CHAPTER 5: Programs
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.
/*Program of finding factorial using recursion*/
#include <stdio.h>
int fact(int n);
int main()
int fact(int n);
int main()
{
int x, i;
printf("Enter a value for x: \n");
scanf("%d", &x);
i = fact(x);
printf("\nFactorial of %d is %d", x, i);
return 0;
}
int x, i;
printf("Enter a value for x: \n");
scanf("%d", &x);
i = fact(x);
printf("\nFactorial of %d is %d", x, i);
return 0;
}
int fact(int n)
{
/* n=0 indicates a terminating condition */
if (n = =1)
/* n=0 indicates a terminating condition */
if (n = =1)
{
return (1);
} else {
/* function calling itself */
return (n * fact(n - 1));
/*n*fact(n-1) is a recursive expression */
}
}
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
Enter a value for x: 4
Factorial of 4 is 24
Explanation:
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 == 1 here;) is a must for a recursive program. Otherwise the program enters into an infinite loop.
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 == 1 here;) is a must for a recursive program. Otherwise the program enters into an infinite loop.
To calculate the factorial value without using recursion.
/*Program of finding factorial without using recursion*/
#include <stdio.h>
int fact(int n);
int main()
int fact(int n);
int main()
{
int x, i;
printf("Enter a value for x: \n");
scanf("%d", &x);
i = fact(x);
printf("\nFactorial of %d is %d", x, i);
return 0;
}
int x, i;
printf("Enter a value for x: \n");
scanf("%d", &x);
i = fact(x);
printf("\nFactorial of %d is %d", x, i);
return 0;
}
int fact(int n)
{
int f=1,j;
for(j=n;j>=1;j--)
{
f=f*j;
}
return(f)
}
Output:
Enter a value for x: 4
Enter a value for x: 4
Factorial of 4 is 24
2. Write a program to check whether the given number is even or odd.
Program:
#include <stdio.h>
int main()
{
int a;
printf("Enter a: \n");
scanf("%d", &a);
/* logic */
if (a % 2 == 0)
int a;
printf("Enter a: \n");
scanf("%d", &a);
/* logic */
if (a % 2 == 0)
{
printf("The given number is EVEN\n");
}
else
printf("The given number is EVEN\n");
}
else
{
printf("The given number is ODD\n");
}
return 0;
}
printf("The given number is ODD\n");
}
return 0;
}
Output:
Enter a: 2
The given number is EVEN
Enter a: 2
The given number is EVEN
Explanation 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).
Note That: % (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.
(Asked in Robert Bosch,Infosys,Samsung)
Swapping interchanges the values of two given variables.
Logic:
step1: temp=x;
step2: x=y;
step3: y=temp;
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;
step2: x=y=8;
step3: y=temp=5;
Thus the values of the variables x and y are interchanged.
Program:
#include <stdio.h>
int main() {
int a, b, temp;
printf("Enter the value of a and b: \n");
scanf("%d %d", &a, &b);
printf("Before swapping a=%d, b=%d \n", a, b);
/*Swapping logic */
temp = a;
a = b;
b = temp;
printf("After 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
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.
(Asked in Robert Bosch,Infosys,Samsung)
(Asked in Robert Bosch,Infosys,Samsung)
Swapping interchanges the values of two given variables.
Logic:
step1: x=x+y;
step2: y=x-y;
step3: 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;
step3: x=11-7=4;
Thus the values of the variables x and y are interchanged.
Program:
#include <stdio.h>
int main() {
int a, b;
printf("Enter values of a and b: \n");
scanf("%d %d", &a, &b);
printf("Before swapping a=%d, b=%d\n", a,b);
/*Swapping logic */
a = a + b;
b = a - b;
a = a - b;
printf("After 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 <stdio.h>
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)
---------
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 <stdio.h>
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
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.
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 <stdio.h>
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("\nGreatest 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<10; i++)
{
if(a[i]>greatest)
{
greatest= a[i];
}
}
{
if(a[i]>greatest)
{
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 <stdio.h>
main() {
int n, i, c = 0;
printf("Enter any number n: \n");
scanf("%d", &n);
/*logic*/
for (i = 1; i <= n; 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
Enter any number n: 7
n is Prime
Explanation with examples:
consider a number n=5
for(i=0;i<=n;i++) /* for loop is executed until the n value equals i */
i.e. for(i=0;i<=5;i++) /* here the for loop is executed until i is equal to n */
consider a number n=5
for(i=0;i<=n;i++) /* for loop is executed until the n value equals i */
i.e. for(i=0;i<=5;i++) /* here the for loop is executed until i is equal to n */
1st iteration: i=1;i<=5;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;
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<=5;i++
here i is incremented 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;
here i is incremented 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<=5;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 ic incremented, but 5%3!=0 and so c is not incremented, c remains 1
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 ic incremented, but 5%3!=0 and so c is not incremented, c remains 1
c=1;
