C Language Basic Question

7/24/2019 C Language Basic Question

1/64

Da

1 Draw th

Compute

1)

2) I

3)

4)

1)

IN(Mous

rshan Inst

e block dia

r is made up

entral proce

nput section

utput sectio

torage devic

entral Proc

Central pr

It contain

It also co

It is also

CPU consi

.1) Arith

It

It

It

Itm

.2) Contr

It

It

It

.3) Prima

It

T

UT SECTIO, Keyboard, e

itute of En

Computer

ram of co

f mainly four

sing unit (CP

s

ssing Unit (

ocessing unit

electronics

trols the flo

nown as brai

sts of,

etic Logic U

performs all

can perform

calculates ve

takes dataemory.

ol Unit (CU)

controls all o

manages all

reads instru

ry Memory:-

is also know

e processor

Ntc...)

gineering

Programmi

A) Co

puter arch

components,

)

CPU):-

is a main par

ircuitry that

of data in th

n of the com

it (ALU)

arithmetic cal

add, subtract

ry fast.

from memor

ther units in

operations

tion and data

as main me

r the CPU di

Cen

SE

(Har

PR

& Technol

1

ng and Util

mputer Fu

itecture an

t of the comp

rocesses the

e system.

uter.

lculations and

, multiply, co

y unit and r

he computer

from memor

mory.

ectly stores

ral Processi

ONDARY M

disk, Pen dri

ALU + C

IMARY ME

RAM, ROM, et

ogy

ization (CP

damentals

explain ea

uter system.

data based o

takes logical

mpare, count

eturns data

system.

y.

nd retrieves i

ng Unit

EMORY

ve, etc)

ORY

c)

) 11000

ch block.

n instructions

decision.

, shift and ot

o memory u

information fr

(

yllabus for 1

.

er logical act

nit, generall

om it.

OUTPUT SE

Monitor, Print

stmidsem ex

ivities.

primary

CTION

er, etc)

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2/64


3/64

Da

s

s

c

Limitati

t

t

3 Describ

Compute

1)

2)

3)

rshan Inst

ame form. It

utomation:

ingle click w

omputer, it c

ns

ake of intel

hink about p

hink about th

nable to Co

r incorrect d

various ty

r languages

Machine le

Compute

has its o

ADVANT

1.

It is

2. It do

3. Tran

4. Suit

DISADV

1. Prog

2. Deb

3. It is

4.

Prog

Assembly la

Every m

describe

System

language

Example

ADVANT

1. Prog

2. Prog

3.

Prog

DISADV

1. It is

2. Prog

3. It re

Higher level

We can

Program

itute of En

works 24 ho

Once the on

enever we w

omputes the i

igence: It c

operness or

e correctness

rrect Mista

ta then it pro

pes of com

ay be classifi

el language

r directly un

n machine l

GES:

very fast in e

es not requir

slation is not

ble for low v

NTAGES:

rams are long

gging is very

not portable.

rammer requi

nguage:-

chine langu

function of in

annot under

to machine l

: 8086 Instru

GES:

rams are eas

rams are sma

rams can be

NTAGES:

not portable.

rammer shoul

uires assem

language:-

rite program

er can perfo

gineering

rs without an

e task is cre

ant. For exa

nterest by se

nnot think w

ffect of wor

of these inst

es: It canno

duces wrong

uter langu

ed in three c

or Low lev

erstands this

nguage.

ecution

any extra sy

required.

lume applica

and difficult

difficult task.

res detailed k

ge instructio

struction.

tand this lan

anguage. Thi

ction Set

to understa

ller in size co

ntered quickl

ld know struc

ler as a tran

s in English li

rm complex t

& Technol

3

y problem as

ted in a co

ple, once th

nding one co

ile doing wo

it is doing.

uctions.

correct the

results or pe

ages and m

tegories,

l language:

language. It

stem or soft

ions.

to write and

.

nowledge of

n is assigned

guage directl

translator is

d compared

mpared to m

y using alpha

ure of assem

lator.

e manner an

ask by using

ogy

it does not f

puter, it can

software for

mand.

k. It does no

t can only e

mistakes by i

forms wrong

ention its a

is a languag

are to run th

nderstand fo

architecture o

to English

y so we requi

called assem

o machine le

chine level l

numeric keyb

bly language

d it is more c

high level lan

el tiredness.

be repeated

banking app

have natura

ecute the in

tself. So if w

calculations.

dvantages

of 0s and

e program.

r human.

f microproce

ord MNEMO

ire translator

bler.

vel language.

nguage.

oard.

of microproc

onvenient to

guages with l

yllabus for 1

ly performed

lication is ins

l intelligence.

tructions but

have provid

nd disadva

s (binary).

sor.

IC such that

that convert

ssor.

use.

ess efforts.

stmidsem ex

again by

alled in a

It cannot

it cannot

ed wrong

ntages.

very CPU

it should

assembly

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4/64

Da

4 Why C i

C is calle

I

I

I

5 Write a

A set of i

called sof

System

System s

for runni

"low-leve

System s

1)

t

2)

I

3)

Applicat

Applicati

some sp

suites, G

Applicati

1)

rshan Inst

It is simil

Example:

ADVANT

1) Easi

2)

Req

3) Provi

4) Easi

5) It is

DISADV

1) It re

2) Prog

3)

It is

called mid

middle level

yntax and k

t gives advan

t gives acces

oreover it d

e can devel

ssembly leve

t is not hard

short note

instruction in

tware.

oftware

oftware is de

g applicatio

l" software.

oftware can

perating sys

o user. It is a

ystem suppo

/O devices or

ystem develx: Editor, pr

ion softwar

n software is

cific task (a

aphics softw

n software is

eneral purp

rocessing, w

itute of En

lar to natural

: C, C++, Jav

GES:

r to learn.

ires less time

ides better do

r to maintain

portable.

NTAGES:

uires compil

rammers nee

bit slow comp

dle level la

language be

ywords of C

tages of high

to the low le

es support th

op applicatio

l language to

are or syste

n types of

a logical ord

igned to ope

software. Si

e classified in

em: It contr

bridge betw

rt software:

routine for s

pment softw-processor, c

designed to

counting, tic

re and medi

classified int

se software:

b browser, e

gineering

language and

a, etc

to write.

cumentation.

