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C Programming 185151Computer Practice I

cseannauniv.blogspot.com S.K. Vijai Anand

Ex. No. 3C PROGRAMMING

C is a powerful, portable and elegantly structured programming language. It combines

features of a high-level language with the elements of an assembler and therefore, suitable for

writing both system software and application packages. It is the most widely used general-

purpose language today. C has been used for implementing systems such as operating

systems, compilers, linkers, word processors and utility packages. It is a robust language

whose rich set of built-in functions and operators can be used to write any complex program.

Programs written in C are fast and efficient.

Turbo C IDE

Turbo C IDE facilitates editing, debugging and execution of applications written in C.

C programs are saved with .c extension. Some of the shortcut keys are:

Ctrl + F1 HelpF2 Save

F3 OpenCopy Ctrl + Ins

Cut Shift + DelPaste Shift + Ins

Clear Ctrl + Del

Alt + F9 CompileCtrl + F9 Execute

Alt + F5 User ScreenF7 Trace Into

Alt + F3 CloseAlt + X Quit

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Ex. No. 3.1AREA OF A CIRCLE

Aim

To write a program to compute the area and circumference of a circle.

Algorithm

Step 1 : Start

Step 2 : Define constant pi = 3.14

Step 3 : Read the value ofradius

Step 4 : Calculate area using formulae pi*radius2

Step 5 : Calculate circumference using formulae 2*pi*radius

Step 6 : Print area and circumference

Step 7 : Stop

Flowchart

Read radius

pi = 3.14

area = pi * radius2

circumference = 2 * pi * radius

Print area, circumference

Start

Stop

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Program

/* Area and circumference of a circle */

#include #include

#define pi 3.14

main()

{

int radius;

float area, circum;

clrscr();

printf("\nEnter radius of the circle : ");

scanf("%d",&radius);

area = pi * radius * radius;circum = 2 * pi * radius;

printf("\nArea is %.2f and circumference is

%.2f\n",area,circum);

getch();

}

Output

Enter radius of the circle : 8

Area is 200.96 and circumference is 50.24

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Ex. No. 3.2BIGGEST OF TWO NUMBERS

Aim

To write a program to determine the bigger of two numbers using ternary operator.

Algorithm

Step 1 : Start

Step 2 : Read values of a and b

Step 3 : If a = b then print "Both are equal"

Step 3.1 : Else ifa > b then print "A is big"

Step 3.2 : Else print "B is big"

Step 4 : Stop

Flowchart

YN

Y

N

Read a, b

Start

Stop

a=b

?

a>b

?

Print "Both

are equal"

Print "A is

big"

Print "B is

big"

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Program

/* Biggest of 2 Nos using Ternary Operator */

#include #include

main()

{

int a,b;

clrscr();

printf("Enter A value : ");

scanf("%d",&a);

printf("Enter B value : ");

scanf("%d",&b);

(a==b) ? printf("\nBoth are equal") : a>b ? printf("\nA

is greater") : printf("\nB is greater");getch();

}

Output

Enter A value : -2

Enter B value : -5

A is greater

Enter A value : 4

Enter B value : 7

B is greater

Enter A value : 11

Enter B value : 11

Both are equal

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Ex. No. 3.3LEAP YEAR

Aim

To write a program to check whether the given year is a leap year.

Algorithm

Step 1 : Start

Step 2 : Read the value of year

Step 3 : If yeardivisible by 400 then print "Leap year"

Step 3.1 : Else ifyeardivisible by 4 and not divisible by 100 then print "Leap year"

Step 3.2 : Else print "Not a leap year"

Step 4 : Stop

Flowchart

YN

Y

N

Read year

Start

Stop

year%400=0

year%4 = 0 and

year%100 0

?

Print

"Leap"

Print

"Leap"

Print "Not

Leap"

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Program

/* Leap Year */

#include #include

main()

{

int year;

clrscr();

printf("\nEnter the year : ");

scanf("%d",&year);

if (year%400 == 0)

printf("\n%d is a Leap year",year);else if (year%100 != 0 && year%4 == 0)

printf("\n%d is a Leap year",year);

else

printf("\n%d is not a Leap year",year);

getch();

}

Output

Enter the year : 2000

2000 is a Leap year


1800 is not a Leap year


2004 is a Leap year

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Ex. No. 3.4SIMPLE CALCULATOR

Aim

To write a menu driven calculator program using switch statement.

