Cp Lab Manual 2010

8/3/2019 Cp Lab Manual 2010

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Exercise 2

2s complement of a number is obtained by seaming it from right to left and complementingall the bits after the first appearance of a 1. These 2s complement of 11100 is 00100.Write a C program to find the 2s complement of a binary number.Algorithm: main()

1. Input binary no of length 16 bit binary.2. Copy binary to dup.3. test = check(binary).

4. If test ==0 then (i) else (iii).(i) Print Invalid No.(ii) goto step 5.(iii) Call compl(binary, a).(iv) Print 2s complement , a.

5. Stop.Algorithm: Input(binary)

1. Input binary.2. return.

Algorithm: check(binary)

1. l = strlen(binary).

2. for i =0 to l.(i) if !(binary[i] = 0 or binary[i] = 1) then (ii) else (iii).(ii) goto step 2.(iii) x = 0.

3. return.Algorithm: compl(binary, a)

1. l = strlen(binary).2. for i = l 1 to 0 (Repeat step 3).3. If (binary[i]=0 ) then (i) else (iii).

(i) a[i] = 1.(ii) goto step 2.

(iii) a[i] = 0.(iv) goto step 2.4. for i =l 1 to 0.5. If (a[i] = 0) then (a) else if (b) else (c).

(a) a[i] = 1.(b) a[i] = 0.(c) if (check = 1 && a[i] = 0) then (i) else (iv).

(i) a[i] = 1.(ii) check = 0.(iii) goto step 4.(iv) if (check == 1 && a[i] == 1) then (a) else goto step 4.

(a) a[i] = 0.(b) check = 1.6. return

Program:

#include#include#include


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#include#define L 16main(){

int i,test;char binary[L],dup[L],a[L];void input(char binary[]);int check(char binary[]);void compl(char binary[],char a[]);

clrscr();printf("Enter Binary no of Max length 16\n");input(binary);strcpy(dup,binary);test=check(binary);if(test==0){

printf("\nInvalid Binary number.");exit(1);

}

printf("\nOriginal Binary No:%s",dup);compl(binary,a);printf("\n2's complement is %s",a);getch();

}void input(char binary[]){

printf("Enter only 0's and 1's without spaces up to Max 16 Bits\n");scanf("%s",binary);return;

}

int check(char binary[]){

int i,l,x=1;l=strlen(binary);for(i=0;i


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l=strlen(binary);for(i=l-1;i>=0;i--){

if(binary[i]=='0')a[i]='1';elsea[i]='0';

}for(i=l-1;i>=0;i--)

{if(i==l-1){

if(a[i]=='0')a[i]='1';

else{

a[i]='0';check=1;

}

}else{if((check==1)&&(a[i]=='0')){

a[i]='1';check=0;

}

else if((check==1)&&(a[i]=='1')){

a[i]='0';

check=1;}}

}a[l]='\0';return;

}


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No

False

True

Yes No

Yes No

No

Yes No

Yes

Output:

Enter Binary no of Max length 16Enter only 0's and 1's without spaces up to Max 16 Bits1111

Original Binary No:11112's complement is 0001

a[i] = 0 a[i]=1

fori=l-1 to 0

if i=l-1

If a[i] =0

If check==1 &&a[i] = 0

a[i] = 1

a[i] = 0

check = 1

a[i] = 1

check = 0

If check== 1

&& a[i]= 1

a[i] = 0check = 1

a[l] = \0

return


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Exercise 3

a. Write a Program in C to find the sum of individuals Digits of a positive Integer.Solution: -

Algorithm

1) Input a number n.2) Check n is +ve, if not goto step 6.3) S=0,m=n.4) Repeat the following until n=0.

i. r=n%10.ii. s+=s+r.iii. n=n/10.

5) print s, m.6) stop.