.

r or interpre

to learn str

ared to low l

guage?

ause

re just like hi

r level langu

vel memory

e Low Level

n specific pr

give more sp

dependent.

software.

r to perform

ate the com

nce system

to three cate

ls hardware

en computer

It makes wor

cket progra

are: It proviompiler, inter

elp the user

et booking

players.

two categor

It is used

xcel, etc It i

& Technol

4

mathematic

er to convert

cture of high

vel and medi

igher level la

age through f

hrough Point

rogramming

grams in C

eed and effici

Hence porta

a meaningf

uter hardwar

oftware runs

gories

as well as int

hardware an

king of hard

ming, etc

es programpreter, loade

to perform g

). Example:

ies.

widely by m

s designed o

ogy

l notation.

higher level l

level langua

um level lang

guage (Engli

unction, mod

rs.

i.e., Assembl

and at the

ency

le programs

l task is call

e efficiently.

at the most

eracts with u

user. Ex: Wi

are more ef

ing developr, etc..

neral tasks (

Enterprise s

any people

vast concep

anguage to

e.

uage.

h).

ular program

Language.

ame time w

an be writte

d program a

t provides an

basic level o

ers, and pro

indows XP, Li

iciently. For

ent environ

word processi

ftware, Acco

or some co

so many pe

yllabus for 1

achine langu

ing and bre

e can use f

with C com

nd a set of p

d maintains

computer, i

vides differen

ux, UNIX, et

example driv

ent to prog

ing, web bro

unting softw

mon task,

ple can use i

stmidsem ex

age.

kup.

atures of

iler.

rogram is

platform

is called

t services

c

rs of the

rammers.

ser ) or

re, Office

like word

.

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5/64

Da

2)

t

s

6 Define t

1.

2.

3.

4.

5.

6.

I

t

7. I

I

rshan Inst

pecial purpo

ax calculatio

pecial requir

he followin

rogram

set of instru

oftware

set of instr

rogram is cal

ardware

hysical parts

perating sy

perating sys

rovides inter

ssembler

ssembler is s

ompiler

t translates

ime.

nterpreter

t translates p

itute of En

e software: I

software, ti

ment.

g terms:

ction in a logi

uction in a l

led software.

of computer i

stem

em is system

ctive platfor

ystem softwa

rograms of h

rograms of hi

gineering

t is used by

ket booking

cal order to p

ogical order

s known as H

software whi

to user.

re which con

igher level la

gher level lan

& Technol

5

limited peopl

software, ba

erform a me

to perform a

ardware. Use

ich works as

erts program

nguage to m

guage to ma

ogy

for some sp

nking softwar

ningful task i

meaningful

r can see and

n interface b

s of assembl

achine langu

hine languag

ecific task lik

e etc It is

s called progr

task is calle

touch the ha

etween hard

language to

ge. It conver

e. It converts

yllabus for 1

e accounting

designed as

am.

program, a

rdware comp

are and user

machine lan

ts whole pro

program line

stmidsem ex

software,

er users

nd set of

onents.

and

uage.

ram at a

by line.

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6/64

1

Dars

What is

them.

Flowchar

Flowch

differe

togeth

solutio

Advan

Ea

Ea

Ea

Ea

Disad

Ti

Di

Ve

Algorithm

An Alg

in the

Advan

Ea

H

Al

Disad

Di

Di

Ju

Comparis

Flowcha

It is a pic

Solution i

Easy to u

Easy to s

Flowchar

an Institu

C

lgorithm?

art is a picto

nt symbol a

er with arro

n of the give

tages

sy to draw.

sy to underst

sy to identify

sy to show br

antages

e consumin

ficult to modi

ry difficult to

orithm is a fi

natural langu

tages

sy to write.

man readabl

orithms for

antages

ficult to debu

ficult to sho

mping (goto)

n:

rt

orial represe

s shown in gr

nderstand as

ow branchin

for big probl

te of Engi

mputer Pr

hat is Flo

rial or graphi

d contains

s showing t

problem.

and logic.

mistakes by

anching and l

.

fy.

draw flowcha

ite sequence

ges like Engl

techniques

ig problems

g.

branching a

makes it har

ntation of a p

aphical forma

compared to

and looping

m is impract

eering &

gramming

B) Flowc

chart? Wri

cal represent

short descr

e process fl

non compute

ooping.

rt for big or c

of well defin

ish.

o understand

an be writte

d looping.

to trace so

rocess.

t.

algorithm.

.

ical

Technolog

and Utiliza

art & Algo

te down th

ation of a pr

iption of the

w direction.

person.

omplex probl

d steps for s

logic.

with modera

e problems.

Algorithm

It is step w

Solution is

It is some

Difficult to

Algorithm

y

ion (CPU)

ithm

advantage

cess. Each s

process ste

This pictorial

ms.

lving a probl

te efforts.

ise analysis o

shown in non

hat difficult t

show branchi

an be writte

Syll

110003

s and disad

ep in the pr

. The flow

representati

em in system

f the work to

computer la

o understand

ng and loopin

for any prob

abus for 1st

vantages.

cess is repre

hart symbol

n can give

atic manner.

be done.

guage like E

.

g.

lem

idsem exam

ompare

sented by a

are linked

tep-by-step

It is written

glish.


7/64

2

W

3

Dars

Explain v

ite an alg

To find w

Algorithm

Step 1 :

Step 2 :

Step 3 :

Step 4 :

Step 5 :

Flowchar

an Institu

arious sym

rithm an

ether given

:-

Input no.

If no mod

Print give

Print give

Stop.

:-

te of Engi

ol used in

Draw a fl

number is

2=0, goto 4.

no is odd, g

no is even.

Print no

eering &

lowchart.

owchart

ven or odd.

to 5.

is Even

Yes

Technolog

Start /

Input /

(Read

Proces

Decisi

Subro

Arrows

st

R

Is No

y

Stop

Output

/ Print)

s

n making

tine

op

start

ead No

mod 2=0?

Syll

Pri

abus for 1st

No

nt no is Odd

idsem exam


8/64

4

Dars

To enter

negative

Algorithm

Flowchar

Step 1 :

Step 2 :

Step 3 :

Step 4 :

Step 5 :

Step 6 :

Step 7 :

Step 8 :

Step 9 :

Step 10 :

Step 11 :

Step 12:

an Institu

umber still

nd zero ent

:-

:-

Initialize p

Read no

If no>0, g

If no0?

