Yashvant Kanetkar

Why C? .) C

is simple

.) C

is fast

+ C is small

+ C offers better interaction wlth hardware

Historical Development of C 1960 International Committee

Mach~neOriented , Ex. Assembly,

KlClT I

C I Lecture 1

Ex. Basic, Fortran, Pascal,Cobot

I

In The Beqinnina

...

Alphabets, Dlglts

Symbols Dlgltr, SP. Alphabets,

Words, Numbers

Constants. Variables,

Alphabets, Diaits,

-

Constants and Variables

KlClT I C i Lecture I

Character1

1

String

1I

Structure Union Enurn etc.

i

lnteqer Constants

.

Rules:

/-

1. No decimal point.

1

a

ic

2. May be +vs or -ve. Default: +vs

3. No comma or

4

*

spaces 4. Valid Range: -32768 to +32767

Real Constants EX. 427.62

+24.297

.

-0.00254

Rules: 1. Must contain a decimal point. 2. May be +ve or -ve. Default: +ve

3. No comma or spaces. 4. Valid Range: - 3 . 4 ~t030 to +3.4 x

KlClT I C 1 Lecture 1

I

]

I

V

+

m

@&=J

many types? As many as the types of constants.

HOW

Is it necessary to identify types?

7 Memory y=4

z=x+y print z

z

+ An entity whose value can change .) A name given to a location in

memory

KlClT I C I Lecture 1

1 I

--

Integer Constant

3.0 Real Constant $3' ---t Character Constant a -7 b - ? c + ?

C's Wav of ldentifyinq Variables int a a=3

float b b = 3.0

char c c = '3'

$h pt.3 @*tdbin$h t

Rules for Buildinq Variable Names First character must be an alphabet, rest can be . @ h a t s , or underscores.

@a,

Ex, pop98

a

r

si-int

sl-int

Length <= 8 (Us ally)

Usl-~nt

m

No commas or spaces

I-p

Variable names are Ex. abc ABC Abc aBc AbC

KICIT I C I Lecture 1

C Keywords .)

How many? What are Keywords?

I)

What are Reserved words?

Would This Work? integer a El ink '

1

real b

11db

character c

~ h &c

And How About This... int f l o a e float = 3

char float = 3.1

Where Uo We Stand?

P' 91

Alphabets, Digits, Sp. Symbols

~ o n s t m t sVariables. , Keywords

51

Statements or Instructions

KlClT I C I Lecture I

1

The First C Proclram

float p, r, si

int n p = 4000.50 n=3

r=15.5 sl=p*n*rll00

Tip: All variables must be declared.

1

float p, r, sI int n p = 1000.50 n=3

r=15.5 sl=p'n+r1100 printf ( "%fl, si )

KlClT I C I Lecture I

*fim

Terminoloay Matters ()

Parentheses

{ ) Braces [ 1 Brackets

General Form of ~ r i n t f ( printf ( "format string", list of variables )

'I b can . contain , <+ -

,

-

5

%i integer

4,L

-

i.?!

-

%c character

%f float

Ex, prlntf ( "%f %I %f %fl,p, n, r, si )

) tlw

awd ulric,bb in mdQ/r d dC Q I & & ~ -

I I amStatement Terminators p, a boy I go to school

noat int n

r, si

-

float p, r, si ; int n ;

c t a t y ~t erminato)

: colon

; semicolon

KlCIT I C I Lecture 1

Free borm Lanauaae float p, r, si ; int n ; is same as

9 Q

floatp,r,si; intn;

What I o Execute ,-main(

n=3: r=15.5; si=p+n'rIlOO: printf "%fl, si ) ;

maln( )

e{

float p, r, si ; Int n ; . p=IOOO.SO;n = 3 ; r = 1 5 . 5 ;

si=p*n'rllOO; prlntf "OhP', si ) ;

I!)

KlClf I C I Lecture I

I' Calculation of slrnpltr Interest ' I main( ) { float p, r, si ; Int n ; p=1000.50:n=3;r=15.5; si=p'n'rllOO; printf ( ""/of', si ) ;

& &'

1

Any number of comments anywhere I' formula ' 1 si = p

•

Multiline comments

......*I

n * r 1100 ; I*

-

r .............................

............................... 'I

Nested comments

-

r ..............I* .......' I ........' I

A I' Calculation of simple interest *I main( ) { float p, r, sl ; int n ; prtntf ( "Enter values of p, n and r scanf j "%f %i%t", sl=p'n*r1100; printf "%f", sl ) ;

) ,

1

KlClT I C I Lecture I

");

1

Forms of Real Constants

-

427.62 ----+ 4.2762E2 +24.295 -4 2.4295~1 -0.00254 -2.54e.3

Fractional Form

Exponential Form

==a -3

-.,

CI

4.2762: E&T -

->

Exponent

Mantissa

Character Constants '3'

'm'

Rule : A single character enclosed

within a pair of '

'.

Are Thev OK?

'Nagpu-

I>cj

C Variables

Integer

Polnter

Character

Strlng Structure Unlon ' Enum etc.

KlClT I C I Lecture I

1

.k Turbo C k QuickC

* Microsoft C j,

Artech C

Jr Zortech C tattlce C .rt Watcom C 3r GreenLeaf C

*

* Vitamin C

1 J

l ntesrated Develooment Environment --cde~k?m n

r*

4p

Editor and Com~iler Edltor

-

Helps In typing I edltlng of a program

* Compiler - Converts C language program to machine language program

KlClT I C I Lecture 2

ommands

Editin Cursor Movement

Daletlon

+ Up Arrow 4 Down Arrow + Rlght Arrow +Len Arrow

Backspace Ctrl+T CtrltY

Home

End

PgUp

PgDn

Computers are fun

b o b

*+ ir\.ln\ Some ~ hrub. k More Commands

~lsinhciid

-6'

;?$ Perm

* '

Save As Exit

' Do* Shell

Tempora

.-]

CtrltFg Compile AlttF5 OutputAIi +F3 CIosm wlndow

RLt t X

e~i*

Tlp: Use rneanlngful filenames. Ex. CHtPR1.C

1

lnterchanclina Contents of Two ~ariab1.s maln( )

I

h

~ h e + h f m d+&SW

Int c.d: . .

prtnti ( "Enter values of c and d" ) ;

KICIT I C I Lecture 2

d

I

(9j

~P,D/

~wmh ,

f i b o b +Fvr~v q 1i ~ h t

lnterchanaina Contents of Two Variables e mar( )

I

Inlc,d,t; printf ( "Enter values of c and d" 1 ;

rcsntl "XI%I",LC,&d ) ; t-c;

C

~1~~

dwt:

c

1

One More Way c=c+d; d=c-d; c=c.d;

:

Sum of Diaits main( ) Int n ; printf ( "Enter a flve dlnlt no." ) ;

.....0.