Flowchart

Yes No

Yes

No

Input n

If

n0){

ch=2;}else

ch=3;printf("A=%6.2f\nB=%6.2f\nC=%6.2f\n",a,b,c);printf("Discriminate D=%6.2f\n",d);switch(ch){

case 1:printf("Equal Roots");

x1=x2=-b/(2*a);printf("x1=x2=%6.2f\n",x1);break;

case 2:printf("Roots are Real and inequal");x1=(-b+sqrt(d))/(2*a);x2=(-b-sqrt(d))/(2*a);

printf("x1=%6.2f\n",x1);printf("x2=%6.2f\n",x2);break;

case 3:

printf("Imaginary roots \n");printf("x1=%6.2f+i%6.2f\n",-b/(2*a), sqrt(-d)/(2*a));printf("x2=%6.2f-i%6.2f\n",-b/(2*a), sqrt(-d)/(2*a));break;

}getch();

}Flowchart:

Yes No

Start

Enter values a, b, c

d= b2 - 4 ac

Ifd=0

Ifd>0ch = 1


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No Yes

1 2 3

Output:

1. Enter a,b,c values5 2 1

A= 5.00B= 2.00c= 1.00

Desciminat D=-16.00Imaginary rootsx1= -0.20+i 0.40x2= -0.20-i 0.40

2. Enter a,b,c values2 4 2

A= 2.00

ch = 3 ch = 2

Print a, b, c, d

Swi

tch

cas

ech

Print Equal roots

x1 = x2 = - b / 2a

Print x1, x2

Print roots are real &unequal

x1 = - b+d/2a

x2=-b-d/2a

Print x1, x2

Print Imaginaryroots

Print x1 iscomplex

Print x2 iscomplex

Stop


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B= 4.00c= 2.00Desciminat D= 0.00Equal Roots x1=x2= -1.00

Exercise 5

a. The total distance travelled by vehicle in t seconds is given by distance =ut+1/2at2 whereu and a are the initial velocity(m/sec) and acceleration (m/sec2). Write C Program tofind the distance travelled at regular intervals of time given the values of u and a. The

program should provide the flexibility to the user to select his own time intervals and repeatthe calculations for different values of u and a.

Algorithm:

1. Begin.

2. Input initial velocity u.

3. Input Acceleration a.

4. Input starting time and final time i & n.

5. Input time interval it.

6. Print Time Distance U m/sec A m/sec2 .

7. For (t = i ; t


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8. Stop.

Program:

#include#include#includemain(){

float u, a;float d;int t, i, n, it;clrscr();

printf("Enter Initial velocity\n");scanf("%f",&u);

printf("Enter Acceleration");scanf("%f",&a);

printf("Enter starting time/final time");scanf("%d%d",&i,&n);

printf("Enter time interval");scanf("%d",&it);

printf("Time\t Distance\t U m/sec \t a m/sec2)\n");printf(\n----------------------------------------------\n);for(t=i;t


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}getch();

}

Flowchart:

No

Yes

Start

Input Initial velocity u

Input acceleration a

Input start time i

Input ending time n

Input Interval it

Repeat t=i ton with step

it

d=ut+1/2 at2

Print t, d, u, a


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Input:

Enter Initial velocity4Enter Acceleration: 5Enter starting time/final time: 1 10Enter time interval: 1

Output:

Time Distance U m/sec a m/sec2)1 6.50 4.00 5.002 18.00 4.00 5.003 34.50 4.00 5.004 56.00 4.00 5.005 82.50 4.00 5.006 114.00 4.00 5.007 150.50 4.00 5.008 192.00 4.00 5.009 238.50 4.00 5.0010 290.00 4.00 5.00

Stop


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b) Write a C Program, which takes two integers operands and one operator from the user,performs the operation and prints the result. (Consider the operators +,-,*,/,% and use SwitchStatement )Algorithm:

1. Begin

2. Enter a, b values.

3. Print MENU.

(i) Print + Addition.(ii) Print - Subtraction.(iii) Print * Multiplication.(iv) Print / Division.(v) Print % Remainder.(vi) Print E Exit.

4. Print Enter your choice.

5. If op==E then goto step 8 otherwise follow the below steps

6. Switch(op)

a. case +:

i. Print Addition.ii. c=a+b.iii. Print Sum=c.

iv. breakb. case -:

v. Print Subtraction.vi. c=a-b.vii. Print Difference=c.viii. break

c. case *:

ix. Print Multiplication.x. c=a*b.

xi. Print Product=c.xii. breakd. case /:

xiii. Print Division.xiv. c=a/b.xv. Print Quotient=c.xvi. break


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e. case %:

xvii. Print Remainder.xviii. c=a%b.xix. Print Remainder=c.xx. break

f. default:

xxi. Print Invalid Option.xxii. break

7. while(1) then goto step 3.

