1. Write a program to find factorial of the
given number.
Recursion: A function is called' recursive 'if a
statement within the body of a function calls the same function. It is also called' circular definition '. Recursion is
thus a process of defining something in terms of itself.
Program: To calculate the factorial value using
recursion.
#include
int fact(int n);
int main(){
int x, i;
printf("En ter a value for x: \n");
scanf("%d" ,&x);
i = fact(x);
printf("\n Factorial of %d is %d", x,
i);
return 0;
}int fact(int n){
/* n=0 indicates a terminatin g condition */
if (n
return (1);
}else{
/* function calling itself */
return (n * fact(n - 1));
/*n*fact(n -1) is a recursive expression */
}
}
Output:
Enter a value for x:
4
Factorial of 4 is 24
Explanatio n:
fact(n) = n * fact(n-1)
If n=4
fact(4) = 4 * fact(3) there is a call to fact(3)
fact(3) = 3 * fact(2)
fact(2) = 2 * fact(1)
fact(1) = 1 * fact(0)
fact(0) = 1
fact(1) = 1 * 1 = 1
fact(2) = 2 * 1 = 2
fact(3) = 3 * 2 = 6
Thus fact(4) = 4 * 6 = 24 Terminating condition( n infinite loop.
2. Write a program to check whether the given
number is even or odd.
Program:
#include
int main(){
int a;
printf("En ter a: \n");
scanf("%d" ,&a);
/* logic */
if (a % 2 == 0){
printf("Th e given number is EVEN\n");
}
else{
printf("Th e given number is ODD\n");
}
return 0;
}
Output:
Enter a: 2
The given number is EVEN
Explanatio n with examples:
Example 1: If entered number is an even number
Let value of'a'entered is 4 if(a%2==0) then a is an even number, else odd.
i.e. if(4%2==0) then 4 is an even number, else
odd.
To check whether 4 is even or odd, we need to
calculate (4%2).
/* % (modulus) implies remainder value. */
/* Therefore if the remainder obtained when 4 is
divided by 2 is 0, then 4 is even. */
4%2==0 is true
Thus 4 is an even number.
Example 2: If entered number is an odd number.
Let value of'a'entered is 7
if(a%2==0) then a is an even number, else odd.
i.e. if(7%2==0) then 4 is an even number, else
odd.
To check whether 7 is even or odd, we need to
calculate (7%2).
7%2==0 is false /* 7%2==1 condition fails and
else part is executed */
Thus 7 is an odd number.
3. Write a program to swap two numbers using a
temporary variable.
Swapping interchanges the values of two given
variables.
Logic:
step1: temp=x;
step2: x=y;
step3: y=temp;
Example:
if x=5 and y=8, consider a temporary variable
temp.
step1: temp=x=5;
step2: x=y=8;
step3: y=temp=5;
Thus the values of the variables x and y are
interchanged.
Program:
#include
int main(){
int a, b, temp;
printf("En ter the value of a and b:
\n");
scanf("%d %d",&a,&b);
printf("Be fore swapping a=%d, b=%d
\n", a, b);
/*Swapping logic */
temp = a;
a = b;
b = temp;
printf("Af ter swapping a=%d, b=%d",
a, b);
return 0;
}
Output:
Enter the values of a and b: 2 3
Before swapping a=2, b=3
After swapping a=3, b=2
4. Write a program to swap two numbers without
using a temporary variable. Swapping interchanges the values of two given
variables.
Logic:
step1: x=x+y;
step2: y=x-y;
step3: x=x-y;
Example:
if x=7 and y=4
step1: x=7+4=11;
step2: y=11-4=7;
step3: x=11-7=4;
Thus the values of the variables x and y are
interchanged.
Program:
#include
int main(){
int a, b;
printf("En ter values of a and b:
\n");
scanf("%d %d",&a,&b);
printf("Be fore swapping a=%d, b=%d\n",
a,b);
/*Swapping logic */
a = a + b;
b = a - b;
a = a - b;
printf("Af ter swapping a=%d b=%d\n",
a, b);
return 0;
}
Output:
Enter values of a and b: 2 3
Before swapping a=2, b=3
The values after swapping are a=3 b=2
5. Write a program to swap two numbers using
bitwise operators.
Program:
#include
int main(){
int i = 65;
int k = 120;
printf("\n value of i=%d k=%d before
swapping", i, k);
i = i ^ k;
k = i ^ k;
i = i ^ k;
printf("\n value of i=%d k=%d after
swapping", i, k);
return 0;
}
Explanation:
i = 65; binary equivalent of 65 is 0100 0001
k = 120; binary equivalent of 120 is 0111 1000
i = i^k;
i...0100 0001
k...0111 1000
---------
val of i = 0011 1001
---------
k = i^k
i...0011 1001
k...0111 1000
---------
val of k = 0100 0001 binary equivalent of this
is 65
---------( that is the initial value of i)
i = i^k
i...0011 1001
k...0100 0001
---------
val of i = 0111 1000 binary equivalent of this
is 120
--------- (that is the initial value of k)
6. Write a program to find the greatest of three
numbers.