Algorithm

Step 1 : Start

Step 2 : Display calculator menu options

Step 3 : Read the operatorsymbol and operands n1, n2

Step 4 : If operator= + then calculate result= n1 + n2

Step 4.1 : Else ifoperator= then calculate result= n1 n2

Step 4.2 : Else ifoperator= * then calculate result= n1 * n2

Step 4.3 : Else ifoperator= / then calculate result= n1 /n2

Step 4.4 : Else ifoperator= % then calculate result= n1 % n2

Step 4.2 : Else print "Invalid operator" and go to step 6

Step 5 : Print result

Step 6 : Stop

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Flowchart

Other%/*+

Display Calculator menu

Start

Stop

operator

?

Print res

Read operator, n1, n2

res=n1+n2 res=n1n2 res=n1*n2 res=n1/n2 res=n1%n2

Print "Invalid

operator"

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Program

/* Simple Calculator */

#include

#include #include

main()

{

int n1, n2, result;

char op;

clrscr();

printf("\n Simple Calculator\n");

printf("\n + Summation");

printf("\n - Difference");

printf("\n * Product");

printf("\n / Quotient");printf("\n % Remainder\n");

printf("\nEnter the operator : ");

op = getchar();

printf("Enter operand1 and operand2 : ");

scanf("%d%d",&n1,&n2);

switch (op)

{

case '+':

result = n1 +n2;

break;

case '-':result = n1 - n2;

break;

case '*':

result = n1 * n2;

break;

case '/':

result = n1 / n2;

break;

case '%':

result = n1 % n2;

break;

default:printf("Invalid operator");

exit(-1);

}

printf("\n%d %c %d = %d", n1, op, n2, result);

getch();

}

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Output

Simple Calculator

+ Summation

- Difference* Product

/ Quotient

% Remainder

Enter the operator : -

Enter operand1 and operand2 : 2 4

2 - 4 = -2

Simple Calculator

+ Summation

- Difference

* Product

/ Quotient

% Remainder

Enter the operator : *


3 * 2 = 6

Simple Calculator

+ Summation

- Difference

* Product

/ Quotient

% Remainder

Enter the operator : %


5 % 2 = 1

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Ex. No. 3.5BINARY TO DECIMAL

Aim

To write a program to convert binary number to its decimal equivalent using whileloop.

Algorithm

Step 1 : Start

Step 2 : Read the binary value

Step 3 : Initialize dec and i to 0

Repeat steps 47 until binary > 0

Step 4 : Extract the last digitusing modulo 10

Step 5 : Calculate decimal = decimal + digit* 2i

Step 6 : Recompute binary = binary / 10

Step 7 : Increment i by 1

Step 8 : Print decimal

Step 9 : Stop

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Flowchart

N

Y

Read binary

decimal = i = 0

binary>0

digit = binary % 10

decimal = decimal + digit * 2 i

binary = binary / 10

i = i + 1

Print decimal

Start

Stop

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Program

/* Binary to Decimal */

#include #include

#include

main()

{

int dec=0,i=0, d;

long bin;

clrscr();

printf("Enter a binary number : ");

scanf("%ld",&bin);

while(bin){

d = bin % 10;

dec = dec + pow(2,i) * d;

bin = bin/10;

i = i + 1;

}

printf("\nDecimal Equivalent is %d", dec);

getch();

}

Output

Enter a binary number : 1101011

Decimal Equivalent is 107

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Ex. No. 3.6PRIME NUMBER

Aim

To write a program to check the given number is prime or not using for loop.