8. Stop.

Program:

#include#include#includemain(){

int a, b, c;char op;clrscr();printf("Enter a and b:");scanf("%d%d",&a,&b);do{

printf("\n\nMENU\n");printf("+ Addition\n");printf("- Subtraction\n");printf("* Multiplication\n");

printf("/ Division\n");printf("% Remainder\n");printf("E Exit\n");printf(Enter your choice);getchar();op=getchar();if(op=='E'||op=='e')

exit(1);switch(op){

case '+':

printf("Addition\n");c=a+b;

printf("Sum=%d\n",c);break;

case '-':printf("Subtraction\n");c=a-b;


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printf("Difference=%d\n",c);break;

case '*':printf("Multiplication\n");c=a*b;

printf("Product=%d\n",c);break;

case '/':

printf("Division\n");c=a/b;

printf("Quotient=%d\n",c);break; case '%':

printf("Remainder\n");c=a%b;

printf("Remainder=%d\n",c);break;

default:printf("Invalid Option\n");

break;} /*end of switch statement*/}while(1); /*End of while*/

}/*End of main function*/

Flowchart:

Start

Input a,b

Swi

tch(OP)

c = a + bPrintsum c c = a b

PrintDiff c c = a * b

Printproductc c = a / b

PrintQuaff c c = a % b

PrintremainderStop


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Yes

No

+ - * / %

Input:

Enter a and b: 5 3Output:

MENU+ Addition- Subtraction* Multiplication/ Division% RemainderE Exit+

Menu+ Addition- Subtraction*Multiplication/ Division% RemainderE Exit

Input choice op

If op =E


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AdditionSum=8MENU+ Addition- Subtraction* Multiplication/ Division% RemainderE Exit

*MultiplicationProduct=15

MENU+ Addition- Subtraction* Multiplication/ Division% Remainder

E ExitgInvalid Option

MENU+ Addition- Subtraction* Multiplication/ Division% RemainderE Exit

E (Exited)Exercise 7

Write a C program that uses functions to perform the following operations.

i) To insert a sub-string into given main string from a given position.

Algorithm:

1. Enter the main string: s

2. Determine length of string: s i.e. l

3. Print the main string: s

4. Input position of the substring to be inserted: p

5. if p is ve or p is greater than length of main string (l), then goto step 6 otherwise goto step 7 .

6. print ie substring out of the main string position, goto step 13.


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7. input the substring, s1.

8. l1=string length of s1.

9. if l + l1 > L then go to step 10 otherwise go to step 11.

10. substring cant be inserted, because it is too long goto step 13.

11. call insert(s, p, s1)

12. print After string insertion the main string is: s.

13. stop.

Program:

#include#include#include#define L 80#define P 40main(){

int p,l,i,l1;char s[L],s1[P];void inst(char s[],int p,char s1[]);clrscr();

printf("Enter the Main string:\n");scanf("%[^\n]",s);l=strlen(s);

printf("The ma in string is \n%s",s);printf("\nPosition of the substring to be inserted:\n");scanf("%d",&p);

if((pl))printf("Substring position is out of the main string.");else { getchar();


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printf("Enter substring:\n");scanf("%[^\n]",s1);l1=strlen(s1);if((l+l1)>L)

printf("\nstring can't be inserted because too long.");else{inst(s,p,s1);

printf("\nAfter string insertion:\n");

printf("%s",s);}

}getch();

}/*End of main functioin*/

Algorithm: Insert(s, p, s1)

1.l = length of s.

2. i=p;

3. k=length s1.

4. l1 = k

5. last = k+l+1

6. s[last] = \0

7. last = last 1.

8. p1 = l (p - 1)

9. s[last] = s[l]

10.last= last 1

11. l = l 1

12. p1 = p1 1

13. if(p!=0) then repeat 6 through 14.

14. k = 0

15. repeat i = p to i


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i) s[i] = s1[k]

ii) k = k + 1.