0,zero0

neg+1

stop

y

display cou

No

No

nt toe no?

g, zero

Is no


9/64

5

Dars

To print

Algorithm

Step 1 :

Step 2 :

Step 3 :

Step 4 :

Step 5 :

Step 6 :

Step 7 :

Step 8 :

Step 9 :

Flowchar

Prin

an Institu

aximum nu

:-

Read a ,b,

If a>b, go

If b>c, go

Print c is

If a>c, go

Print c is

Print a is

Print b is

Stop.

:-

Yes

t max = a

te of Engi

mber from a

c

to 5

o 8

aximum got

o 7.

aximum, go

aximum, go

aximum

Yes

Isa>c?

P

eering &

given 3 nu

9

o 9

o 9

rint max =

No

Technolog

bers.

Read a, b, c

Isa > b?

c

start

stop

y

Print max

Yes

No

Syll

Isb>c?

= b

abus for 1st

Print m

No

idsem exam

x= c


10/64

6

Dars

To find th

Algorithm

Flowchar

Step 1 :

Step 2 :

Step 3 :

Step 4 :

Step 5 :

Step 6 :

Step 7 :

an Institu

e factorial o

:-

:-

Initialize c

Read no

Calculate f

Increment

If countb)?a

>b)

x=a;

x=b;

wise Operat

ise operator

ouble.

bitwis

bitwis

bitwis

shift l

shift ri

cial Operat

Addre

Point

Com

zeof It ret

mem

mem

conditiona

rovides speci

itute of En

s value of ri

1 is same as

is same as

1 is same as

is same as

1 is same a

Decrement

l operators in

ents value b

s postfix, the

10; b=a++;

s prefix, the v

10; b=++a;

ments value

postfix, the e10; b=a--; a

prefix, the va

10; b=--a; a

erator

or is known a

if exp1 is tru

:b; which is s

ors

s are used to

AND

OR

exclusive O

ft ( shift left

ght ( shift rig

ors

ss operator, i

r operator, it

a operator. I

rns the num

er selection

er selection

l operator (t

al conditional

gineering

ht side to lef

a = a + 1

a = a - 1

a = a * 1

a = a / 1

a = a % 1

Operators

C which are

y 1.

expression is

after this sta

alue is incre

after this sta

y 1.

xpression is efter this stat

lue is decrem

fter this stat

s Conditional

e then execu

ame as

perform oper

eans multip

ht means divi

t is used to d

is used to de

t is used to li

er of bytes t

perator, use

perator, use

ernary oper

operator ( ? :

& Technol

6

side

generally not

evaluated fir

ement, a= 1

ented first a

ement, a= 1

valuated firstment, a= 9,

ented first an

ment, a= 9,

Operator.

e exp2 other

ation bit by bi

ly by 2)

de by 2)

etermine add

clare pointer

k the relate

e operand o

in structure.

in pointer t

ator) with e

) to evaluat

ogy

found in othe

t and then t

1, b = 10.

nd then the e

1, b = 11.

and then theb = 10.

d then the ex

b = 9.

ise exp3

it. Bitwise op

ess of the va

variable and

expressions

cupies.

.

structure.

xample.

conditional

r languages.

e value is inc

xpression is e

value is decr

pression is e

rators may n

riable.

o get value f

together.

xpression in

yllabus for 1

remented.

valuated.

emented.

aluated.

ot be applied

om it.

single line.

stmidsem ex

to float

m


18/64

Da

It i

or

Sy

Fir

ex

Ex

9 What i

Wh

as

Ty

Implici

Explici

Exampl

rshan Inst

s also known

xpressions.

tax: expr1?

t of all expr

r3 is evaluat

mple: c = a>

s type conv

en an operat

ype casting

ecasting is m

t Type Casti

C automatic

evaluated wi

This automa

The lower ty

result is hig

Example:

o If o

o If o

Thus conver

conversion

Type Casti

Sometimes

The process

The general

Where type-

an expressi

e:

#include

short

ConversionHierarchy

gineering

erator becau

pr3

, if it is non

f expr2 and

a is greater

in implicit

ds of differen

sion.

ble of one da

ny intermedi

ny significan

is known as

ically conver

int and other

float and oth

from low dat

own below.

ce a type co

l conversion

s: (type

f the standa

unsi

int

char

& Technol

7

se this is the

zero (true) t

xpr3 is evalu

than b then a

nd explicit

types, they

a type to act

ate values to

e.

implicit type

ed to the hig

is float then

er is long dou

type to high

version in a

or casting is

-name) expre

d C data typ

unsigned

long int

gned int

ogy

only operator

en the expr

ated.

is assigned t

ype convers

are converted

like another

the proper ty

conversion.

her type befo

int will be co

ble then float

data type so

ay that is di

nown as expl

ssion.

. The expres

long int

float

doubl

lon

in C which r

ssion expr2

o c otherwise

ion with ex

to a commo

ata type suc

pe so that th

e the operati

verted to flo

will be conve

that informa

ferent from

licit casting.

ion may be c

double

yllabus for 1

equires three

is evaluated

b.

mple.

type, this is

as an int to

expression

on proceeds.

t.

rted to long

ion is not los

utomatic con

onstant, vari

stmidsem ex

operands

otherwise

known

float.

an be

The

ouble.

. The

version.

ble, or

m


19/64

Da

10 Explai

Pre

Op

(5

We

op

Ass

sa

Op

8 -

We

firs

mu

Foll

Ra

Prece

1

2

3

4

5

6

rshan Inst

void main()

{

int

floa

avg

prin

avg

prin

}

operator p

cedence of a

rator preced

3) * 2, givi

say that the

rator (+), so

ociativity is t

e precedenc

rator associ

(3 - 2), givin

say that the

t. When we c

st use associ

owing table

k 1 indicates

dence As

Le

Ri

Le

Le

Le

Le

itute of En

um=47, n=1

avg;

sum / n;

tf(Result of I

(float)sum /

tf(Result of

recedence a

operator is i

ence is why t

g 16.

multiplication

the multiplic

e left-to-righ

e.

tivity is why

g 7.

subtraction o

an't decide by

tivity.

rovides a co

highest prec

sociativity

t to Right

ht to Left

t to Right

t to Right

t to Right

t to Right

gineering

0;

mplicit Type

(float)n;

xplicit Type

nd associati

s priority in

e expression

operator (*)

tion must be

t or right-to-l

he expressio

erator (-) is

operator pre

plete list of

dance level

Operat

()

[]

+, -

++, --

!