Algorithm

Step 1 : Start

Step 2 : Read the value ofn

Step 3 : Initialize i to 2 and flg to 0

Repeat steps 4 and 5 until i n/2

Step 4 : Ifn is divisible by i then assign 1 to flg and go to Step 6


Step 6 : If flg = 0 then print "Prime"

Step 6.1 : Else print "Not prime"

Step 7 : Stop

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Flowchart

YN

N

N

Y

Read n

i = 2, flg = 0

i n/2

Start

n%i = 0 flg = 1Y

i = i + 1

Print "Prime"

Stop

flg = 0Print "Not Prime"

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Program

/* Prime number */

#include

#include

main()

{

int i,n,flg=0;

clrscr();

printf("\nEnter a number : ");

scanf("%d",&n);

for(i=2; i

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Ex. No. 3.7ARRAY MINIMUM/ MAXIMUM

Aim

To write a program to find the largest and smallest of an array

Algorithm

Step 1 : Start

Step 2 : Read the number of array elements as n

Step 3 : Set up a loop and read array elements Ai, i = 0,1,2,n1

Step 4 : Assume the first element A0 to be min and max

Step 5 : Initialize i to 1

Repeat steps 68 until i < n

Step 6 : Ifmax < Ai then max = Ai

Step 7 : Ifmin > Ai then min = Ai


Step 9 : Print max, min

Step 10 : Stop

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Flowchart

N

Y

Y

N

Read n

max < Ai

Start

Stop

=

Read Ai

i

min = max = A0

i = 1 to n1

max = Ai

min > Ai min = Ai

i

Print min, max

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Program

/* Maximum and Minimum of an array */

#include

#include

main()

{

int a[10];

int i,min,max,n;

clrscr();

printf("Enter number of elements : ");

scanf("%d",&n);

printf("\nEnter Array Elements\n");

for(i=0; i

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Ex. No. 3.8MATRIX MULTIPLICATION

Aim

To write a program to perform matrix multiplication based on their dimensions.

Algorithm

Step 1 : Start

Step 2 : Read the order of matrix A as m and n

Step 3 : Read the order of matrix B as p and q

Step 4 : If n p then print "Multiplication not possible" and Stop

Step 5 : Set up a loop and read matrix A elements Aij, i = 0 to m-1 and j = 0 to n-1

Step 6: Set up a loop and read matrix B elements Bij, i = 0 to p-1 and j = 0 to q-1

Step 7 : Initialize i to zero

Step 8 : Initialize j to zero

Step 9 : Assign 0 to Cij

Step 10 : Initialize kto zero

Step 11 : Compute Cij = Cij + Aik * Bkj

Step 12 : Increment kby 1

Repeat steps 11 and 12 until k< n

Step 13 : Increment j by 1

Repeat steps 913 until j < q


Repeat steps 814 until i < m

Step 15 : Set up a loop and print product matrix Cij, i = 0 to m-1 and j = 0 to q-1

Step 16 : Stop

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Flow Chart

Y

N

Read m, n

Start

i = 0 to m1

Read Aij

j

=

n=p

X

Read p, q

Print min, max

i

2

=

= 0 to 1

Read Bij

j

i

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2

Stop

=

j

=

i

Cij = 0

k = 0 to n1

Cij = Cij + Aik * Bkj

k

i = 0 to m1

Print Cij

j

= 0 to 1

i

X

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Program

/* Matrix Multiplication */

#include

#include

main()

{

int a[10][10], b[10][10], c[10][10];

int r1, c1, r2, c2;

int i, j, k;

clrscr();

printf("Enter order of matrix A : ");

scanf("%d%d", &r1, &c1);

printf("Enter order of matrix B : ");

scanf("%d%d", &r2, &c2);

if (c1 != r2)

{

printf("\nMatrix multiplication not possible");

getch();

exit(0);

}

printf("\nEnter matrix A elements\n");

for(i=0; i

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for(k=0; k

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Ex. No. 3.9ALPHABETICAL ORDER

Aim

To write a program to arrange names in their alphabetic order using bubble sortmethod.

Algorithm

Step 1 : Start

Step 2 : Read number of name as n

Step 3 : Set up a loop and read the name list in name array

Step 4 : Assign 0 to i

Step 5 : Assign i+1 to j

Step 6: If namei > namej then swap the strings namei and namej

Step 7 : Increment j by 1

Repeat steps 6 and 7 until j < n


Repeat steps 58 until i < n-1

Step 9 : Set up a loop and print the sorted name array

Step 10 : Stop

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Flow Chart

Y

N

Read n

namei > namej

Start

Stop

=

Read namei

i

=

Swap strings

namei , namej

i

i = 0 to n2

j

i = 0 to n1

Print namei

i

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Program

/* Sorting strings */

#include

#include #include

main()