16. return.

void inst(char s[],int p,char s1[]){

int l1,last,p1,i,j,l,k;l=strlen(s);i=p;k=strlen(s1);l1=k;last=k+l+1;s[last]='\0';last--;

p1=l-(p-1);

do{

s[last]=s[l];last--;l--;

p1--;}while(p1!=0);

k=0;for(i=p;i


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Flowchart:

Yes No

Yes No

Start

Input main string: s

l=strlen(s)

print sub string:

s1

Input substring p

If pl

Print subs teing of

a main string

Input substring s1

l1=strlen(s1)

If l

+l1>l

Print string cant

be inserted

Insert(s, p,

s1)

Print after

insertion: s

Stop


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Java is oops languagePosition of the substring to be inserted:8Enter substring:an

After string insertion:Java is an oops language

ii) To delete n Characters from a given position in a given string.

Algorithm:

1. Input text line.

2. Determine length of line l.

3. input position of starting character pos.

4. input No of characters to be deleted nc.

5. na = l pos.

6. if nc > na then perform the following otherwise goto step 7.


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i) print text deleted from pos to end.

ii) line[pos]=\0.

7. p = pos + nc.

8. k = p.

9. for i=1 to l k perform the following

i) line[pos]=line[p].

ii) pos=pos+1

iii) p = p + 1.

10. line[pos]=\0.

11. print After deleting line is line,s.

12. stop.

Program:

#include#include#include#define ML 80main(){

char line[ML];int p,l,n,pos,nc,na,k,i;clrscr();

printf("Enter the line of text:");scanf("%[^\n]",line);

l=strlen(line);

printf("Enter the position of the starting character\n");scanf("%d",&pos);

na=l-pos;printf("Enter No of characters to be deleted ie should be


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if(nc>na){

printf("The text is deleted from %d till the end",pos);line[pos+1]='\0';

printf("After deletion the string :%s",line);}else{

p=pos+nc;

k=p;for (i=1;i na

Print the text is deletedfrompos to end

P=pos + nc

K=p

For i=1 to

l - k

Line[pos] =line[p]


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Output:

1) Enter the line of text: ABJ KALAM IS A FORMER PRECIDENT

Enter the position of the starting character14Enter No of characters to be deleted ie should be


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Exercise 8

Write a C Program that uses functions to perform the following operations usingStructure:

1. Reading a complex number.

2. Writing a complex number.

3. Addition of two complex numbers.

4. Multiplication of two complex numbers.

(Represent complex no using structs).

Algorithm: main()

1. Ch =menu().

2. On ch goto 3, 10 ,17

3. Input comples c1 ie c1 = input().

4. Input complex c2 ie c2 = input().

5. C = add (c1, c2).

6. Print c1, ie output(c1).


8. Print c, ie output(c).

9. Goto step 16.

10. Input comples c1 ie c1 = input().

11. Input complex c2 ie c2 = input().



14. C = add (c1, c2).

15. Print c, ie output(c).

16. Repeat 1 through 16 until ch = 3.

17. Stop.


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Algorithm: input()

1. Input c.rpart, c.ipart.2. return c.

Algorithm: output()

1. print c.rpart, c.ipart.2. return.

Algorithm: add(comp c1, comp c2)

1. c.rpart = c1.rpart + c2.rpart.

2. c.ipart = c1.ipart + c2.ipart.

3. return c.

Algorithm: mul(comp c1, comp c2)

1. c.rpart = c1.rpart*c2.rpart c1.ipart*c2.ipart.2. c.ipart = c1.rpart*c2.ipart + c1.ipart*c2.rpart.3. return c.

Program

#include#include#include#includestruct compl{

float rpart;float ipart;

};typedef struct compl comp;main(){

comp c1,c2,c;int ch;int menu(void);comp input(void);comp add(comp c1,comp c2);comp mul(comp c1,comp c2);void output(comp c);do{

ch=menu();if(ch==3){

clrscr();printf("\nThank you\n");break;


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}switch(ch){case 1:

printf("Complex Addition\n");printf("Enter complex 1");c1=input();

printf("Enter complex 2");c2=input();

c=add(c1,c2);printf("\nc1=");output(c1);

printf("\nc2=");output(c2);

printf("Sum is c=");output(c);

break;case 2:

printf("Complex Multiple\n");

printf("Enter complex 1");c1=input();printf("\nEnter complex 2");c2=input();c=mul(c1,c2);

printf("\nc1=");output(c1);

printf("c2=");output(c2);c=mul(c1,c2);

printf("\nOutput/Product c=");

output(c);break;}

}while(1);/*End of do-while*/}

comp input(void){

comp c;printf("Enter Real part & Imaginary part\n");scanf("%f%f",&c.rpart,&c.ipart);return(c);