I

KIClT I C I Lecture 2

d

t

d

t

mmm

.-

prlnU( "XI XI". c, d )

t

BBLl c d t

c-d;

prlntf ( "XI Xi", c. d ) ;

d

Calculations

: 2691 - Division I - Modular division 2691 3 % 10 1 7 3 26913 1 10

I O ~ Z 6 ~ 1 3 / 2 6 9 ~

The Whole Picture main( )

n

s =dl +dZ+d3+d4+d5; printf ( "%I", s ) ;

0"-'

&*

i modulus i R e a v Useful *

*

Leap year or not Odd 1 Even Prime or not -s

KiClT I C I Lecture 2

diw&bk by Ior ib$.

1

Where Are We .... Alphabets, Digits, Sp.Symbols

1Constants, Variables, Keyword& I

J


C Statements I Instructiong

*

Type Declaratlon lnstructlon Ex. Int i, 1, k : float a, b, c ; char ch ;

Tvae Declaration, A few Subtleties Int a ; a-5;

1s same as

2

int a = 5 ;

*inta=S,b=lO, c=a+ba5Y*2: Here order Is Important

KlClT I C I Lecture 2

Is This Ok? It. tnt a, b, e, d ;

a=b=c=d=5;

&??I

C~3tatementsllnstructions k Type Declaration Instructions

* Arithmetic Instructions Ex.

s=dl+d2*d3;

si=j*n"r1100; c=5.019'f-32; +, -,', I , %

-

Arlthrnetic Operators For exponentlatlon Use pow( )

-

Arithmetic Instructions. A Few Small Issues

KlClT I C I Lecture 2

T V Dof~ Arithmetic lnstruction~ +

Integer mode Al

* RmtmoQo Al

+ Mixad modo AJ Ex. i n t m = 3 , b * 4 , c ; c-agb+5%8-b+l4;

Leaal Arithmetic Operations 1

Onerandl float

I

Omrand2 float

Tty This Int a ; a=512; ao5.012; 4 Q a=512.0; - 2 a = 5.0 12.0 ; + ar215; -0 a82.015; --*D a 2 1 5.0 ; ---+ @ a = 2.0 I 5.0 ; L 0

2

KlClT I C I Lecture 2

1

Result float

I

Try This float a ;

float c, f

= 21 2.0 ;

KlClT I C I Lecture 2

r" G % P - @ b ~ - ihn t

vanicJda

C Instructions

* Type Declaration Instructions

*

Arithmetic lnstrudlonr

* Input I Output Instructions printf() scanf( )

- Output - Input

General Form of prinffLS prlntf ( "lormat string", list of rarlabler ) ; prlntf I "Enter values of c and d" ) ; List d varlab4es is optional.

4

printf ( "%I %i %i",a, 3 9 2 + 8 % 3 ) ; -/ ~l~tcancontalnvarlables,cons~nts or expressions.

printi( ) prlntf ("format atrlr~g",llrt of varlmbl~s) ;

a

KlClT I C I Lecture 2

I

1

Numberina Svstems Declrnal Octal Hsxadsclmal (0.S)

(0.7)

0

0 I

t

(0.W.A-F) 0

Binary (0.1) 0

I

1

10

1Wo

IW

1W

...

FF 100

....

Conversr on^ Oeel 473

4~1O2+7*10~+3*IOP~473

krrr 11 d e c l m a L l * ~ g ~ + l ~ l ~=~17 !

Bloaryl? - $ * 2 1 + 1 ' 2 0

\L*

.

\*a' + \ A % '

More Conversions

It=~ a i n d P / v .

KlClT I C I Lecture 2

=3

q

-.-

q0d = &JpJmd P ~ ~ v ~ v R L P ~ % O =& & q w l i v M

Xr=

gB y

~

~

~

-

\n Newllne Irr \t -Tab $r prlntf ( '%d\n%d\t%dt', 10,20,30 )

:

v Other Characters

* prlntf ( ''Enter valuer of c and d" ) ; It prlntf ( "Hello" ) ; g p r l n " ( "\nSlmple Interest rn n \

KlClT I C I Lecture 2

-

~

- hkaq.bhd

p r ~ n t f l Makes ) It Handy

$r

~

Rs. %f', $1) ;

-3

-

e

d

t

u

r

u

I

.*scanf ( "format string", list of variables j ; scanf ( "%d %f %c", &l,&a, bch ) ;

*

C Instructions

KlClT I C I Lecture 2

+ Type Declaratlon lnstructlons + Arithmetic lnstructlons + Input 1 Output Instructions

* Control Instructions

Control Instructions + +

What are they? Control the sequence of execution of lnstructtons What different types? Sequence Decision ~e~etition(hp> Case

-

-

int .....; float ...... ;

...... ......

prlrltf ( ); scanf ( ); a = .... ; b = ....; prlntf (: .......);

1

: CI Is the default CI I I 1 ~ i pSequence

KlClT I C I Lecture 3

I*Calculatlon of total expenses *I main( )

{ lnt qty ; float prlce ; printf ( "Enter quantity and prfce" ) ; scanf ( "%d%f", &qty, &price ) ;

I' Calculatlon of total expenses ' I

main( ) {

lnt qty ; float price ; Int dis ; float iotexp ;

dis = 0 ;

-

totexp = qty ' price qty * price ' dis I100 ; printf ( "Total expenses = Rs %f", totexp 1 ;

1

Tips. + if, else Keywords + General form: if ( condltlon )

statement1 ; else statement2 ; + Cond. is usuallv built uslng <,

a==-

2

KlClT I C I Lecture 3

>, c=, >=, = =, I=

(@ Would This Work? Int a = 5, b = 4, c, d ; Jprlntf ( "Xd", a + b ) ; / J prlntf ( "%d", a r = b ) :& J

/~=a'b; j ' p r l n t f ( "Y,d", c j

Tip:

-

. )

. & .----

:

Condition -True - Replaced by 1 Condition False Replaced by 0

-

-

Int a ; printf ( "%d", a ) Tip: Unless specifleally initialised a variable contains abarbage value.

r

Calculation of Total Expenses *I

main( ) {

Int qty ; float price ; int dls = 0 ; float totexp ;