~

*

&

sizeof

(type)

*

/

%

+

-

>

=70)

pr i nt f

se i f ( avg>

pr i nt f

se i f ( avg>

pr i nt f

se i f ( avg>

pr i nt f

se

pr i nt f

with examp

ecisions are i

n-1is true th

n Statement-

n-1is false t

i t i on- 1)

est- condi t i

St at ement

St at ement

ment - 3;

gineering

nt based on

, m4, m5;

r m1, m2, m3,

%d%d" , &m

3+m4+m5) / 5;

( Di st i nct

=60)

( "F i r s t cl

=50)

( "Second c

=40)

( "Pass cl a

( " f ai l " ) ;

le and draw

volved, we

en test-condi

2will be exe

en Statemen

on- 2)

-1 ;

-2 ;

& Technol

4

verage mark

m4, m5: ") ;

, &m2, &m3,

) ;

ass") ;

l ass" ) ;

s s " ) ;

flowchart.

ay have to u

tion-2is eval

uted.

t-3will be ex

ogy

s.

, &m4, &m5)

se more than

ated. If it is

ecuted.

;

one ifelse s

rue then Sta

yllabus for 1

tatement in

ement-1will

stmidsem ex

ested form

be executed,

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25/64

Da

Flow

Exam

S

rshan Inst

hart:

ple: Find ma

#i ncl ude

voi d mai

{

i p

s

i f

{

}

e

{

}

}

False

atement-3

itute of En

imum numb

n()

t a, b, c;i nt f ( "Ente

anf ( "%d%d%

( a>b)

i f ( a>

el se

se

i f ( b>

el se

Cond

gineering

r from given

r val ue of

d", &a, &b,

)

pr i nt f ( "a

pr i nt f ( "c

)pr i nt f ( "b

pr i nt f ( "c

Entry

estition 1?

& Technol

5

three numbe

a, b, c: " ) ;

&c);

i s max") ;

i s max") ;

i s max") ;

i s max") ;

True

True

TCondi

State

Next st

State

ogy

s.

;

sttion 2?

ent-1

atement

ent-x

False

yllabus for 1

tatement-2

stmidsem ex m


26/64

Da

5 Expl

T

I

e

E

I

I

T

Exam

rshan Inst

in switch-c

he switch sta

tests wheth

eneral formswi t ch( e

{

c

c

c

d

}

xpressionin

ach case is la

a case matc

alue of all ca

none of the

he break stat

ple:

swi t ch (

{

c

c

c

c

c

d

}

itute of En

se stateme

tement is a m

r an expressi

f switch-casxpr essi on)

se const - e

se const - e

se const - e

f aul t :

state

witch should

beled by one

hes the expre

e expression

cases are ma

ement cause

grade)

se 1:

pr i nt f

break;

se 2:

pr i nt f

break;

se 3:

pr i nt f

break;

se 4:

pr i nt

break;

se 5:pr i nt f

break;

f aul t :

pr i nt f

break;

gineering

nt with exa

ulti-way deci

on matches

is as below,

xpr 1: st

br

xpr 2: st

br

xpr 3: st

br

ent s

be integer or

or more inte

ssion value t

s must be diff

ched then de

an immedia

( "Fal l ( F )

( "Bad ( D) \

( "Good ( C)

( "Very Goo

( "Excel l en

( "You have

/ /

& Technol

6

ple.

sion making.

ny one of th

t ement 1;

ak;

t ement 2;

ak;

t ement 3;

ak;

character ex

er-valued co

en execution

erent.

fault case is

e exit from t

\ n " ) ;

n" ) ;

\ n " ) ;

d ( B) \ n") ;

t ( A) \ n") ;

i nput t ed f

br eak i sn t

ogy

constant val

ression. Floa

nstant.

starts at tha

xecuted. def

e switch.

al se gr ade

necessary

ues or not.

t or any othe

case.

ult case is o

n" ) ;

her e

yllabus for 1

r data type is

tional.

stmidsem ex

not allowed.

m


27/64

Da

Rules

6 Stat

If

al

If

If

Fl

b

Write a C Pr

1 Find

#i ncl

voi d

{

}

2 Find

#i ncl

voi d

{

rshan Inst

for switch st

The switc

Case label

Case label

Case label

The break

The defaul

Nesting of

difference

statement is

ernatives.

can have val

implements

at, double, c

used in if co

ogram to.

out whethe

l ude

ent er no: ")

d", &no);

=0)

i nt f ( "no i

i nt f ( "no i

umber from

h>

c;

ent er a, b, c

d%d%d" , &a,

gineering

ust be integ

stant or con

que.

ith colons.

optional.

ent is option

ent is allow

lse and swi

among two

constraints.

ther data typ

umber is od

;

s even") ;

s odd") ;

given 3 nu

: " ) ;

b, &c);

& Technol

7

al type. Floa

tant expressi

l.

d.

ch-case

Th

mul

Swi

Swi

es can Onl

blo

d or even.

bers.

ogy

or other dat

on.

switchstate

ltiple alternati

tchcan have

tchimplemen

y int and cha

k.

types are n

switch

ent is used

ves.

values based

ts Binary sea

data types c

yllabus for 1

t allowed.

o select amo

on user choi

rch.

an be used in

stmidsem ex

ng

e.

switch

m


28/64

Da

}

3 whic

#i ncl

voi d

{

}

rshan Inst

i f ( a>b&&

p

el se i f (

p

el se

p

h asks day

l udec)

i nt f ( " a i s

b>c)

i nt f ( " b i s

i nt f ( " c i s

umber and

h>

ent er day

d", &day) ;

ay)

se 1:

pr i nt f

break;

se 2:

pr i nt f

break;

se 3:

pr i nt f

break;

se 4:

pr i nt f

break;

se 5:

pr i nt f

break;

se 6:

pr i nt

break;

se 7:pr i nt f

break;

f aul t :

pr i nt f

gineering

maxi mum") ;

maxi mum") ;

maxi mum") ;

prints corre

umber ( 1- 7

( "sunday")

( "monday")

( " t uesday"

( "wednesda

( "t hur sday

( " f r i day")

( "saturday

( "wr ong i n

& Technol

8

ponding da

\ n" ) ;

;

;

) ;

y " ) ;

" ) ;

;

" ) ;

put" ) ;

ogy

y name usin

g switch-ca

yllabus for 1

e.

stmidsem ex m


29/64

Darshan Institute of Engineering & Technology Syllabus for 1stmidsem exa

1

Computer Programming and Utilization (CPU) 110003

E) Loop Control Structure (for, while, dowhile)

1 Explain loops available in C with example

Loops are used to repeat execution of a block of code.

