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Table of Contents
Aclotowledgments
2. Pointers atrd Arrryr ...
Jntrcduc|ion Io Second Edhiot
IDtroductio[ To Poitrters,,,,,....,..;...............,.......-.......1
The & and , Operctors, 2
Poifier Explessions, 4
The Jargon of Pointers, 9
char, int andfloat Poihters, 9
Passing Addrcsses to Functiohs, 12
Fuictions Retuming Pointers, 15
Solyed Problems, 17
Exercise, 30
Pointers and Stings, 142
The const Qualifier, 143
Retumirrg eonst Yaluef, 146
Two Dirtensional A|ray ofCharacters, 147
Array ofPoi\rers to Strings, 149
Lirnitario4 ofArray ofPointers to Strings, I 52
Solved Problems, 154
F,xerclse. 184.
ttv,'
4. PolnterrandStructures ......,..........------------.,.........189
An Array of gructures, 191
Mole about Structures, 192
Structure Poi ters, I94
Offsers ofstructufe Elemehts, 196
Linked Lists, 200
Stacl$ and Queuet, 220
Doubly Lhked Lisk, 2ii
Solved Prcblens, 243
Etercise,270
5. Pohters rlrd Dats Structurcs ...,......,.,..........,,,,.....,,,...,....279
Mergirrg dLinked Lists, 280
t-
llhat are Arrays, 38
Passing Atay Elernents to a Function.
Poihters and Atays, 43
Pqssing a4 E ttirc Affoy to a Function,
TheRealThing,5l
Morc Thqn One Di ension, 52
Poinlers arrd Tfuo Dimetsional Arrays,
Poi tetto sn Atay, 58
Pdssing 2-D Arroy to a Function, 59
Three Dimensiondl Arrays, 62
Passing 3-D Array to a Functiok, 66
Retuming Aftayfrorn Fuictiot , 69
Relumifig 3-D Aftay from a Function,
Array of Poi ters, 82
Dyuhtic Memory Allocation, 84
Solved Problems, 87
Erercise, 130
Pointers and Stiitrgs
,.....................133
Wat are Slrings, 1j3
*
Li ked Lists and Polynonials, 285
Sotting a Lin*ed List, 297
Circalar Linhed List, 323
Trees, 326
Binary Tre*, j28
41
50
3.
Stqhdard Library String Funcrions, I38
vt
Traversal d a Binary Tree, 33 1
Deletionlrom a Binary Tree, 336
Threaded Binary Tree, 344
Graphs,356
Solved Problems, i69
Exercise. 389
6. PointersMisceIlrtry..,,,.....................................................391
File Poi rets, 391
Pointers to Functions, 394
typedef with Furrction Potuters, 397
srgc and srgv - Arguments to mairt( ), 398
Poin ers afid Ydriable Nurnber of Argunen s' 399
rcor,lat and huge Poinlers, 40i
Which Poi en to Use?, 408
Physicol ,4ddrcss to Segne :qsel 4 I I
The Darcing Dolls, 412
Caps Locked!, 413
How Much Memory Do You Have?, 415
Exercise,4l7
7. Appllcrdorr of Pohters '..,....--........................................423
Exploting the Dith 423
Dictioiary, 434
Manoging Database, 439
The Keybod Queue, 459
InJix to Postfa, 462
Evaluation of Postfa Exprcssion, 467
Locatw Duplicote Filelanes, 469
Eoshing 474 i
Function Calls and Stqck. 480
Solved Probtens, V85
I
tnrex
|
493
Preface to Third Edition
In all walks of life anythiDg can be done better the next lime
around. I realised this more emphatically when I completed lhe
third edition oftlds book. Ihus, Ore so-cJlled ,perfect' c'hapters of
the second edition had to be redone for more clarity and simplicity.