If qty >= 1000) dis= 1 0 ; totexp = qty " price qty price * dls 1 100 ; prlntf ( "Total expenses = Rs %f",totexp ) ;

-

1 Tip: else block is optional

KICIT 1 C I Lecture 3

( - -' float bs, hra, ca, da, 0s ; prlntl( "Entrr bask walary" ; *cant ( "W',&bs ) ;

ca=bs*IZ1100; else

-

If block

hn-br'lSI100;

ea 200 ; ga bs + hta da + printf ( "Grolr salary

balongs to

= Ra. sf',~r ) :

k

t bm. hta, ca, da, gs ;

-

dam

pdntli "Entar bmlc salary" ) : mcanf (*%r',&bs ) :

6 )

...;bra= ...; c a m . ..;

else

a

'

Y )

dam

... ;bra- ... ; c # - , , , ;

gs=bs+d*+hra+ca;

prlntl( "Grorr firlary

Rs Xf", gs )

1

:

One More Form If ( condllbn ) I

\

ITlp: - Dafault scopa of If and o I m Ia'only the next strtmrnent after tham.

-To ovorrlde *I#dotaust scope use { )

KlClT I C I Lecture 3

w

Leap Year or Nof

Int y : prlntf ( "Enter year" ) ; scanf ( "%d", By J ;

Laap

Leap

rna\n( ) (

Nwted If-else

... !f(y~.~oa--aj

lf(y%40Om-0)

(

prlntf ( "Lmap" 1 ;

11*1

prlntt ( "Not l e a p "

1

69

olio prlnff ( "Not Leap* ) ;

1

First Day of Anv Year main(

1

(

Int y ; prlntf ( "Enter year" ) ;

scanf ( "%d", &y )

.. .

KlClT I C I Lecture 3

;ax>

11111

dMonday

8t $11

<' Monday

151111

C~Manday

221 11 1

Monday

281 11 1

Mandry

... ... ...

1111 1 to3111Zl 1887

i ~ i t i v ~C@ e

scanf ( "Xd". &y ) : normaldays = ( y 1 ) 365 ; leapd~ys-(y-1)14-(y-1)/100+(y-1)/400;

30

totaldays = normaldays + leapdayr firstday * toUldays % 7 : If ( firstday 0) prlntf ( "Monday" ) ;

--

4

%-OAoo+'&

7 $7

't-C%-\))~oo 3

h.

~00iod t

2000,

cy-Q&

c w d i n q T-otd no. qj d&,

main( ) {

tnt y, firstday, leapdays ;

Long Int normaldays, totaldays ; ,, ,,

-

normaldays = ( y I) '365 ; ,, ,,

1 -32768 to +32767

long int -2147483648 to +2147483647

KlClT I C I Lecture 3

a

a

I

What's Wrona Here'r

-

normaldays = ( y I) ' 365 ;

c? float a ;

; float a ;

-

Soln: normaldays = ( y Soln: normaldays = ( y 11 ) 365 ; 4 -Soln: long Int y :

l

The sizeoft ) O~erator

six O-A u (J

int i ; long int j ; prlntf ( "%d % d l sizeof ( I ), sizeof ( j ) ) ;

-L\w

WU~U bib

4

rwm9

w&&J

Keyword

Are You Sure? 4 % 3 - r n

,

3 % -3 % 3 % -3

u

4 4 4 -

L 4

ip: Sign of remainder Is me as slgn of numerator

2 3

4ea

int a = 5 , b = S , c = S ; i((a==t]==c) prlntf ( -Hello':)

;

KlClT I C I Lecture 3

;

e h

cOLei~ be.

.

What Will Be The O u t ~ u t

I

-F\O

rn

'U Joe

WI

-ad ? v r n rkm'b & !- p h d d J

b=30; printf ( "%dm,b ) ;

nh.

I

What Will Be The O u t ~ u t main( )

I

Tip : Truth In C is nonzero, whereas,falslty Is zero.

int a = 10 ; if ( a = 5 ) printf ( "Hi" ) ;

6 9

if(a=5)+

if(=)+

if(5)

main()

:--

Int m l , m2,m3, m4,rn5 ; lnt per ; prlntf ( "Enter marks In five subjects" ) ; scanf ( "\n%dY.d%d%d%dn, &ml, &m2, &m3, &m4, &mS ) per=(ml +m2tm3+m4+m5)15; lf(per>=60) prlntf ( "Flrst divislon" ) ;

rrse

I f ( per>= 5 0 ) prlntf ( "Second dlvlslon" ) ;

;Ise

if ( per >= 40 ) printf { "Third dlvislon" ) ; else printf { "Fall" ) :

1 )

KlClT I C I Lecture 4

sdo4em.d.

Lo&

Ak "d

main( J

I Int rnl, m2,m3, m4. m5,per ; prlntf ( "Enler marks In five mubjeets") ; scant ( "%dY.d%dY*d%d", Lml, &m2, am3. bml, h m 5 ) ; p#ra(mZ+mZ * m 3 + m 4 + m 5 ) / 5 ;

If(per*-W ) prlntf ( " F l n t dlvlslon" ) ; If ( p a r > - 5 0 ) prlntf ( "Second dtvlslon" ) ; prlnti ( "Thlrd dlvlslon" 1 :

prlntf ( "Fail" ) :

I----

Inlrnl. m2, rn3, m4, m5, per ;

prlnti ("Enter mark* In five subjects'' ) : scanf ( "*hdXd%d%dY.d, L m l , Bm2, Lm3. &m4. am5 ) ; prr=(ml tml * m 3 + m 4 * m 5 ) / 5 ; If (per >= 60 ) prlnll( "Flrwt dlvlrlon" ) ; l l ( per 50 printt ( "Second dlvlaion" ) ; If ( pat ?= 40 && per < 50 ) prlntf 1 "Thlrd dlvlrion" ) ;

*=

111m

prlntf ( "Fall" ) i

1

(

c;>

Int m i , m2,rn3, m4, m5,per ; printf ( "Enter marks in five cuh~ods" );

runt ( "Xd%d%d%d%d",&mi,&m2,am3, &ml,LmS ) ; per=(ml+mZ +m3+rn4+m5)/5; if ( per prhtf >= 60F i )r s t d~vlslon"1 ; if ( per >= 50 a& per c 60 ) printf ( "Second division" ) ; I f ( per - 4 0 L& per c 5 0 ) prlntf "Thlrd dlvlslon" ) : If ( per r 40 ) printf ( "Fall" ) ;