During looping, a set of statements are executed until some condition for termination is encountered.

Generally, looping process would include the following four steps

1) Initialization of a counter

2) Test for a termination condition

3) Loop body statements

4) Increment the counter

C supports three types of looping

while loop

The simplest of all looping structure is while statement.

The general format of the while statement is:

I ni t i al i zat i on;

whi l e ( t est condi t i on)

{

body of t he l oop ;

i ncrement or decr ement ;

}

Test condi t i onis evaluated and if the condition is true then the body of the loop is executed.

After the execution of the body, the t est condi t i onis once again evaluated and if it is true, the body

is executed once again.

This process is repeated till the t est condi t i on is true. When it becomes false, the control is

transferred out of the loop.

On exit, the program continues with the statements immediately after the body of the loop.

While loop is also known as entry control loop because first control-statement is executed and if it is true

then only body of the loop will be executed.

Exampl e: To pr i nt f i r st 10 posi t i ve i nt eger numbers

voi d mai n( )

{

i nt i ;

i = 1; \ \ i ni t i al i zat i on of i

whi l e(i


30/64


2

dowhile loop

In contrast to while loop, the body of the dowhile loop is executed first and then the loop condition is

checked.

The body of the loop is executed at least once because dowhile loop tests condition at the bottom of

the loop after executing the body.

dowhile loop is also known as exit control loop because first body statements are executed and then

control-statement is executed, thus condition checking happens at exit point.

The general format of the dowhile statement is:

i ni t i al i zat i on;

do

{

st atement ;

i ncrement or decr ement ;

}

whi l e( t est- condi t i on) ;

Exampl e: To pr i nt f i r st 10 posi t i ve i nt eger number s

voi d mai n( )

{

i nt i ;

i = 1; \ \ i ni t i al i zat i on of i

do

{

pr i nt f ( \ t %d, i ) ; \ \ st at ement execut i on

i ++; \ \ i ncrement of cont r ol var i abl e

} whi l e(i


31/64


3

Exampl e: / / The f ol l owi ng i s an exampl e t hat f i nds t he sum of t he f i r st t en posi t i ve

i nteger number s

voi d mai n( )

{

i nt i ; / / decl are var i abl e

i nt sum=0;

f or ( i =1; i < = 10; i ++) / / f or l oop

{

sum = sum + i ; / / add t he val ue of i and st or e i t t o sum

}

pr i nt f ( %d, sum) ;

}

We can include multiple expressions in any of the fields of for loop provided that we separate such

expressions by commas. For example: for( i = 0; j = 100; i < 10 && j>50; i++, j=j-10)

f or Loop whi l e Loop dowhi l e Loop

f or( i =1; i < = 10; i ++)

{

sum = sum + i ;

}

i =1;

whi l e( i


32/64


33/64

1

Dar

What i

An

An

Th

Th

Syntax

Exampl

Th

Th

In

Advan

Yo

An

An

Examp

Types1) Sin

2) Tw

3) Mul

han Insti

s an array?

array is a fix

array is deriv

individual el

subscript fo

: dat

e : i nt

data_types

sizeindicate

he example,

ages:

can use one

array is very

array makes

le:

#i ncl ude)

a[ 5] = {5,

i =0; i


34/64

2

Dar

Explai

Single

An arra

Syntax

Exampl

An

g =

Mo

will

We

sig

Th

an

For

var

for

Ab

ad

tot

Th

up

Initiali

The ge

There a

1. i nt

will

2. i nt

will

3. i nt

firs

ele

han Insti

initializati

Dimensiona

y using only

: dat

e : i nt

individual arr

marks [60];

e generally

take the val

can store val

like marks[

ability to re

efficient pro

example we

iable that is

i=0; i


35/64

3

Dar

Two di

Tw

Tw

Firs

Syntax

Exampl

He

thr

A t

35.

Initiali

1.

i nt

will

col

2. i nt

her

col

3. i nt

init

Explai

C has s

For Exa

Funct

strlen

strcm

han Insti

mensional a

dimensional

dimensional

t subscript d

: dat

e : i nt

e m is decla

ugh 19). Th

o dimension

5 & second el

mar

35.

mar

66.

mar

85.

mar

25.

zation of tw

t abl e[ 2] [

initialize 1st

mn to 6 and

t abl e[ 2] [

e, 1stgroup i

mn element

t abl e[ 2] [

ializes as abo

various str

everal inbuilt

mple: char s

ion

s1)

(s1,s2)

ute of Eng

rrays:

arrays are al

arrays have

notes the nu

a_t ype arr a

marks[ 10] [

ed as a mat

first elemen

al array mark

ement is mar

ks [0][0]

ks [1][0]

ks [2][0]

ks [3][0]

o dimension

3] = {1, 2, 3

ow 1stcolum

so on.

3] = {{1, 2,

for 1strow a

is 4, 2ndrow

3] ={{1, 2},

ve but missin

ing handlin

functions to

[]=Their, s

Returns le

l = strlen(s

Compares

It returns

ineering

so called tabl

wo subscript

ber of rows

y_name[ r ow

20] ;

ix having 10

of the matri

s[4][3] is sh

ks[0][1] and

mark

40.5

mark

55.5

mark

78.5

mark

35.2

al array:

, 4, 5, 6};

element to

3}, {4, 5, 6}

nd 2ndgroup

rdcolumn el

{4}}

g elements w

operations

perate on str

2[]=There;

gth of the st

1); it returns

two strings.

egative valu

Technolo

3

or matrix.

.

and second s

s i ze] [ col u

rows (numb

is m[0][0]

wn below. T

contains 40.5

[0][1]

[1][1]

[2][1]

[3][1]

, 1strow 2nd

;

is for 2ndrow.

ment is 6 so

ill be initialize

available in

ing. These fu

ing.

5

if s1s

S

tes the numb

o 9) and 20

w last colum

nt is given b

][2]

][2]

][2]

][2]

1strow 3rdcol

tcolumn ele

mple.

own as strin

and zero if

llabus for 1st

er of columns

columns (nu

is m[9][19]

marks[0][0]

umn to 3, 2nd

ent is 1, 2nd

handling fu

1=s2.

midsem exa

.

bered 0

contains

row 3rd

row 1st

ctions.