}

void output(comp c){

printf("r%6.2f+i%6.2f",c.rpart,c.ipart);return;


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}

comp add(comp c1,comp c2){

comp c;c.rpart=c1.rpart+c2.rpart;c.ipart=c1.ipart+c2.ipart;return(c);

}

comp mul(comp c1,comp c2){

comp c;c.rpart=(c1.rpart*c2.rpart)-(c1.ipart*c2.ipart);c.ipart=(c1.rpart*c2.ipart)+(c1.ipart*c2.rpart);return(c);

}int menu(void){

int ch;while(1){

printf("1. Addition of Complex Numbers");printf("2. Multiplication of Complex Numbers");printf("3. Exit");printf("Enter your choice:");scanf("%d",&ch);if(ch3){

printf("Choice should be between 1 and 3.");

continue;}else

break;}return(ch);

}

Flowchart: main() Start

ch = menu()

Print Thankou


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Yes

No

1 2

True False

Flowchart: Input() Flowchart: output(comp c)

Start

Input c.rpart,c.ipart

return c

Start

print c.rpart, c.ipart

return

If ch== 3

Swi

tch

ch

Input complexc1

c1 = input()

While(t)

Input complexc1

c1 = input()

Input complexc2

c2 = input()

Input complexc2

c2 = input()

c = add(c1,c2)

c = mul(c1, c2)

Print c1

output(c1)

Print c1

output(c1)

Print c2

output(c2)

Print c2

output(c2)

Print c

output(c)

Print c

output(c)

Stop


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Flowchart: add(comp c1, comp c2) Flowchart: mul(comp c1, comp c2)

Algorithm: menu()

1. print MENU.

2. Print 1. Addition

3. Print 2. Multiplication

4. Print 3. Exit

5. Input ch

6. If ch < 1 or ch > 3

goto step 1, else 7.

7. return(ch)

Flowchart: menu()

Start

c.rpart = c1.rpart + c2.rpart

c.ipart = c1.ipart + c2.ipart

return c

Start

c.rpart = c1.rpart * c2.rpart c1.ipart * c2.ipart

c.ipart = c1.rpart * c2.ipart + c1.ipart * c2.rpart

return c

Start


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True

Yes

No

Output:

1. Addition of Complex Numbers2. Multiplication of Complex Numbers3. ExitEnter your choice:1Complex AdditionEnter complex 1Enter Real part & Imaginary part1 1Enter complex 2Enter Real part & Imaginary part1 1

c1=r 1.00+i 1.00c2=r 1.00+i 1.00Sum is c=r 2.00+i 2.001. Addition of Complex Numbers2. Multiplication of Complex Numbers3. ExitEnter your choice:3

While (1)

If ch

3

return ch

MENU1.Addition2.Multiplicatio

n3.Exit

Input ch


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1. Addition of Complex Numbers2. Multiplication of Complex Numbers3. ExitEnter your choice:2Complex MultipleEnter complex 1Enter Real part & Imaginary part2 2

Enter complex 2Enter Real part & Imaginary part1 1

c1=r 2.00+i 2.00c2=r 1.00+i 1.00Output/Product c=r 0.00+i 4.001. Addition of Complex Numbers2. Multiplication of Complex Numbers3. ExitEnter your choice:3

Exercise 10

a). Write a C Program to find the factorial of given number use both recursive and

non-recursive functions.

i) To find the factorial of a given integer.

ii) To find the GCD (greatest common divisor) of two given integers.

iii) To solve Towers of Hanoi problem.

Non-Recursive Approach:Algorithm:

(1) Input a +ve integer No n.

(2) Check whether n is +ve or not.

i) If n is ve print cant be evaluated the factorial goto step (4).

ii) Else calculate fact i.e., n! call function fact(n).

(3) Print factorial of n.

(4) Stop.

Algorithm for fact():

(1) f = 1

(2) Repeat i=1 to n times.

f = f * i.


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(3) Return f.