1

KlClT I C I Lecture 4

Is This

Loqical Operators

Useful in

j

'iP

C'

'

checking? Ranges

/

;h

I Problem

prlnn ( "Enter age. sex. marital status" ) ; scanf ( "%d%t%c", Bags, as, f ( ms = = ' m ' ) prlntt ( "Insured ) ;

L

gl...,m~,

else printf ( "Not In

if ( age r 30 1 -pint! ( "lnsured" ) ; else .. 1

printf ( "Not I n s u r e d )

1

;Ir

I

if age > 25 ) prinlf ( " l n s u r e d ) ;

Int age ;

chars, ms ; prlntf ( "Enter age, sex, marital status" scanf ( "%d%c%c", &a e, as, &ins ) if(Ims9.'m'lll(ms 11 [me= = 'u' &&

B

s = 'f

prlntf ( "Insured") ;

Q

else prlntf ( " ~ o itn s u r e d ) ;

KlClT I

)XI

C I Lecture 4

88 ape > 25 ))

QL mh o m ecltEhtw ~t b ~ b i & 11534 +3bt+k b b b e b .

Workinq of && And

False ,

i

True False 6

False False True

I

False True True

False

,

False False

hula3

0

0

0

0

1

1 I

0

0 0

U

OR

0 \

I

I

I

Arlthmellc Arithmsllc

+c

T

=.

C-1

<= >=

Relational

[=

Exchancinq Blocks rnaln( )

main( )

I inta=3, b = 4 ;

inta=3,b=4;

if ( a < = b ) printf ( "A" ) ;

if(arb) printf ( "B" ) ; else printf ( "A" ) ;

else printf ( "0" j ;

1

1 Output:

A

KlClT I C I Lecture 4

Output:

A

main( )

main( )

ff(a<=b) printf ( "A" ) ;

i f ( l a > b) printf ( "A" ) :

else

else

I

:.

net&=

p

pr~ntf( "B" ) ;

printf ( "B" ) ;

1

output:

A

output:

:.ButptCt 0 .

anta=3,b=4;

lf(lIa>b)) printf ( " A ) ; else prinif ( "B" ) ;

1 output:

A

Yet Another Way main( )

main( )

I inta=3,bE4; it(!(ac=b)) printf ( "0") ;

Inta=3, b = 4 ; if ( a > b ) printt ("B");

else

else printf ( "A"

1 Output:

1;

fl

KlClT / C I Lecture 4

printf ( "A" 1 ;

1 Output:

4

What Would be The O u t ~ u t main( )

I Int a

-

3, b

= - 4, c ,d ;

cwla; dmlb; printf ( "%d %d %d %d", c, d , a, b ) ;

main( )

I

Int a

= 3, b = 4, c :

printf ( "%d %d", b, c )

1

e m : L ~ o l U e wdJ

main( {

Int a = 3, b = 4, c ; c=!all(b=7); printf ( "%d %d", b, c ) ;

1

KlClT I C I Lecture 4

Yet Another Way

I

inta=3,b=4;

if(a<=b) prlntf ( "A" ) ; else printf ( " B ) ;

Output:

A

1

..

..

a *r b 7 prlnff ( "A" ) : pflntf ( "6" 1 ;

HOW

would

If ( a < = b ) prlntf ( "A" ) ;

I

YOU

Convert

a c= b 7 printf ( " A ) ;

+I

Remove The Error { inta=3, b = 4 ; a <= b 7 printf ( "A" ) :

1

KlClT I C I Lecture 4

-

I

a <= b 7 printf ( " A " ) : '.

1

I Terminator] 3 S0ln: a c= b 3 printf,( "A" ) : printf ( " " 1 ;

aabtgnmd

< - b ? p r i n l l [ l a ~ ) (: . = a ) ;

bnO

+ Use one, use another 4 Dummy

statement

main( ) Int a = 3, b ; ~1<~5?(b=IO):(b=20); prlntf ( "%dl1,b ) ;

1

r prlntf ( "%do',a <= 5 7 10 : 20 ) ;

KlClT I C I Lecture 4

0

fl

int a = 1 0 , b = 2 0 ; float c = 3.14, d = 6.28 ; printf "%d %d %f %f ",a, b, c, d ) ; print! ( "%d %d %f %d ", a, b, c, d ) ; printf ( " " / ~%d f %f %f ",a, b, c, d 1 ;

CIS^ 7 8 1 h D , Q O O / -i.g2?e ~

Conversions Continued ... main( ) {

int i = 65, j = 90 ; char ch = 'A', dh = 'Z'; printf ( "%d %d %c %c", i, j, ch, dh ) ;

printf ( "%c %c", i, j ) ;

printf ( "%d %d", ch, dh ) ;

1

int 1 = 65 ;

Re~resentation float a = 3.14 char ch = 'A'

7 ii I

B ~ n a r yEqv.

KlClT I

Binary Eqv.

C I Lecture 5

ASCII code

5

ASCII Codes

\

I

inary Is Difficult .-

I

I ' :

Character

Decimal

A

Of 0000D1

B

01000010

C

01000011

D

i E

I

I

I

I

I

"

01000100 01000~01 ""

68 '

6i

I

..

Methods Are Different char ch = 'A' = 65 ;

int I

pr~ntf( "%cT7.i ) ; 4 printf ( " % d U , ch ) ;

KlClT I C I Lecture 5

14

+ 65

;

I

Characters to be Kenreranted

.-

+ 26 +2

+fO + 32 + 34

I

Capital letters ( A- Z ) Srn.lh.*. 1.11.~ ( 1) cb Digits ( 0 - 9 ) Spwlai Symbol8 ( + 1 . : ; etc. ] Non-prhkblr cham control chars Graphlc charactem

-

+ 128

256 00000000 To l l t l l l l l

€9+ b b ,,kt h r n( e . ~b l e .) ,,A p m b a ehcye

What About More Than 255

Whv Unreliable Conversions float I = 3.14 ; printf ( "%em,i ) ; prlntf ( "Kd", i ) ;

i / ~ ~ a n d l s ~ ~ s m 0 nd md l lsa l~0 m a n d l a

Int I = 35 ; prinlf ( "Wf". I ) ;

i 0sandI1 I ~ s n n d t s l

KlClT I C I Lecture 5

I

1

Sur~rised?

rain' '

float a = 0.7 ; if ( a c 0.7 ) printf ( "A" ) ; else printf { "B" ) ;

1

= 04d=A .