For quite some time now I have beeo getting requests fiom
readers that the chapter on .pointers and Oata Structures, be
made more elaborate and exhaustivg. I have completely
overhaule.d this chapter. Now it also includes binary treis ani
graphs,
While reading the fust draft ofthis edition I felt that something
is missin-gin this book. It was only when I was ttuougtr *ittr rtr!
final draft I.realisgd that the missing link (poinrer) wa-s a chapter
on applications of pointers in real-world programming. After
all, no amount of theory is useful unless you cun poi it into
practice. So the final draft became semifinai since I decided to
add this chapter to the book. And now I am presenting you the
edition which I feel is complete in all respects, lili I "-
pmying tlat I won't be required to wrire aflother editiol of this
book, Any author would testiS/ that writing a new edition is
always more palD in the neck than writing a fresh title.
tt
Poihten ca be tuade to work ifyou frddle ,a,ith them lo g
e ough. Ifyou Jiddle with anything long enough you wilt uttimately
fiess it
An Introtuctionto
Qointrs
f f Tbich featue of C do beginners find most difficulr to
VV
understand? The answer is easy: pointers. Other
v V languages have pointeE but few use tbem so freely, with
such abandon, as C does. Aod why rct? Ir is C,s clever use of
pointeN that mak€s it the excellent language it is.
The difficulty begiDners have with pointers has much to do with
C's pointe, terminology than lhe actual concept. For instance.
when a C progmmmer says that a cedain variable is a . pointer..,
what does that mean? It is hard to s€€ how a variable can poinr to
something, or in a certain direction.
2
Understanding Pointers In C
Chapter 1: An Introduction to Pointers
3
It is difficult to get a grip on poioters just by listening to
programmer's jargon, In our discussion of C pointers, therefore,
we will try to avoid this difliculty by explaining them in terms of
simple programming concepts. The fi$t thing vi,e want to do is
e\plain the rationale ofC's pointer notalion.
We oan print this address through 1Le
/- Program 1 */
main0
{
inti=3;
following program:
The& and * Operators
Consider the declamtion,
This declaration tells the C oompiler to:
(a) Reserve space in memory to hold the integer value.
(b) Associate the name i with this memory location.
(c) Srore the value I ar this location.
We may rcpresent i's looation irl the memory by the following
memory map:
i-__}
El--*
6485+
Location Dnme
I-ocation no. (Ad&ess)
Figure l.l
We see that the computer has selected memory Iocation 6485 as
the place to store the value 3. This location number 6485 is not a
number to be relied upon, because some other time drc computer
ult
r\,
may ohoose a different location for storing th€ value 3.
impofiant point is, l's address in memory is a number.
printf ( "\nAddress of i= %u^,
prlntf('hvalue of i = %u", i)
)
l'he output ofthe above program would be:
Addrcss ot i= 6485
Value of i= 3
&i) ;
inti=3;
l,ook at the fi$t printf( ) statement carefully. The .&, opeBtor
[sod in this statement is C's 'address of operator. The expression
&i retums the address ofthe vagi4llglwhich in this case happens
lo be 6485
pointer operator avaikble in C is '*'. called 'value at
erator. It retums the vaiiE stored at a particdlalEddfess.
also called an 'indirection'
at address' operator is
l,hlene carefully the output olthe following programj
/' Program 2 '/
nrain0
I
inti=3;
printf ( "\nAddress of i = %u', &i) ;
%d-, i) ;
%d",.(&i));
printf ( \nvalue ofr =
other
ess o
e'value
orator.
printi ("\nvalue of i=
)
Ttre
IJnderstanding Poinlers In C
Chapter I : An Introduction to pointers
4
5
The output ofthe above prog€m would be:
Addressoli=6485
Valueofi=3
Value oli = 3
Note that pdnting the value of *( &i ) is
ofi.
As you can see, i,s value is 3 andj,s value is i,s address.
"l:1^1"]! ". :an'r use j jn a program wilhout dectaring it. And
8rn.e j rs a \arjable. which conlains lhe addre., of i, it is-aeilurea-
0r.
lnl 'j ;
'lltis declaralion tells the compiler that I will be used lo store the
rddress ofan integer value _ in o*re. *Jro.; poino ,o _ iri"g;;
Ilow do wejustify tlie usage of * in the declaiation,
lnt'j;
1;d ,t g:-Py the meaning of +. It stands for ,value at address,,
I hus,.irt *J would mean, the value at the addre* ""","rn"ji, ij,
same as printing the value
Pointer Expressions
Let us now see what are pointers and how they can be used in
various expresJions. We hav9 seen in the previous section that the
expression &i retums the address ofi. If we so desire, thig address
can be collected in a variable by saying,
But rcmembcr that J is not an ordinary variable like any other
int€ger variable. lt is a variable, which contains the address of
llcre is a program that dcmonshates the relationships we have
t ccn discussing.
another variable (t in this case)
Since J is a variable the compiler must provide it space in mernory'
Once again, the following mcmory map would illusts?te the
contents of I and J.