Flowchart: Flowchart: fact(n)

No

Yes No Yes

Program:

#include#includemain(){

int f,n;int fact(int n);clrscr();

printf("Enter a +ve no. to find fact");scanf("%d",&n);if(n


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1. Enter a +ve no. to find fact6Factorial of 6 is 720

2. Enter a +ve no. to find fact3Factorial of 3 is 6

Recursive Approach

Algorithm:

FlowChart:

Flowchart: fact(n)

Start

Input n

Print n,

fact(n)

Stop

fact(n)

If

n


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i. /*To find factorial of given integer using recursive */

#include#includelong int fact(int n);main(){

int n;long int fact(int n);clrscr();

printf("n= ");scanf("%d",&n);printf("n!=%1d\n",fact(n));getch();

}long int fact(int n){

if(n


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ii. Write a C Program to find GCD (Greatest Common Divisor) of two given integers

use both recursive and non-recursive functions.

Recursive Approach

Algorithm:

Flowchart: main()

Flowchart: hcf(p, q)

No

Yes

Program:p

Start

Input 2 Nos,

a, b

gcd=hcf(a, b)

Print a, b,

gcd

Stop

hcf(p, q)

r = p-(p/q2)

If

r=

=0

return(q)

return(r)

hcf(p, q)


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int a,b,gcd;int hcf();clrscr();

printf("Enter in any 2 numbers whose GCD is to be found\n");scanf("%d%d",&a,&b);

gcd=hcf(a,b);printf("GCD of %4d and %4d is %4d \n",a,b,gcd);getch();

}int hcf(p,q)int p,q;{

int r;r=p-(p/q*q);if(r==0)

return(q);elsehcf(q,r);

return(r);}

Output:

1. Enter in any 2 numbers whose GCD is to be found46 35GCD of 46 and 35 is 11

2. Enter in any 2 numbers whose GCD is to be found25 56GCD of 25 and 56 is 25

iii. Write a 'C' Program to solve tower of HANOI problem using bothRecursive and Non-Recursive.

Recursive Approach

Algorithm:

Flowchart:

Start


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No

Yes

Program:


int nvalue;char snvalue='L',invalue='C',dnvalue='R';void hanoi();clrscr();

printf("Enter number of disks: ");scanf("%d",&nvalue);

printf("\nTower of Hanoi problem with %d disks \n,nvalue");hanoi(nvalue,snvalue,invalue,dnvalue);

snvalue=Linvalue=C

dnvalue=R

Input

Disks, n

Print Tower of Hanoi problem with

disks, n

hanoi(n, snvalue, invalue,

dnvalue)

Stop

hanoi(n, snvalue, invalue,

dnvalue)

If n!=0

hanoi(n-

1,sndl,dndl,indl)

Print Move disk n from sndl

to dndl

hanoi(n-1, indl, sndl,

dndl)

return


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printf("\n");getch();

}void hanoi(int n,char sndl,char indl,char dndl){

if(n!=0){

/*Move n-1 disks from starting needle to intermediate needle*/hanoi(n-1,sndl,dndl,indl);

/*Move disk n from start to destination*/printf("Move disk %d from %c to %c \n",n,sndl,dndl);/*Move n-1 disks from intermediate needle to destination needle*/hanoi(n-1,indl,sndl,dndl);

}return;

}

Output:

Enter number of disks: 3Tower of Hanoi problem with 3 disksMove disk 1 from L to RMove disk 2 from L to CMove disk 1 from R to CMove disk 3 from L to RMove disk 1 from C to LMove disk 2 from C to RMove disk 1 from L to RExercise 11

b) Write a C function that uses functions to perform the following:

i. Write a C program that displays the position or index in the string S where thestring T begins or -1 if S doesnt contain T.

Algorithm:

1. Input first string s.

2. Input second string t ( to be searched).

3. Search t is ie found = strstr(s, t).

4. If found then (i) else (iii).

(i) Print t is found is s.

(ii) Goto step 5.

(iii) Print t is not found.