I

Aanearances Are Misleadinq tloat a = 0.7 ; printf ( " X d K d , slzeof ( a ), stteof ( 0,7 ) ) ;

1

Binary of A Float

Truncated after 64-blt

KlClT I C I Lecture 5

I

Local vls S l U vls 1SU Calls i

printf ( "in I am In main" 1 ;

CdlD c & )

-+K q c n u I )

mdm

---3 Communication

(n-oem

main( )

(

~ l s c&) a

* ,b&

Int a = 10, b = 20, c = 30 ; Int s ; calsum ( ) ; printf ( "%dU, s ) :

>calsurn( I )

<

L

v

Passinq Values main( )

I

1

calsum ( int x, Int y, int z ) int

I

W

&w&@I iM

lnt a, b, c, r ; s=a+b+c: prlntf ( "Kd", s ) ;

(

I

s,

Arguments

prlntf ( YO^", s ) ; I

KlClT I C I Lecture 8

d o w t z & h &ky)

a

=-

~

~

Returninq Values r = calsum ( a, b, c ) ; prinv ( "%dm,r )

9

calsum ( lnt x, Int y, Int

t:

Int ss ; ss=x+y*x; return ( ss ) ;

z

)

return ( 80 ) ;

---= 3 Are These Calls OK? calsum ( a, 25, d ) ;. calsum ( ID

+ 2 , 2 5 x 3, d 1 ;

calsum ( a, calsum ( 25, 10,4 ), d ) ;

.

int ss :

u=x+y+z; return ( 8s 1 ;

1 Nested calls are legal. Call wlthln 417 expression Is legal.

Returninq More Than 1 Value main( )

Ivalue at a tlme

I

Int a = 1 0 1 b = 2 0 . c - 3 0 ; int 8 , p : 3 s, p = sumprod { a, b, c ) ; prlnn ( ll%d%dy*.r. p ) ;

1

sumprod ( Int

I

*, int y, Int x )

Int

IS,pp ; ss=x+y+z; pp'x'y'z: return I ss,PP 1

;x

KlClT I C I Lecture 8

8

sqaw&

/&clmati~r

=\me

>

1

calsum ( int x, int y. int z )

I

j 5 ( = . ~

Ye25 ~ = d , y=n5%? r = d , ) = vahQ 2-r

9c&~su~~'[~s,~~,Q)

'

main( ) {

-

One More Try

-

10, b = 20. c 30 : int s. p : I sumprod ( a, b, c ) ; p sumprod ( a, b, c ) ; prlnn ( "%d~/.d",5, p ) ;

Int a

-

1

rumptod ( lnt x, Int y, lnt z ) {

int ss,pp ;

::':':':.-

,.

return ( ss 1 ; turn ( PP ;

4

durn0-b

hrr, (~1;i); $g

c!mBPh

)

The Only Wav Out maln( )

s sumprod ( a, b. c, I ) : printf { "%d%dn, s, p ) ;

n

1

sumprod ( Int w, Int y, Int z. Int code ) {

roturn ( IS ) ; else return ( pp ) ;

time

,=

,I

hw ~r)be

&-'d m

&t&.

1

II

A Better Wav

;urnprod ( lnt a, int y, Int z, Int code) -

int IS,pp ; pp-x'y'r;

return ( pp ;

KICIT I C I Lecture 8

I

%

i

d

\

q

no.

4 -J?4

1;

calsum ( int x, int y, int r )

I

N ~ H O V O JS k m d ~ d

NSI VIS K & R ,.

q

-

calsurn ( x, y, z )

Alvl~-tm

l

Kornighan B.

I Int x, y, z ;

I ) '

I

Int y ; prlntf ( "Enter year" ) ; scanf ( "Y.d", &y ) ; romanize ( y ) ;

} romanlze ( Int yy J

int n. i ; n n yy11000: tor(I.1 ; I < = n : l + + ) prlntf ( "m" ) :

1

A More General Gall maln( )

I

Int y ; prlntf ( "Enter year" ) ; scanf ( "%d", By ) ; y = romanlse ( y, 1000, 'rn' ) ;

1

romanire ( lnt yy, lnt j , char ch) ( int n, i ;

n=yylI; lor(l=l ;Ic=n;I++) prlnlf ( "%c", ch ) return(yy%]);

1

KlClT I C I Lecture 9

:

...

y = romanlse ( y, 1000, 'm' ) ; ,-2 y = romanlse ( y, 500. 'dl ) ; _3 y = romanlse ( y, 1W. 'c' ) ,;

y=romanlro(y,50,r1w): y=mrnsnlsa(y,~~,~x~)~-

,

%03

Int n, I ; n-yytj ; for(Ia1 ;I<mn;i++) prlnff ( "xc", ch ) ;

Advanced Features of Functions 4 Returning a non-integer value

+

Call by value I Call by reference

+

Recursion

main( )

float a, b, c ;%square (float) ; a=squan(2.0); Function b = square ( 2.5 ) ; Prototype c-squam ( 1 . 5 ) : printl ( " Sf 161 W ", a, b, c, 1 :

C

2.0

2.5 \ - 5 + a * D&\aa

(bmd b

C-

d*OF r.34 2 9 s

oat square ( float x )

I

flaat y : yax-x;

bcl Jalnll~P- n

9,mn

\ ~ ! w ~ n ~ f i r

float y ; Y RSqUaro [ a ) ; prlntf ( "%1". y ) ;

float square ( float x )

What Would Be The Output mrln( ) (

Int a = 10 ; prinn ( *ma = %don " Ia ) ; a=fO; prlntl ( "\n a = % d , I1 ;

.

1 f(

-

J

{

9 d y

printf ( "Hello I" ) ;

1

Int r

10 ; printl ( k l \ ~ a = % d o . , a J'O );

f0;

prlntf ("\n am Y.d", a ) :

A \\3 Returned value is Ignored.

prlntf ( "Hello I" ) ;

KlClT I C I Lecture 9

olution main( )

I

Int a = l O ; prlntf ( 'Ina = X d " ,a ) ; f( 1 ; prlntf ( " h a = % d " , a ) ;

Advanced Features of functions

+ Returning a non-integer v a l u e d + Call by Value I Call by Reference + Recursion

prlntf ( " \n address of 1 a %d" , &I ) ; printf ("ktvalueof I = ' A d " , ' ( & I ) ) ;

KICIT I C I Lecture 9

-x

Would lhls Work ... Int I = 10 : j=&i;. j= jEg(i434);

/--

a m : ----

wad

LO&

prlntf ( "%do, 'j ) ;

variable, constant

"

lnt I 10 ; Int 'j ; Int '*k ; 4 B BOtO 9 1 2 prlntf ( " \ n v 3 l u a o f I = % d " . 1 1 : v n o~ ~ p 1..