E'.-l
6485
3276
Figure 1.2
j=&i;
I Program 3./
maino
(
inl i= 3;
int 'j ;
j=&i;
pdntt ( '\nAddE$ of i = %u', di ) :
printf ('\nAddEss of i = %u'. i ) :
Printf ( '\nAddress ofj = o/ou', ai) ;
pdntt("\nvalueotj=%d'. i ) :'
pdntf ( 'hvatue of i = %d", i i :
pdntf ( 'hvalue of j = o/od,. .{ &i ) I .
Printf ('\nvalueof i = %d' l) "'
6
(Jnderstanding Pointers In C
Chapter 1 : An Introdaction to pointers
7
The output ofthc above program would be:
AddEssoli=6485
Addressofi=6485
AddEssofl=3276
Value ofl = 6485
Valu6 ofl = 3
Valueoli=3
Valu6 of i= 3
Wo* thrcugh thc above program caretulln taking hclp of the
memory locations of t and J shom ea isr. This program
summarires everything we have discusscd so far. If you don't
understand thc'program's output, or the meoning of ihe
oxpregsions &1, &J, rl ard *( &l ), re'read the last few pages.
Evcidring wc say about poilters ftom hcr€ onwards will dcpcNd
on your understanding ofthcse exprcssions thoroughly.
Look at the following dcctarations,
lnt'alpha;
ohar 'ch ;
lloat 's :
Hcrc, rlphr, ch and I ale declarcd as pointer variablel, i.c'
variableicapable of holding addtesses. RfiIember that, addrgsses
(location nos,) ar€ always going to be whole oumbe!$, thorcfore
pointers ahvays contain whole numbers. Now we call put these two
iacts togethir and say-poiotels are variables that contain
since addrcsses ai;;lwaysE;G;mBEiies
!way
would always co!!Bi!
concept of pointer caD be further extended. pointer we know is a
vadable, which contairc address of another variaUte, Niw thii
variable itself could be another pointer. Thus, we now have a
pointer., whicb.contains another pointer.s address. The following
example should make this point clear.
f Prcgram 4'l
maino
{
iiqt i= 3;
int'j:
lnt nt;
j= &;
k=&j;
pinf ('bAddr€ssof I = %u., & ) ;
Ptinu ( trAddcss oti= oi6u', j ):
pdntf ( lnAddEss of i = %u., ik ) :
Pdntf ( '\nAddrsss ofj = %u', &j i ;
pdnf ( lnAddrEss ofj= %u', k ):
prinf ( tAddrBss ofl( = %u', &k ) ;
prinf ( lnhvdus ofj = %u', j);
pdnf ( Walus of k r %u', k );
printf ( '\nvalue oti = %d', t)j
pdntf( lnvaluo ofi =/od',.(&j))j
pdnf (hvalue of i = %d", J );
pdnf( lnvalueofi = %d',,* );
The declaration llgat *Ejqes not mean lhat s is going [o cooBln a
fl oati-ng-point valqq. @
--aterisi-oIa'hoating-point value. Similarly, char *ch means that
ch is going to oontiin the address of a char value. Or irr other
words, the value at ad&ess stored in ch is going to be a ch'r' The
The output ofthe above program would be:
Addressoti=6485
Addre6sori=6485
AddEssofi=6485
AddEssofj=3276'
8
Understanding Pointers In C
Address ofj= 3276
Addressolk=7234
Value ofj= 6485
Valueofk=3276
Valueoli=3
Valueofi=3
Value of i= 3
Valu6ofi=3
The following memory tDop would help you ir tmcing out how the
prcgram prints dle above outpul.
Figurc 1.3
Obs€rvc how the variables l, J and k have been declarcd,
inti;
lnl -l
i
int -k i
this, one mrely requires to
just in case...
extend the definition of a Pointer' But
H€re, I is an ordinary ldt, I is a pointer to ar lnt, whsreas k is a
poiflrer to a pointer. ln pdnciple, there could bG a pointer to a
;ointer to a pointer, or a pointer to a pointer to a pointer to a
pointer. There is no limit on how far can we go on extending this
a€finition. Possibly, titl the point we can csrFehetrd it' And that
point of oomprehensiotr is usually a pointe, to a pointer. Beyotd
char, int zad.Jlod, Pointers
Considei the followiog program:
f