{

char name[20][15], t[15], srch[15];

int i,j,n,upper,lower,mid;

clrscr();

printf("Enter number of students : ");

scanf("%d",&n);

printf("\nEnter student names\n");

for(i=0; i

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Output

Enter number of students : 10

Enter student names

Raghu

PrabaGopal

Anand

Venkat

Kumar

Saravana

Naresh

Christo

Vasanth

Alphabetical Order

Anand

ChristoGopal

Kumar

Naresh

Praba

Raghu

Saravana

Vasanth

Venkat

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Ex. No. 3.10STRING REVERSE

Aim

To write a program to reverse the given string without using library functions.

Algorithm

Step 1 : Start

Step 2 : Read string str

Step 3 : Assign 0 to len

Step 4 If lenth character '\0' then go to step 5 else step 6

Step 5 : Increment len by 1 and go to step 4

Step 6 : Assign 0 to i and len-1 to j

Step 7: Swap the ith

and jth

character


Step 9 : Decrement j by 1

Repeat steps 79 until i < len/2

Step 10 : Print reversed string str

Step 10 : Stop

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Flow Chart

Y

Y

N

Read str

Start

Stop

T = i

i = j

j = T

Print str

len = 0

lenth char '\0'

len = len + 1

i = 0j = len - 1

i < len/2

i = i + 1

j = j 1

N

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Program

/* User-defined String reverse */

#include

#include

main()

{

char *str;

int i,j,len=0;

char t;

clrscr();

printf("Enter a String : ");

gets(str);

/* String Length */while (str[len] != '\0')

len++;

/* String Reverse */

for(i=0, j=len-1; i

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Ex. No. 3.11SINE SERIES

Aim

To write a program to generate sine series sin(x) =

!7!5!3

753 xxxx value using

function.

Algorithm

Step 1 : Start

Step 2 : Read sine degree as deg

Step 3 : Convert degree to radians using the formula x = deg * / 180

Step 4 : Call sinecalc function with parameter x

Step 5 : Print computed value of sine series

Step 6: Stop

sinecalc Function

Step 1 : Assign x to term and sum

Repeat steps 3 and 4 until term < 0.00001

Step 2 : Compute new term as term = term * x2 / (2i (2i+1))

Step 3 : Alternatively subtract and add term to sum

Step 4 : Return sum

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Flow Chart

Y

N

Read deg, n

Start

sgn = -sgn

term = term * x2

/ (2i (2i+1))

sum = sum + sgn * term

sgn = 1sum = term = x

Print sinevalue

Stop

sinecalc (x)

x = deg * 3.14 / 180

Return (sum)

term> 0.00001

Call sinevalue = sinecalc(x)

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Program

/* Sine series */

#include

#include

#define pi 3.14

float sinecalc(float, int);

main()

{

int deg,n;

float x,sineval;

clrscr();

printf("Enter Sine degree : ");

scanf("%d",&deg);printf("Enter no. of terms : ");

scanf("%d",&n);

x = deg*pi/180;

sineval = sinecalc(x);

printf("\nGenerated series value : %.4f", sineval);

printf("\nBuilt-in function value : %.4f", sin(x));

printf("\nComputational error\t : %.4f", sin(x)-sineval);

getch();

}

float sinecalc(float x)

{

int i=1,sgn=1;

float sum,term;

sum = term = x;

while (1)

{

if (term < 0.00001) break;

printf("\nThe term is %.4lf and sum is %.4lf",

term, sum);

sgn = -sgn;term *= x*x / (2*i * (2*i+1));

sum += sgn *term;

i++;

}

return (sum);

}

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Output

Enter Sine degree: 73

The term is 1.2734 and sum is 1.2734


The term is 0.0279 and sum is 0.9572The term is 0.0011 and sum is 0.9561


The value of Generated Sine series is 0.9561

Value using built-in function is 0.9561

Computational error is 0.0000

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Ex. No. 3.12RECURSIVE FUNCTION

Aim

To write a program to find factorial value of a given number n! = n * (n-1)! usingrecursion.