36/64

Dar

strcpy

strcat

strchr

strstr(

strrev

strlwr

strupr

strncp

strnca

strnc

strrch

han Insti

(s1,s2)

s1,s2)

(s1,c)

s1,s2)

(s1)

s1)

(s1)

y(s1,s2,n)

t(s1,s2,n)

p(s1,s2,n)

r(s1,c)

ute of Eng

printf(%d

Output : -

Copies 2nd

strcpy(s1,s

s2 remains

Appends 2n

strcat(s1,s

becomes

Returns a

printf(%s

Output : ir

Returns a

printf(%s

Output : h

Reverses g

strrev(s1);

Converts s

printf(%s

Converts s

printf(%s

Copies firs

s1=; s2=

strncpy(s1,

printf(%s

Output : T

Appends fi

strncat(s1,

printf(%s

Output : T

Compares

strcmp() f

printf(%d

Output : 0

Returns th

printf(%s

Output : e

ineering

, strcmp(s1,

string to 1sts

2) copies the

unchanged.

dstring at th

); a copy o

heirThere

ointer to the

,strchr(s1,i)

ointer to the

,strstr(s1,h

ir

iven string.

makes string

ring s1 to lo

, strlwr(s1));

ring s1 to up

, strupr(s1));

n character

There;

s2,2);

,s1);

st n characte

s2,2);

, s1);

eirTh

first n chara

nction.

, strcmp(s1,

last occurre

,strrchr(s2,e

e

Technolo

4

2));

tring.

string s2 in t

end of 1stst

string s2 is

first occurre

);

first occurre

));

s1 to riehT

er case.

Out

per case.

Out

f string s2 t

r of string s2

cter of strin

2,3));

nce of a give

));

gy

o string s1 so

ring.

appended at

ce of a given

ce of a given

ut : their

ut : THEIR

string s1

at the end of

s1 and s2

character in

S

s1 is now T

the end of

character in

string s2 in s

string s1.

and returns

a string s1.

llabus for 1st

ere.

tring s1. No

he string s1.

tring s1.

similar resu

midsem exa

s1

lt as


37/64

Da

1 What

A

T

T

T

It

P

It

f

While

1

2

rshan Ins

is user defi

function is a

he functions

he functions

he functions

has a uniqu

rameter or a

is optional t

nction, their

using functio

.

Function D

Like v

The fu

four p

1

2

3

4

Synta

Examp

In this

functio

. Function D

Functi

It has

Fu

Fu

.

Function

Functi

Functi

enclos

itute of En

Comput

ned functio

block of code

hich are cre

hich are in-b

hich are imp

name and it

rgument pas

o return a v

return type is

n, three thing

eclaration

riables, all th

nction declar

rts,

) Function t

) Function n

) Parameter

)

Terminatin

: Fu

m( i nt , i n

ction return t

e integer.

s also called

rts.

: It is same

: It is actual

from main fu

led by simpl

eses.

& Technol

1

ing and Ut

G) Functi

ith example

s a specific ta

mmer are ca

r are known

eader librari

. it can be c

n is optional.

lling progra

nt

ust be declar

nown as fun

pe).

ct i onName

) ;

ype is int, na

unction impl

as function d

logic or codi

ction or oth

using a fun

ogy

ilization (C

ns

. Define the

sk.

lled user-defi

as system fu

s are known

lled from an

. Function w

ed before the

tion prototyp

( Argument 1,

me of functio

mentation.

claration but

g of the func

r function th

ction name f

PU) 1100

syntax of fu

ned functions

ctions.

as library fun

part of a pro

hich is not r

y are used.

e or function

Argument 2

n is sum, 2 p

with argume

tion

t is known as

ollowed by a

yllabus for 1

3

ction in C.

.

tions.

gram.

turning any

signature. It

, Argument

rameters ar

nt name.

function call.

list of actual

stmidsem ex

value from

consists of

) ;

passed to

.

argument


38/64

Da

Synta

An ex

2 Expl

Functi

whet

1. F

2. F

3. F

4.

F

5. F

1. F

rshan Ins

x or general

Funct i

at ement 1;

at ement 2;

at ement 3;

tion

st di o. h>

nt , i nt ) ;

( )

t a, b, ananf ( %d%d,

s = sum( a,

i nt f ( Answ

i nt x, i nt

t r esul t ;

sul t = x +

turn (resu

categories

lassified in on

returned or n

no argument

no argument

arguments a

arguments a

return multip

o argument

ction has no

es not return

re is no data

gineering

Function

i onName ( Ar

;&a, &b) ;

b ) ;

r = %d, a

y)

y;

t ) ;

f functions.

e of the follo

ot.

and no retu

and return

d no return

d one return

le values usin

nd no return

argument, it

a value, the

ransfer betw

& Technol

2

gument 1, A

ns);

wing categor

n value

value

alue

value

g pointer

value.

does not rec

calling functio

en the callin

ogy

r gument 2,

\ Functi on

\ Functi on

\ Functi on

based on w

\ void printli

\ int printlin

\ void printli

\ int printlin

\ void printli

ive any data

n does not re

function an

r gument 3)

Decl arat i

Cal l i ng

Def i ni t i o

ether argum

e(void)

(void)

e(int a)

(int a)

e(int a)

from calling f

ceive any dat

called functi

yllabus for 1

n or Si gna

ents are pres

nction.

a from the ca

on.

stmidsem ex

t ur e

ent or not,

lled


39/64

Da

Exam

2. F

Exam

3. F

E

rshan Ins

l e:

#i ncl ude

voi d pr i

voi d mai{

c

pr

pr

}

voi d pr i

{

i n

f o

{

}

}

nction with

When a fu

When a fu

l e:

#i ncl ude

i nt get _

voi d mai{

i n

m

pr

}

i nt get _

{

i n

pr

sc r e

}

nction with

When a fu

When it do

function.

ampl e:

itute of En

st di o. h>

t l i ne(voi d

( )

r s c r ( ) ;

i nt l i ne( ) ;

i nt f ( " \ n G

t l i ne(voi d

t i ;

r ( i =0; i

umber ( voi d

( )

t m;

get _number

i nt f ( "%d",

umber ( voi d

t number ;

i nt f ( "ente

anf ( "%d", &t urn numbe

rguments an

ction has ar

es not return

gineering

; / /

U \ n") ;

i ++)