Program:
#include
int main(){
int a, b, c;
printf("Enter a,b,c: \n");
scanf("%d %d
%d",&a,&b,&c);
if (a>b&&a>c){
printf("a is Greater than b and c");
}
else if (b>a&&b>c){
printf("b is Greater than a and c");
}
else if (c>a&&c>b){
printf("c is Greater than a and b");
}
else{
printf("all are equal or any two values
are equal");
}
return 0;
}
Output:
Enter a,b,c: 3 5 8
c is Greater than a and b
Explanation with examples:
Consider three numbers a=5,b=4,c= 8
if(a>b&&a>c) then a is greater
than b and c
now check this condition for the three numbers
5,4,8 i.e.
if(5>4&&5>8) /* 5>4 is true but
5>8 fails */
so the control shifts to else if condition
else if(b>a&&b>c) then b is
greater than a and c
now checking this condition for 5,4,8 i.e.
else if(4>5&&4>8) / * both the
conditions fail */
now the control shifts to the next else if
condition
else if(c>a&&c>b) then c is
greater than a and b
now checking this condition for 5,4,8 i.e.
else if(8>5&&8>4) / * both
conditions are satisfied */
Thus c is greater than a and b.
7. Write a program to find the greatest among
ten numbers.
Program:
#include
int main(){
int a[10];
int i;
int greatest;
printf("Enter ten values:");
//Store 10 numbers in an array
for (i = 0; i<10; i++){
scanf("%d" ,&a[i]);
}
//Assume that a[0] is greatest
greatest = a[0];
for (i = 0; i<10; i++){
if (a[i]>greatest){
greatest = a[i];
}
}
printf("\n Greatest of ten numbers is
%d", greatest);
return 0;
}
Output:
Enter ten values: 2 53 65 3 88 8 14 5 77 64
Greatest of ten numbers is 88
Explanation with example:
Entered values are 2, 53, 65, 3, 88, 8, 14, 5,
77, 64
They are stored in an array of size 10. let a[]
be an array holding these values.
/* how the greatest among ten numbers is found
*/
Let us consider a variable'greatest' . At the
beginning of the loop, variable'greatest' is assinged with the value of first element in the array greatest=a [0]. Here
variable'greatest' is assigned 2 as a[0]=2.
Below loop is executed until end of the
array'a[]';.
for(i=0; i
{
if(a[i]>gr eatest)
{
greatest= a[i];
}
}
For each value of'i', value of a[i] is compared
with value of variable'greatest' . If any value greater than the value of'greatest' is encountered, it would be
replaced by a[i]. After completion of'for'loop, the value of variable 'greatest' holds the greatest number in the
array. In this case 88 is the greatest of all the numbers.
8. Write a program to check whether the given
number is a prime.
A prime number is a natural number that has only
one and itself as factors. Examples: 2, 3, 13 are prime
numbers.
Program:
#include
main(){
int n, i, c = 0;
printf("Enter any number n: \n");
scanf("%d" ,&n);
/*logic*/
for (i = 1; i
if (n % i == 0){
c++;
}
}
if (c == 2){
printf("n is a Prime number");
}
else{
printf("n is not a Prime number");
}
return 0;
}
Output:
Enter any number n: 7
n is Prime
Explanation with examples:
consider a number n=5
for(i=0;i
i.e. for(i=0;i
1st iteration: i=1;i
here i is incremented i.e. i value for next
iteration is 2
now if(n%i==0) then c is incremented
i.e.if(5%1 ==0)then c is incremented, here
5%1=0 thus c is incremented.
now c=1;
2nd iteration: i=2;i
here i is incremente d i.e. i value for next
iteration is 3
now if(n%i==0) then c is incremented
i.e.if(5%2 ==0) then c is incremented, but
5%2!=0 and so c is not incremented, c remains 1
c=1;
3rd iteration: i=3;i
here i is incremented i.e. i value for next
iteration is 4
now if(n%i==0) then c is incremented
i.e.if(5%3 ==0) then c is incremented, but
5%3!=0 and so c is not incremented, c remains 1
c=1;
4th iteration: i=4;i
here i is incremente d i.e. i value for next
iteration is 5
now if(n%i==0) then c is incremented
i.e. if(5%4==0) then c is incremented, but
5%4!=0 and so c is not incremented, c remains 1
c=1;
5th iteration: i=5;i Here i is incremented i.e. i value for next
iteration is 6 now if(n%i==0) then c is incremented i.e. if(5%5==0) then c is incremented, 5%5=0
and so c is incremented. i.e. c=2 6th iteration: i=6;i here i value is 6 and 6 now if(c==2) then n is a prime number we have c=2 from the 5th iteration and thus n=5
is a Prime number.
