While iterative control structure:
It is the entry controlled iterative control structure.
Condition is checked while entering into the loop.
Body of the loop executes continuously until the condition
is false.
It is the most commonly used control structure in any
language.
 |
Specification1:
Print the natural
numbers from 1 to n
Program:
#include<stdio.h>
|
void main()
{
int n,i;
printf("Enter
the limit:");
scanf("%d",&n);
i=1;
while(i<=n)
{
printf("%10d",i);
i=i+1;
}
}
Execution:
Enter the
limit: 5
1 2 3 4 5
Specification2:
Print the natural numbers from n to 1
Program:
#include<stdio.h>
void main()
|
{
int n,i;
printf("Enter the limit:");
scanf("%d",i);
i=n;
while(i>=1)
{
printf("%10d",i);
i=i-1;
}
}
Execution:
Enter the limit: 5
5 4 3 2 1
Specification3:
Print the natural number with in the range.
Program:
#include<stdio.h>
void main()
{
int ll,ul,i;
printf("Enter the lower limit:");
scanf("%d",&ll);
|
printf("Enter the upper limit:");
scanf("%d",&ul);
i=ll;
while(i<=ul)
{
printf("%10d",i);
i++;
}
}
Execution:
Enter the lower limit: 5
Enter the upper limit: 9
5 6 7 8 9
Specification4:
Print the sum of natural numbers from 1 to n
Program:
#include<stdio.h>
|
void main()
{
int n,i,sum;
printf("Enter
the limit:");
scanf("%d",&n);
sum=0;
i=1;
while(i<=n)
{
sum=sum+i;
i++;
}
printf("The
sum of natural numbers %d",sum);
}
Execution:
Enter the limit: 5
The sum of natural numbers 15
Example explained:
Using i=1, i<=n, i=i+1 we are generating natural numbers
from 1 to n.
In order to find the sum of natural numbers, adding "i"
to another variable "sum" every time the value of "i" is
changed.
Finally variable "sum" has the sum of natural
numbers that we are printing as output.
Specification5:
Accept any number and print its factorial
Program:
#include<stdio.h>
void main()
{
int i,n;
long int fact;
printf("Enter an integer:");
scanf("%d",&n);
fact=1;
i=1;
while(i<=n)
{
fact=fact*i;
i++;
}
printf("Factorial of
%d is %ld",n,fact);
}
Execution:
Enter an integer: 5
Factorial of 5 is 120
Example explained:
Using i=1, i<=n, i++ we are generating natural numbers
from 1 to n.
In order to find the factorial multiplying "i" to
another variable "fact" every time the value of "i" is
changed.
Finally variable "fact" has the factorial that we
are printing.
Specification6:
Print the even numbers form 1 to n
Program:
#include<stdio.h>
void main()
{
int n,i;
printf("Enter the limit:");
scanf("%d",&n);
i=1;
while(i<=n)
{
if(i%2==0)
printf("%10d",i);
i++;
}
}
Execution:
Enter the limit: 10
2 4 6 8 10
Example explained:
Using i=1, i<=n, i++ we are generating natural numbers
from 1 to n.
But printing only when the value of "i" is an even
number by checking if(i%2==0).
Specification7:
Accept any number and print its factors.
Program:
#include<stdio.h>
void main()
{
int i,n;
printf("Enter any integer:");
scanf("%d",&n);
i=1;
while(i<=n)
{
if(n%i==0)
printf("%10d",i);
i=i+1;
}
}
Execution:
Enter any integer: 15
1 3 5 15
Example explained:
Using i=1, i<=n, i++ we are generating natural numbers
from 1 to n.
But printing only when "i" divides "n"
with reminder zero (factor)
Specification8:
Accept any number and print the sum of all the digits.
Logic:
The modulo division operator (%) return the last digit when
we divide by 10;
n=n/10 removes the last digit. For example if n is 345, n/10
is 34.5 and assigning to integer variable "n" results 34 is stored
into "n".
It is only possible if "n" is of integer type.
Example1:
if "n" is a three digit number.
Example2: if "n" is a four digit number.
 |
Summery:
Extracting individual digits one by one using n%10, n=n/10
until "n" becomes 0
while(n!=0)
{
n%10;
n=n/10;
}
Program:
#include<stdio.h>
void main()
{
int n,sum;
printf("Enter any integer:");
scanf("%d",&n);
sum=0;
while(n!=0)
{
sum=sum+n%10;
n=n/10;
}
printf("Sum of all the digits %d",sum);
}
Execution1:
Enter any integer: 345
Sum of all the digits 12
Execution2:
Enter any integer: 4536
Sum of all the digits 18
Specification9:
Accept any number and print it in reverse.
Logic:
rev=0;
while(n!=0)
{
rev=rev*10+n%10;
n=n/10;
}
Program:
#include<stdio.h>
void main()
{
int n,rev;
printf("Enter any number:");
scanf("%d",&n);
rev=0;
while(n!=0)
{
rev=rev*10+n%10;
n=n/10;
}
printf("Reverse number %d",rev);
}
Execution1:
Enter any number: 4567
Reverse number 7654
0 comments:
Post a Comment