rd and 4th, for example, it is

necessary for the new record to point to the 4th record, and the pointer

in the 3rd record must point to the new one. Adding a new record to

the beginning or end of a list are each special cases. Consider what must

be done to add a new record in a doubly linked list.

Entire books are written describing different types of linked lists and

how to use them, so no further detail will be given. The amount of

detail given should be sufficient for a beginning understanding of C and

its capabilities.


One more function, the calloc function, must be mentioned before

closing. This function allocates a block of memory and clears it to all

zeros – which may be useful in some circumstances. It is similar to

malloc in many ways. We will leave you to read about calloc as an

exercise, and use it if you desire.

Exercise 6

Rewrite STRUCT1.C to dynamically allocate the two structures.

Rewrite STRUCT2.C to dynamically allocate the 12 structures.

Page 813-41

Objective 7 After working through this module you should be able to manipulate

characters and bits.

Upper and lower case

The first example program [UPLOW.C] in this section does character

manipulation. More specifically, it changes the case of alphabetic

characters. It illustrates the use of four functions that have to do with

case. Each of these functions is part of the standard library. The

functions are prototyped in the ctype.h file, hence its inclusion in line 2

of the program.

#include "stdio.h"

#include "ctype.h"

void mix_up_the_chars(line);

void main( )


char line[80];

char *c;

do {

/* keep getting lines of text until an empty line is found */

c = gets(line); /* get a line of text */

if (c != NULL) {



} while (c != NULL);


void mix_up_the_chars(line)

/* this function turns all upper case characters into lower */

/* case, and all lower case to upper case. It ignores all */

/* other characters. */

char line[ ];


int index;

for (index = 0;line[index] != 0;index++) {

if (isupper(line[index])) /* 1 if upper case */

line[index] = tolower(line[index]);

else {

if (islower(line[index])) /* 1 if lower case */

line[index] = toupper(line[index]);





It should be no problem for you to study this program on your own and

understand how it works. The four functions on display in this program

are all within the user written function, mix_up_the_chars. Compile and

run the program with data of your choice. The four functions are:

isupper( ); Is the character upper case?

islower( ); Is the character lower case?

toupper( ); Make the character upper case.

tolower( ); Make the character lower case.

Many more classification and conversion routines are listed in your C

compiler’s Reference Guide.

Page 813-42

Classification of characters

We have repeatedly used the backslash n (\n) character for representing

a new line. Such indicators are called escape sequences, and some of

the more commonly used are defined in the following table:

\n Newline

\t Tab

\b Backspace

\ Double quote

\\ Backslash

\0 NULL (zero)

A complete list of escape sequences available with your C compiler are

listed in your C Reference Guide.

By preceding each of the above characters with the backslash character,

the character can be included in a line of text for display, or printing. In

the same way that it is perfectly all right to use the letter n in a line of

text as a part of someone's name, and as an end-of-line, the other

characters can be used as parts of text or for their particular functions.

The next program [CHARCLAS.C] uses the functions that can

determine the class of a character, and counts the characters in each


#include "stdio.h"

#include "ctype.h"

void count_the_data(line)

void main( )


char line[80]

char *c;

do {

c = gets(line); /* get a line of text */

if (c != NULL) {



} while (c != NULL);


void count_the_data(line)

char line[];


int whites, chars, digits;

int index;

whites = chars = digits = 0;

for (index = 0; line[index] != 0; index++) {

if (isalpha(line[index])) /* 1 if line[ is alphabetic */


if (isdigit(line[index])) /* 1 if line[ ] is a digit */


if (isspace(line[index]))

Page 813-43

/* 1 if line[ ] is blank, tab, or newline */


} /* end of counting loop */

printf("%3d%3d%3d %s", whites, chars, digits, line);


The number of each class is displayed along with the line itself. The

three functions are as follows:

isalpha( ); Is the character alphabetic?

isdigit( ); Is the character a numeral?

isspace( ); Is the character any of \n, \t, or blank?

This program should be simple for you to find your way through so no

explanation will be given. It was necessary to give an example with

these functions used. Compile and run this program with suitable input


Logical functions

The functions in this group are used to do bitwise operations, meaning

that the operations are performed on the bits as though they were

individual bits. No carry from bit to bit is performed as would be done

with a binary addition. Even though the operations are performed on a

single bit basis, an entire byte or integer variable can be operated on in

one instruction. The operators and the operations they perform are

given in the following table:

& Logical AND, if both bits are 1, the result is 1.

| Logical OR, if either bit is one, the result is 1.

^ Logical XOR, (exclusive OR), if one and only one

bit is 1, the result is 1.

~ Logical invert, if the bit is 1, the result is 0, and if

the bit is 0, the result is 1.

The following example program [BITOPS.C] uses several fields that

are combined in each of the ways given above.

#include "stdio.h"

void main( )


char mask;

char number[6];

char and,or,xor,inv,index;

number[0] = 0X00;

number[1] = 0X11;

number[2] = 0X22;

number[3] = 0X44;

number[4] = 0X88;

number[5] = 0XFF;

printf(" nmbr mask and or xor inv\n");

mask = 0X0F;

for (index = 0; index <= 5; index++) {

Page 813-44

and = mask & number[index];

or = mask | number[index];

xor = mask ^ number[index];

inv = ~number[index];

printf("%5x %5x %5x %5x %5x %5x\n", number[index],

mask, and, or, xor, inv);



mask = 0X22;

for (index = 0; index <= 5; index++) {

and = mask & number[index];

or = mask | number[index];

xor = mask ^ number[index];

inv = ~number[index];

printf("%5x %5x %5x %5x %5x %5x\n", number[index],

mask, and, or, xor inv);



The data are in hexadecimal format (it is assumed that you already

know hexadecimal format if you need to use these operations). Run the

program and observe the output.

Shift instructions

The last two operations to be covered in this section are the left shift

and the right shift instructions; their use is illustrated in the next

example program [SHIFTER.C].

#include "stdio.h"

void main( )


int small, big, index, count;

printf(" shift left shift right\n\n");

small = 1;

big = 0x4000;

for(index = 0; index <>

printf("%8d %8x %8d %8x\n", small, small, big, big);

small = small <<>

big = big >> 1;



count = 2;

small = 1;

big = 0x4000;

for(index = 0; index <>

printf("%8d %8x %8d %8x\n", small, small, big, big);

small = small <<>

big = big >> count;



The two operations use the following operators:


>> n Right shift n places.

Once again the operations are carried out and displayed using the

hexadecimal format. The program should be simple for you to

understand on your own as there is no tricky or novel code.

Exercise 7

Using the reference manual of your C compiler, describe any operations

on characters and bits that it provides that are not described in this

section, particularly those that are described in the ctype.h header file.