9. Write a program to check whether the given
number is a palindromi c number.
If a number, which when read in both forward and
backward way is same, then such a number is called a palindrome number.
Program:
#include
int main(){
int n, n1, rev = 0, rem;
printf("En ter any number: \n");
scanf("%d" ,&n);
n1 = n;
/* logic */
while (n>0){
rem = n % 10;
rev = rev * 10 + rem;
n = n / 10;
}
if (n1 == rev){
printf("Gi ven number is a palindromi c
number");
}
else{
printf("Gi ven number is not a palindromi c
number");
}
return 0;
}
Output:
Enter any number: 121
Given number is a palindrome
Explanatio n with an example:
Consider a number n=121, reverse=0, remainder;
number=121
now the while loop is executed /* the condition
(n>0) is satisfied */
/* calculate remainder */
remainder of 121 divided by 10=(121%10 )=1;
now reverse=(r everse*10) +remainder
=(0*10)+1 / * we have initialized reverse=0 */
=1
number=num ber/10
=121/10
=12
now the number is 12, greater than 0. The above
process is repeated for number=12.
remainder= 12%10=2;
reverse=(1 *10)+2=12;
number=12/ 10=1;
now the number is 1, greater than 0. The above
process is repeated for number=1.
remainder= 1%10=1;
reverse=(1 2*10)+1=12 1;
number=1/ 10 / * the condition n>0 is not
satisfied,co ntrol leaves the while loop */
Program stops here. The given number=121 equals
the reverse of the number. Thus the given number is a palindrome number.
10.Write a program to check whether the given
string is a palindrome.
Palindrome is a string, which when read in both
forward and backward way is same.
Example: radar, madam, pop, lol, rubber, etc.,
Program:
#include
#include
int main(){
char string1[20 ];
int i, length;
int flag = 0;
printf("En ter a string: \n");
scanf("%s" , string1);
length = strlen(str ing1);
for(i=0;i<length ;i++){
if(string1 [i] != string1[le ngth-i-1]) {
flag = 1;
break;
}
}
if (flag){
printf("%s is not a palindrome \n",
string1);
}
else{
printf("%s is a palindrome \n",
string1);
}
return 0;
}
Output:
Enter a string: radar
"radar"is a palindrome Explanation with example:
To check if a string is a palindrome or not, a
string needs to be compared with the reverse of itself. Consider a palindrome string:"radar",
---------- ---------- -------
index: 0 1 2 3 4
value: r a d a r
---------- ---------- -------
To compare it with the reverse of itself, the
following logic is used:
0th character in the char array, string1 is same
as 4th character in the same string.
1st character is same as 3rd character.
2nd character is same as 2nd character.
. . . .
ith character is same as'length-i- 1'th
character.
If any one of the above condition fails, flag is
set to true(1), which implies that the string is not a palindrome .
By default, the value of flag is false(0).
Hence, if all the conditions are satisfied, the string is a palindrome .
11.Write a program to generate the Fibonacci
series.
Fibonacci series: Any number in the series is
obtained by adding the previous two numbers of the series.
Let f(n) be n'th term.
f(0)=0;
f(1)=1;
f(n)=f(n-1 )+f(n-2); (for n>=2)
Series is as follows
011
(1+0)
2 (1+1)
3 (1+2)
5 (2+3)
8 (3+5)
13 (5+8)
21 (8+13)
34 (13+21)
...and so on
Program: to generate Fibonacci Series(10 terms)
#include
int main(){
//array fib stores numbers of fibonacci series
int i, fib[25];
// initialized first element to 0
fib[0] = 0;
// initialized second element to 1
fib[1] = 1;
//loop to generate ten elements
for (i = 2; i<10; i++){
//i'th element of series is equal to the sum of
i-1'th element and i-2'th element.
fib[i] = fib[i - 1] + fib[i - 2];
}
printf("Th e fibonacci series is as follows
\n");
//print all numbers in the series
for (i = 0; i<10; i++){
printf("%d \n", fib[i]);
}
return 0;
}
Output:
The fibonacci series is as follows
01123581
3
21
34
Explanatio n:
The first two elements are initialize d to 0, 1
respectively. Other elements in the series are generated by looping and adding previous two numbers. These numbers
are stored in an array and ten elements of the series are printed as output.
12.Write a program to print"Hello
World"without using semicolon anywhere in the code.
Generally when we use printf("")
statement, we have to use a semicolon at the end. If printf is used inside an
if Condition, semicolon can be avoided.