5. Stop.


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Flowchart:

No

Yes

Program:

#include#include

#include#define L 80#define M 40main(){

char s[L],t[M];char *found;clrscr();

printf("Enter the string S ie first one\n");scanf("%[^\n]",s);getchar();

printf("Enter the string T to be searched ie end string\n");scanf("%[^\n]",t);found=strstr(s,t);if(found)

printf("%s string found in %s string",t,s);else

printf("Not found:-1");

Start

Input first string s

Input enter string to be searched t

Print stringis found

Print notfound

Stop

Search t in s is found = strstr(s, t)

If found


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getch();

}

Output:

i) Enter the string S ie first oneHELLO HOW ARE YOUEnter the string T to be searched ie end string

LL string found in HELLO HOW ARE YOU string

ii) Enter the string S ie first oneHI HOW ARE YOU, THIS IS PRASAD.Enter the string T to be searched ie end string

NNot found:-1

ii. Write a C program to count the lines, words and characters in a given text.

Algorithm:

1. nw = na = nc = 0.

2. Repeat the following until no character is empty.

i. Input a line of text.

ii. Determine the length of a line is c.

iii. if c = 0 then goto step 3 otherwise goto step iv.

iv. for i = 0 to c repeat the following

a. Check line[i] for space or end of line[\0].

If space or end of line then nw = nw + 1, otherwise goto step 6.

b. nl = nl + 1.

3. Print no of lines nl.

4. Print no of words nw.

5. Print no of characters nc.

6. Stop.


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Program:

#include;#include c

#include#include#define L 80main(){

char line[L];int ch,i,n,nw,nc,nl,c;clrscr();nw=nc=nl=0;

printf("Enter the text, leave one space");printf(" between the words and press enter at the end of the text.\n");

while(1){getchar();printf("enter string OR press enter to stop\n");scanf("%[^\n]",line);c=strlen(line);if(c==0)break;elsenc+=strlen(line);for(i=0;i


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printf("\nNo of charactors:%d",nc);printf("\nNo of words:%d",nw);getch();

}

Flowchart:

True

Yes

No

No

Yes

No

Yes

Start

nw = ne = nl = 0

While (1)

c = strlen(line)

If c=

=0

nc = nc + c

for i = 0 to c

Ifline[i]=spaceor line[i]=\0

nw = nw + 1

nl = nl + 1

Print no of lines nl, no of characters c, no of words nw

Input line


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Output:

Enter the text, leave one space between the words and press enter at the end of the text.

Enter string OR press enter to stopCAN I SPEAK TO Mr.PRASAD.Enter string OR press enter to stop

YA THIS IS PRASAD.Enter string OR press enter to stopMAY I KNOW HOW R UEnter string OR press enter to stopTHIS IS RAMU.Enter string OR press enter to stop

No of lines:4No of charactors:74No of words:18

Exercise 12

a). Write a C program to generate Pascals triangle.

Algorithm

Stop


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Algorithm: main()

1. Input size of the Pascal Triangle rows / columns m.

2. Call the function ps(pas, m) (to generate Pascal Triangle).

3. Print Pascal Triangle.

4. Call prps(pas, m).

5. Stop.

Algorithm: ps(pas, m)

1. For i=0 to m

2. Pas[i][0] = 1.

3. Pas[i][i] = 1.

4. For j = 0 to i 1.

5. Pas[i][j] = pas[i-1][j-1] + pas[i-1][j].

6. Next j.

7. Next i.

8. Return.

Algorithm: prps()

1. For I = 0 to m.

2. For j = 0 to i.

3. Print pas[i][j].

4. Next j.

5. Next i.

6. Return.

Program:


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#include#include#include#define MR 20#define ML 30main(){

int m,n,pas[MR][ML];

void ps(int pas[][ML],int m);void prps(int pas[][ML],int m);clrscr();

printf("Enter No of Rows/Columns");scanf("%d%d",&m,&n);

ps(pas,m);

printf("\nPascal Triangle\n");

prps(pas,m);getch();}/*end of main function*/void ps(int pas[][ML],int m){

int i,j;for(i=0;i


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False

True

False

True

Flowchart: ps(pas, m)

False

True

False

True

Output:

Enter No of Rows/Columns5 5

Pascal Triangle11 11 2 11 3 3 1

Start

for i = 0 to m

pas[i][0] = 1; pas[i][i] = 1

for j = 1 to i -1

Pas[i][i] = pas[i-1][j-1] + pas[i-1][j]

return


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1 4 6 4 1

b). Write a C program to construct a Pyramid of numbers.

Algorithm:

1. Begin.

2. For i=1 to 10 repeat the following steps

(i) K=1

(ii) For s=1 to 10-I repeat the following steps

Print .