( " - " ) ;

and return a

argument, it

urn value, th

;

) ;

) ;

)

number: " )

umber ) ;;

no return v

ument, it rec

a value, the

& Technol

3

No ar gument

value

does not rec

calling funct

/ / No a

;

/ / Ret u

lues

eives data fr

calling functio

voi

{Fu}

void mai{a = Fun1}

ogy

No ret u

ive data from

ion receives

r gument

r n val ue

m calling fun

n does not re

d main()

1()

n()

()

r n val ue

calling functi

ne data from

ction.

ceive any dat

NoInput

NoOutput

NoInput

Functionresult

yllabus for 1

on.

the called fu

a from the ca

void Fun{

}

int Fun1(){

return no;}

stmidsem ex

ction.

lled

()


40/64

Da

4. F

Exam

rshan Ins

#i ncl ude

#i ncl ude

voi d sum

voi d mai{

i n

pr

sc

su

}

voi d sum(

{

i f

el

} / /

nction with

When a fu

When a fu

ple:

#i ncl ude

#i ncl udei nt sum( i

voi d mai

{

i n

cl

pr

sc

x

pr

ge}

i nt sum( i

{

i n

w

{

itute of En

st di o. h>

coni o. h>

i nt , i nt ) ;

( )

t no1, no2;

i nt f ( "ente

anf ( "%d%d",

m( no1, no2) ;

i nt no1, i n

( no1>no2)

pri ntse

pri nt

No return

rguments an

ction has ar

ction has ret

st di o. h>

coni o. h>nt ) ;

( )

t no, x;

r s c r ( ) ;

i nt f ( "ente

anf ( "%d", &

sum( no) ;

i nt f ( "sum=

tch ( ) ;

nt no)

t add=0, i ;

i l e( no>0)

i =no%1

gineering

/ /

no1, no2: "

&no1, &no2)

no2)

( "\ n no1 i

( "\ n no2 i

val ue

one return

ument, it rec

urn value, th

/ / Ar gume

no: " ) ;

o ) ;

d" , x) ;

0;

& Technol

4

Argument

) ;

;

gretest")

gretest")

alue

eives data fr

calling funct

t

void{Fun

}

void main({

b = Fun1(}

ogy

;

;

m calling fun

ion receives

main()

(a)

)

Valar

F

ction.

ny data from

Value ofargument

No returvalue

i{

r

}

ue ofument

nctionresult

yllabus for 1

the called fu

void Fu{}

t fun2(f)

turn(e)

stmidsem ex

ction.

1(int a)


41/64

Da

5. F

Exam

#i

v

v

{

}

v

{

}

3 Expl

A

A

Exam

rshan Ins

}

r e}

nction that r

Function c

To receive

So functio

l e:

i ncl udey)

return

gineering

d+i ;

10;

iple value

er one value

e value from

lled with refe

x, i nt y ,

s, &d) ;

" , s , d) ;

a, i nt b,

formal argu

ction during

efinition of a

/ / Functi

ans;

;

d", ans) ;

) / /

x;

& Technol

5

/ / Ret u

r zero value.

function, we

rence not wit

i nt * s , i nt

i nt *sum, i

ment with e

function calli

function are

on Decl ara

/ / a an

x and y ar

ogy

r n val ue

It cannot ret

have to use

h value

*d);

nt *di f f )

xample.

g are called

alled formal

i on.

d b are act

f ormal ar

urn more tha

ointer.

ctual argum

rguments or

ual ar gume

guments.

yllabus for 1

n one value

nts or param

parameters.

nts .

stmidsem ex

eters.


42/64

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4 Expl

The p

C

C

Call

I

S

E

p

Exam

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voi d

voi d

{

}

voi d

{

}

Outp

Enter

Value

X=5

rshan Ins

e

}

in call by v

arameters ca

all by value

all by referen

y value

call by valu

o the original

ven if a func

arameter, no

l e:

ude


43/64

Da

Value

X=3

Call

I

f

S

C

Exam

Outp

Enter

Value

X=5

Value

X=5

Differ

4 What

rshan Ins

s inside the

=5

y Referenc

call by ref

nction.

o the original

all by referen

ple:

#i ncl ude

voi d swa

voi d mai

{

i n

pr

sc

s

pr

pr

}

voi d swa

{

i n

t e

*x

*y

pr

pr

}

t:

the value of

inside the s

=3

inside the m

=3

ence between

do you me

itute of En

ain function

rence, the a

content of th

ce is used wh

st di o. h>

( i nt *, i n

( )

t x, y;

i nt f ( "Ente

anf ( "%d%d",

ap(&x, &y) ;

i nt f ( \ n V

i nt f ( " \ n x

( i nt *x, i

t t emp;

mp=*x;

=*y;

=t emp;

i nt f ( \ n V

i nt f ( " \ n x

& Y: 3 5

ap function

in function

two program

n by recursi

gineering

ddress of th

e calling func

enever we w

*) ;

t he val ue

&x, &y) ;

l ue i nsi de

%d y=%d",

nt *y)

l ue i nsi de

%d y=%d",

is marked a

ve function

& Technol

7

actual par

ion can be c

nt to change

of X & Y:

t he mai n

x, y) ;

t he swap f

x, y) ;

bold in call

Explain wi

ogy

meters is p

anged.

the value of l

unct i on) ;

unct i on) ;

y reference

h example.

ssed as an

ocal variable

rogram

yllabus for 1

argument to

through fun

stmidsem ex

the called

tion.


44/64

Da

Pictor

rshan Ins

Recursive

If a functi

Recursion

Suppose

write it as

computed

1!.which i

Ex: 4! =4

=

=

=

Terminatin

to infinite

ial presentati

f=1;if(..)..;f=4*fareturn

fact 4

itute of En

function is a

on calls itself

is thus the p

e want to ca

n! = n * (

as (n-1)!

s 1.

*3!

*3*2!

*3*2*1!

*3*2*1

g condition m

umber of pr

n of recursio

t(3)

4;

gineering

function that

then it is kno

ocess of defi

lculate the fa

n- 1) ! . First

( n- 1) * (

ust be there

cess.

n.

Tracing

f=1;if(..)