Program: Program to print something without
using semicolon (;)
#include
int main(){
//printf returns the length of string being
printed
if (printf("H ello World\n")) //prints
Hello World and returns 11
{
//do nothing
}
return 0;
}
Output:
Hello World
Explanation:
The if statement checks for condition whether
the return value of printf("He llo World") is greater than 0. printf function returns the length of the string
printed. Hence the statement if (printf("H ello World")) prints the
string
"Hello World".
13.Write a program to print a semicolon without
using a semicolon anywhere in the code.
Generally when use printf("")
statement we have to use semicolon at the end.
If we want to print a semicolon, we use the
statement: printf(";" ); In above statement, we are using two
semicolons. The task of printing a semicolon without using semicolon
anywhere in the code
can be accomplish ed by using the ascii value of';'which is equal to 59.
Program: Program to print a semicolon without
using semicolon in the code.
#include
int main(void) {
//prints the character with ascii value 59,
i.e., semicolon
if (printf("% c\n", 59)){
//prints semicolon
}
return 0;
}
Output:
;
Explanatio n:
If statement checks whether return value of
printf function is greater than zero or not. The return value of function call printf("%c ",59) is 1. As printf
returns the length of the string printed. printf("%c ",59) prints
ascii value that
correspond s to 59, that is semicolon(.
14.Write a program to compare two strings
without using strcmp() function.
strcmp() function compares two strings
lexicograp hically. strcmp is declared in stdio.h
Case 1: when the strings are equal, it returns
zero.
Case 2: when the strings are unequal, it returns
the difference between ascii values of the characters that differ.
a) When string1 is greater than string2, it
returns positive value.
b) When string1 is lesser than string2, it
returns negative value.
Syntax:
int strcmp (const char *s1, const char *s2);
Program: to compare two strings.
#include
#include
int cmpstr(cha r s1[10], char s2[10]);
int main(){
char arr1[10] ="Nodalo";
char arr2[10] ="nodalo";
printf("%d", cmpstr(arr 1, arr2));
// cmpstr() is equivalent of strcmp()
return 0;
}/
/s1, s2 are strings to be compared
int cmpstr(cha r s1[10], char s2[10]){
//strlen function returns the length of argument
string passed
int i = strlen(s1) ;
int k = strlen(s2) ;
int bigger;
if (i<k){
bigger = k;
}
else if (i>k){
bigger = i;
}
else{
bigger = i;
}
//loops'bigger'times
for (i = 0; i<bigger; i++){
// if ascii values of characters s1[i], s2[i]
are equal do nothing
if (s1[i] == s2[i]){
}
//else return the ascii difference
else{
return (s1[i] - s2[i]);
}
}
//return 0 when both strings are same
//This statement is executed only when both
strings are equal
return (0);
}
Output:
-32 Explanatio n:
cmpstr() is a function that illustrate s C
standard function strcmp(). Strings to be compared are sent as arguments to cmpstr().
Each character in string1 is compared to its
correspond ing character instring2. Once the loop encounters adiffering character in the strings, it would
return the ascii difference of the differing characters and exit.
15.Write a program to concatenat e two strings
without using strcat() function.
strcat(str ing1,strin g2) is a C standard
function declared in the header file string.h
The strcat() function concatenat es string2,
string1 and returns string1.
Program: Program to concatenat e two strings
#include
#include
char *strct(cha r *c1, char *c2);
char *strct(cha r *c1, char *c2){
//strlen function returns length of argument
string
int i = strlen(c1) ;
int k = 0;
// loops until null is encountered and appends
string c2 to c1
while (c2[k] !='\0'){
c1[i + k] = c2[k];
k++;
}
return c1;
}
int main(){
char string1[15 ] ="first";
char string2[15 ] ="second";
char *finalstr;
printf("Be fore concatenat ion:"
"\n string1 = %s \n string2 = %s",
string1, string2);
// addresses of string1, string2 are passed to
strct()
finalstr = strcat(str ing1, string2);
printf("\n After concatenat ion:");
//prints the contents of string whose address is
in finalstr
printf("\n finalstr = %s", finalstr);
//prints the contents of string1
printf("\n string1 = %s", string1);
//prints the contents of string2
printf("\n string2 = %s", string2);
return 0;
}
Output:
Before concatenat ion:
string1 = first
string2 = second
After concatenat ion:
finalstr = firstsecon d
string1 = firstsecon d
string2 = second
Explanatio n:
string2 is appended at the end of string1 and
contents of string2 are unchanged. In strct() function, using a for loop, all the
characters of string'c2'are copied at the end of c1. return (c1) is equivalent to return&c1[0] and it returns
the base address of'c1'.'finalstr' stores that address returned by the function strct().