Algorithm

Step 1 : Start

Step 2 : Read the value ofn

Step 3 : Call factorial function with parameter n

Step 4 : Print factorial value

Step 5: Stop

factorial Function

Step 1 : Ifn = 1 then return 1

Step 1.1 : Else return n * factorial(n-1)

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Flow Chart

Read n

Start

Return (1)

Print factvalue

Stop

factorial(n)

n = 1Y N

Call factvalue = factorial(n)

Return (n * factorial(n-1))

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Program

/* Factorial--Recursion */

#include

#include

long factorial(int);

main()

{

int n;

long int f;

clrscr();

printf("Enter a number : ");

scanf("%d",&n);

f=factorial(n);

printf("Factorial value : %ld",f);

getch();}

long factorial(int n)

{

if (n

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Ex. No. 3.13PASS BY VALUE & REFERENCE

Aim

To write a program that demonstrates passing parameters to a function by value andreference.

Algorithm

Step 1 : Start

Step 2 : Assign 10 to a and 20 to b

Step 3 : Print a and b

Step 4 : Call swapval function with values ofa and b


Step 6 : Call swapreffunction with address ofa and b


Step 8 : Stop

swapval Function

Step 1 : t = a

Step 2 : a = b

Step 3 : b = t

Step 4 : Return

swapref Function

Step 1 : t = *a

Step 2 : *a = *b

Step 3 : *b = t

Step 4 : Return

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Flow Chart

Start

Return

Print a, b

Stop

swapval (a, b)

a = 10b = 20

Print a, b

Print a, b

t = aa = b

b = t

Return

swapref (*a, *b)

t = *a*a = *b

*b = t

Call swapval(a, b)

Call swapref(&a, &b)

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Program

/* Pass by value and reference */

#include

#include

void swapval(int, int);

void swapref(int *,int *);

main()

{

int a = 10, b = 20;

clrscr();

printf("\nValues before function calls\n");

printf("Value of A : %d\n",a);

printf("Value of B : %d\n",b);

swapval(a,b);

printf("\nValues after Pass by Value\n");


printf("Value of B : %d\n",b);

swapref(&a,&b);

printf("\nValues after Pass by Reference\n");


printf("Value of B : %d",b);

getch();

}

/* Pass by value */

void swapval(int a,int b)

{

int t;

t = a;

a = b;

b = t;

}

/* Pass by Reference*/

void swapref(int *x,int *y)

{int t;

t = *x;

*x = *y;

*y = t;

}

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Output

Values before function calls

Value of A : 10

Value of B : 20

Values after Pass by Value

Value of A : 10

Value of B : 20

Values after Pass by Reference

Value of A : 20

Value of B : 10

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Ex. No. 3.14PAYROLL APPLICATION

Aim

To write a program to print employee payroll of a concern using structure.

Algorithm

Step 1 : Start

Step 2 : Define employee structure with fields empid, ename, basic, hra, da, it, gross and

netpay

Step 3 : Read number of employees n

Step 4 : Set up a loop and read empid, ename, and basic for n employees in emp array

Set up a loop and do steps 59 for n employees

Step 5 : hra = 2% of basic

Step 6 : da = 1% of basic

Step 7 : gross = basic + hra + da

Step 8 : it = 5% of basic

Step 9 : netpay = gross - it

Step 10: Print company and column header

Step 11 Set up a loop and print empid, ename, basic, hra, da, it, gross and netpay for

each employee in emp array

Step 12 : Stop

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Flow Chart

Start

Stop

Define emp structure

Read n

= -

Read empi.empid,empi.name, empi.basic

i

i = 0 to n-1

i

empi.hra = 0.02 * empi.basic

empi.da = 0.01 * empi.basicempi.it = 0.05 * empi.basic

empi.gross = empi.basic + empi.hra + empi.daempi.netpay = empi.gross - empi.it

= -

Print empi.empid, empi.name,

empi.basic, empi.hra, empi.da,empi.gross, empi.it, empi.netpay

i

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Program

/* Payroll Generation */

#include

#include

struct employee

{

int empid;

char ename[15];

int basic;

float hra;

float da;

float it;

float gross;

float netpay;

};

main()

{

struct employee emp[50];

int i, j, n;

clrscr();

printf("\nEnter No. of Employees : ");

scanf("%d", &n);

for(i=0; i

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printf("\n\n\n\t\t\t\tXYZ & Co. Payroll\n\n");

for(i=0;i

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