(iii) For j=1 to I repeat the following steps

(a) R=k%10.

(b) Print r.


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(c) K=k+1.

(iv) K=k-1.

(v) For j=1 to I repeat the following steps

(a) K=k-1

(b) If k==-1 then

K=10

(c) R=k%10

(d) Print r.

(vi) Goto next line.

3. Stop.

Program:

#include#include

#includemain(){

int i,j,k,r,s;clrscr();for(i=1;i


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k=1;for(s=1;s


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False

True

No

yes

Output:

1

121123211234321

12345432112345654321

1234567654321123456787654321

123456789876543211234567890987654321

k=k-1

for j=1 to i

k=k+1

If

k==-

1

r=k%10

k=10

Print r

Stop

Print \n


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Exercise 16

a) Write a C program which copies one file to anotherAlgorithm:

1. Input the program source destination file.

As command line prompt (argv)

2. Check no of parameters ie argc

If argc !=3 goto step 10

3. Copy source file name into sfname

4. Copy destination file into dfname.

5. Open the file ie source file in read mode.

6. Check source file opening of file .

If not, print file cant be opened goto step 10, otherwise goto step 7.

7. Perform till the end of the file reading the following

(i) Read a character from a source file.


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(ii) Write to destination file.

8. Close source file.

9. Close destination file.

10. Stop.

Program:

#include#include#include#include

#define L 80main(int argc, char *argv[]){

FILE *sfpt,*dfpt;char c;char sfname[L],dfname[L];clrscr();if(argc!=3){

printf("Invalid No of parameters.");exit(1);

}strcpy(sfname,argv[1]);strcpy(dfname,argv[2]);sfpt=fopen(sfname,"r");if(sfpt==NULL)

printf("File can't be opend %s",sfname);else


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{dfpt=fopen(dfname,"w");if(dfpt==NULL)

printf("File can't be found: %s",sfname);else{

while(!feof(sfpt)){

c=getc(sfpt);

putc(c,dfpt);}

}}fclose(sfpt);fclose(dfpt);getch();

}

Flowchart:

Yes

No

Yes

No

Yes

No

No

Start

C:\> filename sfiledfile

If argc!=3

Copy argv[1] to sfileCopy argv[2] to dfile

Open sfile in readmode

Ifsfpt==NULL

Open dfile

Ifdfpt==NULL

While !feop(sfpt)

Read c from sfptWrite c into dfpt

Close sfile

Stop

Close dfile


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Yes

W12a.Output:


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b) Write a C program to reverse the first n characters in a file. (Note: The file name and n arespecified on the command line)

Algorithm:

1. Input C:\> filename, sourcefile.

2. if argc!=3 then goto step (6)

3. Copy argv[1] to sfname

4. Open sfname in read mode

5. if sfpt = = NULL

6. Print file not opened, otherwise goto step (7)

7. Convert string to integer ie i = a to i(argv[2])

8. Read source file buffer ie sfpt into text up to I no of characters.

9. text[n] = \0( NULL Character assignment)

10.j = 1

11.rev[j] = \0

12.j = j 1

13.for i = 0 to l

a. rev [j] = text[i]

b. j = j l

14. Print reversed string : rev


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15. Close file sfpt

16. Stop.

Program:#include#include#include#include

#include#define L 800main(int argc, char *argv[]){

int l,i,j;char text[L];char rev[L];FILE *sfpt;char c;char sfname[L];int n;clrscr();if(argc!=3){

printf("Invalid No of arguments.\n");exit(0);

}strcpy(sfname,argv[1]);sfpt=fopen(sfname,"r");if(sfpt==NULL){

printf("File not found.\n");exit(1);}else{

i=atoi(argv[2]);n=fread(text,1,i,sfpt);text[n]='\0';l=strlen(text);


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j=l;rev[j]='\0';

j--;for(i=0;i


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Flowchart:

Yes

No

Yes

No

False

True

Start

C:\> filename sfiledfile

if argc!=3

Copy argv[1] to sfname

Open sfname in readmode

ifsfpt==NULL

Convert argv[2] tointeger

Print file notfound

n = fread(text, l,

sfpt)

text[n]=\0

l=length(text)

j=1

rev[j]=\0

j = j + 1

for i=0;i


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Output:

Documents

Cp Lab Manual 2010