I

. . .. .

-. * ( a t ) ;V )m$ 1

prinni l * i n v a ~ u c o f i =YO^

7n

1 1

Jmal;-tadddish&hj.

~ H n f f ( " h a d d r e a r o f I = X d " . 8 1 : .,AAL

-

.

-- ,

,

,

-r

'

I

-- , --. ,

-

-9 6 ~ 0 5119, , BY).hrnJ6o1o.

r---

vnlue

"j

"-'x.Lq

&s

a

,

4

variable

*&variable

-. J

-

"

id *k;

KlClT I C I Lecture 9 J&

c*k=v&

of 4bolo

b &clk iht L C L ~ ~;SC ~ urn9 C*k j * k ~ v ~ d h @A~ am @ 'lea.

0xB8000000 whlle ( 1 ) I s = Ox88000000 ; for ( I = O : ler1999; I + + ) {

( .(s+I)

Z=

'A'

a& *(s+I) u 'zl )

*(s+L) a *(s+i) + 32 ; else ( . 'a' && *(*+I) cif *(s+l) = *(s+g E *(s+/) 32 ;

1

1

1

-

1

KlClT I C I Lecture 9

'2' )

I-

char tar 'r ;Int r ; char far Y ; char sh ; 110x68000000 ; Int c ;

-

-

for ( c = 0 ;e <= 79 ; c++ ) ur ( ch = * ( 8 +c.2); kr(r=l;ra24;r++)

I 1

1

KICIT I C I Lecture 10

?in

(1

/

Bells And Whistles ' ~raaooooooI I Screen I

/

char far 'v ;Lnt r, c ; s = OrB800MHIO :

v l I I J +a

1

c = random ( 79 ) ;

mund I-

, .. Are Thev Same

I

char h r s ' ; char far % ;

char Gr 'a, "v ; char k r 'a

E]

char far *s,far % ; 7 -

:

for ( I = 1 ; I C* 3999 ; i += 2 *(s+l~-color;

KlClT l

C I Lecture 10

2

pr-;m

e Caps Lock Un ,

main( ) {

~oggle Kmys Capr lock. Num lock Scroll lock Inwrt

char far *kb ;

kbrOx417 ;

How would you put off

v&

'~u h et

on't Do Delete chat far 'kb : kbaOx417; prtntl { "PI. press delete key"); *kb = l 2 ;

I

-

rthdav Josh1 7 6 5 4 3 2 1

0

<--

I Our Program

L

Del-Ctrl

KlClT I C I Lecture 10

Alt + Del

['\a+

vim D&

rtD &ah&. h C M *+QJ

3

what would be the o u t ~ u t mln( 1

I

1000 2090 9000 lnti=10;flo8ta=3.~4:charch='z'; j b dh Lnt 7, "k ; float 'b, "c ; char 'dh, "ah j=&i: b-&;dh=Lch; 4000 5000 M O O k-&J;c=&b;eh-ldh; k c oh pdnH ( " % d # d % d M j , b , d h l : , c c z ~ rcc MO 1500 2500 prlntl .Ad %d'Ad", 'k ,'c , 'ah ) :

;-I

m'.

1-c.

m W [ 0 0 1 1 8 0 0 P ]

Lcm

prlnbl("%d'AdW',sltmof(I),sluof(a),rl~f(ch 2 1-1 I

1);

...

[~ontlnumd

dky+Sinm

] 4 0 0 0 1 w 1 prlnH('9hd %d%d",slxooi(J), slzoof( b ) , rlzoof(dh) ) ; pr/ntf ( "%d w

KQ-2-

w",sizeof ( k ), mhd ( c ), rlzed ( ah ) ) ;

prlnff ( '%d %f %en ,"k, -c, "oh } ;

\1292,

c main( )

I

int i * 25 ; int *j ; float a = 3.34 ;float *b ; char far "kb ; kboOx417; j-&i;

KlClT I C I Lecture 10

dteIadn)l&o &

@w +bock M ' E!&

Pointer Sizes main( ) {

float 8 3.14 ; noat *b ; char h r *kb ; kb = Ox417 ; J

b

= &I; = 8.;

Int far *m ; m = 0x413 ;

KlClT I C I Lecture t 0

fi)'

he btomvdios

I

.

C B n w i -~- O r R % c ~ @ o o

C~~XINJ&Eh a UMUS .--

?in(

1

Int a n l o , b = 2 0 ; ,o,~,o pdntf ( "KdYod", a, b ) ; /L C rwapv(a, b 1 ; 10 prinn ( "%d%dn. a, b ) ;

&& &

1 J

x

swapv ( Int x, lnt y )

I

Y

t

110]150]110]

int t ; \O t * x ; x = y ; y - t ; TO prlnti ( "%d%dW,x, y ) ;

x

Y

t

X

Y

t

pi-'pJB

1

pqwm k.s=W

Q=\O

-

int a = I O , b = 2 0 ; prlntf ( "~/ld%d",a, b ) ; 17 ' swapr ( &a, &b 1 ;

1

swapr (lnt 'x, lnt 'y ) I Int t ;

1

I

-

0

X

.

Y

t

'v = t ;

;urnprod ( Int x, Int y , Int z,int 'as, Int 'pp ) { ..rrx+v+z:l a b

KlClT I C I Lecture 10

I

P

n heLSQ be u pk. : .

char 'n[ ] = {

I "Sanjay", "Amol", "Slvaramkrishnan",

1:

Selection

I

..

int I, ] ; char 't ; for(I=0;1<=3;l++) I

1)

\

~O~(I=O;IC=~;I++) printf ( "fds", n[ i ] ) ;

yarn(

1

I

Calendar

1/8/1~9-? 1HI1 Mon

Int m, y ; lnt leapdays ; long In1 normatdays : long int totaldays ; prlntt ( "Enter month and year" ) ; scant ( "%d%d", Lrn, & y )-; -normaldays = ( y - 1 ) * 365 L ; 1--leapdays=(y-1)/4-(y-1)1100 + (y-1)1400: totaldays = normatdays + leapdays ;

1

...Calendar Int int 1, a ; inl firstday ;

..

totaldays

= normaldays t leapdays ;

-

120; for(i=O; 1
:

i / i ~999 i to 311711999

...Lalendar +,

0 'I 1c\ jYL .+- 1 rr i n t d a y s [ ] - { 3 1 . 2 8 . ~ 1 , ~ 0 ~ ~ 1 , 3 0 ~ 3 ~ , 3 0 , 3 1 ,13; 0 , 3 1

inain( )

.. ..

tolaldaya = normaldays

+ leapdays ;

i f ( ( y % d D O = = o ) 11 ( y % i a o i = O & b y % 4 = = 0 ) days[l1=29:

---7

3

1

.-

s=o: for(i=O: l < = m - 2 :I++) s = s + days[ I ] ; totaldays += s ; tirstday = tolaldavs 'b 7 :

..