..;f=3*facreturn 6

fact(3)

& Technol

8

calls itself.

wn as recursi

ing somethi

torial of a gi

e have to fin

-2 ) ! . This p

which can ter

of function f

t(2)

;

ogy

on.

g in terms of

en number t

d (n-1)! an

rocess end w

minates the

ct() for n=4

f=1;if(..)..;f=2*facreturn 2

fact(2)

itself.

en in terms

then multipl

en finally we

hain of proce

(1)

;

yllabus for 1

f recursion w

y it by n. the

need to calc

ss, otherwise

f=1;if(..)..return 1;

fact 1

stmidsem ex

e can

(n-1)! is

late

it will lead


45/64

Da

Exam

Adva

Disad

5 What

Scop

rshan Ins

ple: Find fact

#i ncl ude

#i ncl ude

i nt f actvoi d mai

{

i n

pr

sc

f

pr

}

i nt f act

{i n

i f

{

}

el

{

}}

tages

Easy soluti

Complex p

vantages

Recursive

Terminatin

Execution

is scope, lif

The scope

accessible

In what pa

Local vari

of function

itute of En

rial of a give

st di o. h>

coni o. h>

i nt ) ;( )

t f , n;

i nt f ( "ente

anf ( "%d", &

fact (n ) ;

i nt f ( " \ n f

i nt n)

t f =1;

( n==1)

return

se

f =n*f a

return

on for recursi

rograms can

ode is difficu

g condition is

speed decrea

etime and vi

of variable c

rt of the prog

bles which a

.

gineering

n number usi

number: " )

) ;

ctori al =%

1;

ct ( n- 1) ;

f ;

vely defined

e easily writ

lt to understa

must; other

es because o

sibility of v

an be define

ram the vari

e declared in

& Technol

9

ng recursion.

;

" , f ) ;

olution.

en with less

nd and debu

ise it will go

f function call

riable?

as that a p

ble is accessi

side the bod

ogy

ode.

.

in an infinite

l and return a

art of a pro

ble is depend

of function

loop.

ctivity many

ram where t

s on where th

cannot be ac

yllabus for 1

imes.

e particular

e variable is

essed outsid

stmidsem ex

variable is

eclared.

e the body


46/64

Da

Lifeti

Visibil

rshan Ins

Global var

function in

e

Lifetime is

It is also r

ity

Visibility is

If variable

variable w

itute of En

iables which

a program.

a time limit d

ferred to the

the ability of

is redeclared

ich is redecl

gineering

are declared

uring the pro

longevity of

the program

within its sc

red.

& Technol

10

outside any

gram executi

ariable.

to access a v

pe of variabl

ogy

function de

n until which

ariable from

e, the variab

inition can b

a variable ex

he memory.

le losses the

yllabus for 1

e accessible

ist in a mem

visibility in th

stmidsem ex

by all the

ry.

e scope of


47/64

Dar

What

Decla

Initia

shan Insti

Co

is pointer?

A pointer i

Pointer is

Pointers c

manipulat

void main(

{

int

p

pri

}

Output: 10 p is int

& is ad

* is in

& oper

Va

ration of p

Syntax:

Example:

1)

The as

2) pt_na

3)

pt_na

lization of t

int a=5, x,

p = &a

x = *p;

tute of En

mputer P

How to dec

a variable t

derived dat

ntain mem

data stored

)

a=10, *p;

&a;

ntf(%d %d

10 5000eger pointer

dress of or r

irection or d

ator is the in

riable

a

p

inter,

data_ty

int *p,

erisk (*) tell

e needs a m

e points to

he pointer,

*p; // Decl

// Initia

// p con

ineering

ogrammi

lare and ini

hat contains

type in C.

ry address

in memory.

\\ A

d, a, *p,

variable

ferencing o

ereferencing

verse of * op

Valu

pe *pt_nam

loat *q, cha

s that the va

emory locati

variable of

res pointer

lizes p with

tains addres

5000

10

Technol

1

ng and U

Pointer

tialize it?

address or l

as their v

sign memor

);

erator whic

operator wh

erator ( x =

e

;

*c;

riable pt_na

on to store

type data_ty

ariable p an

ddress of a

s of a and *

gy

ilization (

cation of an

lues, so th

y address of

returns me

ich returns v

a is same as

e is a point

ddress of an

pe

regular var

gives value

CPU) 1

other variabl

y can also

a to pointer

ory addres

alue stored a

x = *(&a))

Addre

5000

5048

er variable

other variabl

iable a and x

stored at th

2nd Mid

0003

e.

be used to

variable p

of variable

t that memo

ss

e

t address.

em Syllab

access an

ry address


48/64

Dar

How

Arr

It r

Me

All

res

It i

Disc

Expla

As we

eleme

type k

shan Insti

ointer is di

ay is a const

efers directl

mory allocati

cates the m

ize or reassi

s a group of

ss relatio

int a[10],

Array nam

an array.

a[0] is sam

So every p

in Array of

have an a

nts of an arr

nown by sin

tute of En

fferent fro

Array

ant pointer.

to the elem

on is in sequ

emory space

ned.

elements.

ship betw

p;

is constant

e as *(a+0)

rogram writt

Pointers

ray of char,

ay will store

le name.

a:

ineering

array?

ents.

ence.

which canno

een array

pointer so a

, a[2] is sam

en using arr

int, float e

the address

a[0]

.

.

.

a[i]

a[1]

.

.

.

.

a[9]

Technol

2

Poin

It r

Me

t Allo

It is

and point

is constant

e as *(a+2),

y can alway

tc, same

values. So,

gy

ter variable

fers address

ory allocati

ated memo

not a group

r.

ointer and it

a[i] is same

be written

ay we can

rray of poin

a:

a+1:

a+i:

a+9:

Pointer

can be chang

of the varia

n is random.

y size can b

of elements.

always poin

as *(a+i)

sing pointer

have an arr

ters is a coll

*(a+0)

.

.

.

*(a+i)

*(a+1)

.

.

.

.

*(a+9)

2nd Mid

ed.

le.

.

resized.

It is single

ts to the firs

.

ay of pointe

ction of poi

2000

2002

2000+i*

2018

em Syllab

ariable.

element of

rs, individua

ters of sam

2


49/64

Dar

Synta

Exam

Now,

for(i=

{

}

By usi

dynam

SwapSwap

#inclu

void s

void

{

}

void s

{

}

shan Insti

:

data_type

le :int *ptr[5];

int mat[5]

he array of

;i


50/64

Dar

Write

#inclu

#inclu

void{

}

shan Insti

a C progra

de

de

ain()

int *p, a[1

p=a;

printf("Ent

for(i=0; i

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