1

August 1999 2 0 4 - - 2 8 i --32- & 38- 4 44.c 50-(: 56 I I I I i l l Mon--.T ue., Wed C

....Calendar

I r-rharf = 'm~nfhsi

J

I

26

20

I

l

32

l

/ "January'. "February,, , , 1 :

totaldays += s : firstday = totaldays % 7 : col = 20 + flrslday ' 6 ; clrscr( 1 ;

I

-

- --

-

- I, y ):

rintt ( "96s %d", months[ m 1 I D , 20 ) ;

gotorc

...

printl ( "Mon Tue Wed Thu Fri Sal Sun" ) ;

KICIT I C I Lecture I7

4

...Calendar

,-

I

10

;lain(

26

20

- Mon

32

I

I

...

Iue Wed

)

Int days[ 1 = ( 31, 28, 31. 30,31. 30, 31, 31. 30,

..j

;

I' print month year ' I I' prjnt days 'I rwr-12: for I = 1

: I <=days[ m - 1 ] : I++ )

golorc ( row, c o l ) : printf ( "Tbd". I ) col = col +- -6- : 11 ( col z 56 ) {

:

-

3

row++ ; cob = 20 :

1 1

c--.n u ) c-l'.tien':~! (i& ~ ~ h y ~e y~

F i11): I

I

idanf "%d%dH,am, & y

COY

...calendar]

) ;

d -

while ( 1 ) I

normaldays = ( y

-1

) ' 365L

:

Glendar

Next year

~ O I C ( 20,35

:

printi ( "Rt-Next mth ..." ) ; ch = getkey() ; switch ( ch )

'

h Kt%"C .

1 % -

t

*

Prev. year

case 77 : m++ ;

.

.

'\( -t+

-

1

1

...Calendar ) I

intdays[]= {31,28,31,30,31,30, 31,31,30,31,30,31 ) : scanf ( "%d%dn, &m, &y ) : whlle ( 1 )

i

normaldays = leapdays = totaldays = ~ f ( y % 4 0 0 = = 0 ) l l ( y % l O O ! = O && y % d = = O ) ) days[ 1 ] = 29 :

else

days[ 1 ] = 28 ;

.. -

-I.-_

1

KICIT I C I Lecture 17

K

,,

n.&

,

Terminology

iame

1 1

e:2

Salary

1

4000.50

Handlrna - Data mein( ) I char n [ ] = {'A', 'X', 'Y', 10') ; int a[ ] = { 23, 27, 28 ) ; float s[ 1 = { 4000.50, 5000.00, 6000.75 ) ; int i ; for( i = O ; i ~ = 2 ; I + + ) printf ( "Xc%d %f", n[ i 1, a[ i 1, s[ I ] ) ;

1

1;

float s ;

struct employee e l = { 'A', 23,4000.50 } ; struct employee e2 = { 'X', 27, 5000.00 ) ; struct empioyee e3 = { 'Y', 28,6000.75 ) ; prlnti ( "%c %d %f", el .n, e l a, e l s ) ; printl ( "%c %d %fW, e2.n, e2.a, e2.s ) ; printf ( "%c %d %f", e3.n, e3.a, e3.s ) ;

w \ ~he

}

n h v d d~timkp KlClT / C I Lecture 18

~+bt;l I jot&

,

& \co*e&

I+'*

Arrav of Structures struct employee I

1;

char n ; int a ; float s ;

struct employee e[ 1s {

rf

( 'A', 23,4000.50 { 'X', 27, 5000.00 ) , { 'Y', 28,6000.75 )

int i ;

I }

for( i = O ; i < = 2 ; i++) printf ( "%c %d %f", e[ i ].n,e[ i ].a, e[ i J.s)

Terminology Structure namettag char n ; int a ;

I

A structure is a collection of dissimilar elements. ructure elements are always stored In memory locations.

1 1

struct employee e[ 3 j = {

1 1 1

1

.. }

;

A 23 400.50 X 27 500.00 Y ] 28 ] 600.75 401 403 qo? 408 41 5

KlClT I C I Lecture 18

1

Arrav of Structures -struct employee e l I = ( ... 1 ; char 'p ; struct employee 'q : struct employee ('r 113)

c . - to~ ?~ array of structures

1

I

Ptr.

'

-7 I 3~ +b c k

struct employes @[ 3 I = I

1

1

.. ) :

-1

!A 123 400.50 X 27 500.00 --401

P'.

tr. to structure

* A&S ; ~

408

415

p=e ; q = e : r=e ; p++ : q t t : I++:

printf ( "YOU", printl ( "Xu",

p); q);

wz

Declarations & Definitions gtruct e r n p l o y e m

float s ;

I

iiruct employee e = { "Rahul", 25,4000.50 1 ;

char n[20] ; in! a ; float

s;

) e = { "Rahul", 25,4000.50

I I

1, x, y

'

;

Copying 1

main( ) I

I

struct ernp 1

char n[20]; int a : float s ;

e2.s = el .s ;

G i v i n g at

e3 = el ;-I-* printl ( "%s %d %1", e3.11, e3.a, e3.s ) ;

u

KlClT / C / Lecture 18

I

'd

p b

b

(93 3 Jh.Jffu@.

QOoJ3

Copyinq Arravs

int a1101 = { 3, 6, 5,... 1 ; int bIl 0] ; f o r ( i = O ; i < = 9 ; l++) b[i]=a[i];

struct z {

1;

int srr(lo] ;

struct z a = { 3, 6,5, .., ) ; struct z b ; b=a;

truct address

; int

I

age ;

i i u c t emp e=( "Rahul", 23, "Ngp", 44010,4000.50

}d

Passina Structure Elements main( ) { struct book

{ 1:

char n[20] ; int nop ; float pr ;

ktruct book b = { "Basic", 425, 135.00 ) ;

1

dlspla ( b.n, b.nop, b.pr ) ; show rb.n, &b.nop, bb.pr ) ;

display (chbr *n, int pg, float p )

' 1

printf ( "%s %d %I", n, pg, p ) ;

;how ( char *n, in1 *pg, float 'p ) I

1

printf

"%s %d %f", n, 'pg, 'p ) ;

KICIT I C I Lecture 18

ho\d

P a s s ~ n gStructures struct book char n[20] ; Int nop ; tloat p r : siruet book b = ( "Baalc", 425, 135.00 ) ;

1

displayl ( b ) : showl ( &b>~-----------) /--

display1 ( struct b 6 k bb ).

I

-

printf ( "%s %d %I3', bb.n, bb.nop, bD.pr ) ;

/rr to

showl ( struct book 'bb)

I

I

printf ( "'As %d %f", (' bb ).n, (' bb ).nap,(' bb ).pr ) printf ( "%s %d %f", bb -> n, bb -> nop, bb -r pr )

Complex Nos. --U rstruct corn

I

[ I

+

OA ~hf--hm+ 9

struct corn a = ( 2.5, i.3 } ; struct corn b = I1.2, 1.7 ) ;

I

struct corn c ; struct corn add (struct corn, struct corn) ;

: 7-7 struct corn z ; y:r ; *xf y.1 ; return z ;

2 r

(mainO

+

a en\Lng

+

Complex Nos.

float a [ ] = ( 2 . 5 , 1.3); float b [ ] = ( 1 . 2 , 1 . 7 ) ; float *c , float ' add ( float *, float * ) ; c=add(a, b ) ; printf ( "%f %f", ' c , ' ( c + 1 ) ) ;

b=&.

I

I

'

-

ons of Structures + Pooltlonlng Cursors + Recelulng aecll and scan codas + Dlsplaying characters Prlntlng on prlnter 4

Mause Programmlng

4

Graphlcs Programmlng All Disk Operstlons etc.

4

Endlsk

II * Faster

KICIT I C I Lecture 18

Iplatter ( aluminium ) )

Parts of platter

of same capacity

KlCIT / C / Lecture 18

07)hi-

%a~

9 air q i a a\(

i+.

Readina Boot Sector #~nclude"dos.h"

Fain{ char arr [ 512 ] ; lnt I ;

1

printf ( "Enter floppy in drive A press any key ..." ) ; getch( ; ebsread ( O , l , 0.arr ) , for( 1 = 0 ; 1 < 5 1 2 ; I++) printi ( "%cV, err 1 I 1 1 ;

KlClT / C 1 Lecture 19

-

& B I J ~&\eb J ~ & s P . F,hcr P c ~ i ar n f L i s ~ , b& 5 %S i+ w t o Oie* &fi0di* ,

~b~wad=

~

d & k. p I k ~ & ~ a]89.

GVctb

@

Boot Parameters No-

Kb

1.2 Ub

1.44 Mb

2

512

512

512

2

1

1

1

720

2400 2880

2

7

9

9 2 0

35 2 0

18 2 0

360

,,gs

Descrlptlon

Jump lnstructlon System 10 NO. of byteslsector No, of seetorslcluster No. of sectors In reserved area No, of copies of FAT . Max. no. of root dlr. entries Total no. o l sectors Medla descriptor No. of sectorslFAT No. of sectorsltrack No, of sides No. of hidden sectors

2

2 2

&ch

h b cm h a (3+ChA0,mam

w a 3 d ~ w h pwacedy n we- w ~ W n qdCf mtaei, ih B struct boot ..

I i

-Im a r lump1 . .*.1 ;

I Descrlptlon

3

IT3

Jump lnstructlon

w

...~ o n( t

include "dos.h" yain( )

I

I

struct boot char jump[ 3 ] ; char sysld[ 8 ] ; int bps ; char spc :

..

',

:

struct boot b ; lnt I ; absread ( 0, 1,0, &b ) ; lor(l=0;Is=2:1++) prlnti ( " % X , b,jurnp[ I ] ) for(1=D;I
-

I

b'

ol8 did it

:

dl&&

8)-&4 f

m

d

p

+ Ach

&@no.

H ~ o cw.

In General absread ( drive no., no. of sectors to read, sector from where reading should beg~n,butrer ) ; absread ( 0,1,0, &b ) ;

sector relorred uuinn

LW-

& no.

Typlcal Obtains Values Values t02

€83190 lBM3.3

Bytedsector 08s Secforslclusler Sectors In reserved area

Maw. root dlr. entries Total no. of sactars Madta descrlator

512 1

.... 20480 0

I

I -24

2 224

1 '": 1 1 0

O

SO

4

dl- boot & WuNb

Sides

-9

nldaen *actors

w

i~L

J C A

so b.~& cl~wmw b@w2a--

Anti-Viral

# Include "dos.h"

main( ) l printf ("insert infected disk, Press any key" )

getchl)

:

-

--

1

fl

l o +"d

Q

.

M-Cm4*

eo08*&kab.

20 880

KIClT 1 C 1 Lecture 19

3

#include "d0s.h"

Better...

char a[ 512 ] : char ch = 'y' ; char 'names[ ] P ("0Jerulalwrn'+, "Yankee Doodle",

"Robinson Cruso", "Eddle Murphy", prlntl( "lnaert unlniectmd dlak, Pteas any key" ] ; getch( ; absread( 0 , 1 , 0, a ) ; whlle ( ch = 'y' ) prlntf ( "Insert Infected dlsk, Press any key" )

... 1 ;

r\

:

abswrlte ( 0,1, 0, a ) ; prlntf ( "Another dlsk yln" ) ; ch = getche( ) :

&

disp\% =kG_~l\l~ ih hamQ

4 bytes

Il Include "do8.h"

yaln(

1

rruct

1:

Printincl Directory

entry char n[ 8 ] ; char ext[ char unused[ 17 ] ; long Int size ;

struct entry e [ 16 ] ; Int I, : prlnll ( "Insert dlsk, Press any key" ) absread ( 0 , I , 19, e ) ; for(i=O:i<=l5;l++)

: getch( ) ;

tor(j=O;j<=7;j++) prlntf ( "%c", e[ 1 ].n[ j 1 ) ; lor(j=O;j<=2;j++) printt ( "%c0, e[ I ].ext[ j ] ) : printf ( "Xld", e[ i ].size ) ;

y'iq %C

CIT C / Lecture 19

natdfr &WQ

&

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