Loops :
Loop
is defined as a block of statements and it is executed repeatedly until the
condition becomes false.
(or)
A loop is defines as
a block of statements which are executed repeatedly for certain number of times.
Steps involved in a loop:
1) Loop variable: Declaring a variable which
is used in the loop.
2) Initialization: it is the first step in
which starting and final value is assigned to the loop variable .each time the
updated value is checked by it self.
3)
Test condition: Condition which is to be checked until it becomes false.
4) Iteration (Incrementation/Decrementation):
It is a Numerical value either it is increased or decreased the value of the
variable.
C supports 3 types of loops.
They are:
- While loop
- for loop
- do while loop
While loop
|
For loop
|
Do while loop
|
1.While is an entry control loop.
2. In while loop statements are executed after the test
condition is checked.
|
1. For is an entry control loop.
2.In for loop statements are executed after the test
condition is checked.
3.In while loop Initialization, test condition and
iteration parts are placed in different places where as in for loop Initialization,
test condition and iteration parts are placed in single statement.
4.some time forgetting Initialization, or iteration
parts are in while loop where as in for
loop there is no chance to forgetting.
|
1. Do while is an exit control loop.
2. In do while loop the test condition is checked after
statements are executed once.
3.do while loops are exit control loops as the
condition checks at end even though the condition is false but do while loops
are executed at once.
4.do while loops are used to menu given
programmes.
|
/*1.write a program to demonstrate while loop. */
|
/*1.write a program to demonstrate for loop.*/
|
/*1.write a program to demonstrate do-while loop.*/
|
#include<stdio.h>
#include<conio.h>
void
main()
{
int i;
clrscr();
i=1;
while(i<=5)
{
printf(“\nc-language”);
i++;
}
getch();
}
Result:
Output:
c-language
c-language
c-language
c-language
c-language
|
#include<stdio.h>
#include<conio.h>
void
main()
{
int i;
clrscr();
for(i=1;i<=5;i++)
{
printf(“\nc-language”);
}
getch();
}
Result:
Output:
c-language
c-language
c-language
c-language
c-language
|
#include<stdio.h>
#include<conio.h>
void
main()
{
int i;
clrscr();
i=1;
do
{
printf(“\nc-language”);
i++;
}
while(i<=5);
getch();
}
Result:
Output:
c-language
c-language
c-language
c-language
c-language
|
i=1
1<=5(T)
display c-language
i=2
2<=5(T)
display c-language
i=3
3<=5(T)
display c-language
i=4
4<=5(T)
display c-language
i=5
5<=5(T)
display c-language
i=6
6<=5(F)
Exit
from the loop
|
i=1
1<=5(T)
display c-language
i=2
2<=5(T)
display c-language
i=3
3<=5(T)
display c-language
i=4
4<=5(T)
display c-language
i=5
5<=5(T)
display c-language
i=6
6<=5(F)
Exit
from the loop
|
i=1
display c-language
i=2
2<=5(T)
display c-language
i=3
3<=5(T)
display c-language
i=4
4<=5(T)
display c-language
i=5
5<=5(T)
display c-language
i=6
6<=5(F)
Exit
from the loop
|
/*2.C program to display c-
language n times using while
loop */
|
/*2.C program to display c-
language n times using
for loop */
|
/*2.C program to display c-
language n times using
do-
while loop.*/
|
#include<stdio.h>
#include<conio.h>
void
main()
{
inti,n;
clrscr();
printf(“enter
the value of n:”);
scanf(“%d”,&n);
i=1;
while(i<=n)
{
printf(“\nc-language”);
i++;
}
getch();
}
Result:
input:
enter
the value of n:5
Output:
c-language
c-language
c-language
c-language
c-language
|
#include<stdio.h>
#include<conio.h>
void
main()
{
int i;
clrscr();
printf(“enter
the value of n:”);
scanf(“%d”,&n);
for(i=1;i<=n;i++)
{
printf(“\nc-language”);
}
getch();
}
Result:
input:
enter
the value of n:5
Output:
c-language
c-language
c-language
c-language
c-language
|
#include<stdio.h>
#include<conio.h>
void
main()
{
int i;
clrscr();
printf(“enter
the value of n:”);
scanf(“%d”,&n);
i=1;
do
{
printf(“\nc-language”);
i++;
}
while(i<=n);
getch();
}
Result:
input:
enter
the value of n:5
Output:
c-language
c-language
c-language
c-language
c-language
|
i=1
1<=5(T)
display c-language
i=2
2<=5(T)
display c-language
i=3
3<=5(T)
display c-language
i=4
4<=5(T)
display c-language
i=5
5<=5(T)
display c-language
i=6
6<=5(F)
Exit
from the loop
|
i=1
1<=5(T)
display c-language
i=2
2<=5(T)
display c-language
i=3
3<=5(T)
display c-language
i=4
4<=5(T)
display c-language
i=5
5<=5(T)
display c-language
i=6
6<=5(F)
Exit
from the loop
|
i=1
display c-language
i=2
2<=5(T)
display c-language
i=3
3<=5(T)
display c-language
i=4
4<=5(T)
display c-language
i=5
5<=5(T)
display c-language
i=6
6<=5(F)
Exit
from the loop
|
/*3.C program to display 'n'
Natural Numbers using while
loop */
|
/*3.C program to
display 'n'
Natural Numbers using
for loop
*/
|
/*3.C program to
display 'n'
Natural Numbers using
do-
while loop.*/
|
#include<stdio.h>
#include<conio.h>
void
main()
{
inti,n;
clrscr();
printf(“enter
the value of n:”);
scanf(“%d”,&n);
i=1;
while(i<=n)
{
printf(“\t%d”,i);
i++;
}
getch();
}
Result:
input:
enter
the value of n:5
Output:
1 2
3 4 5
|
#include<stdio.h>
#include<conio.h>
void
main()
{
int i;
clrscr();
printf(“enter
the value of n:”);
scanf(“%d”,&n);
for(i=1;i<=n;i++)
{
printf(“\t%d”,i);
}
getch();
}
Result:
input:
enter
the value of n:5
Output:
1 2 3
4 5
|
#include<stdio.h>
#include<conio.h>
void
main()
{
int i;
clrscr();
printf(“enter
the value of n:”);
scanf(“%d”,&n);
i=1;
do
{
printf(“\t%d”,i);
i++;
}
while(i<=n);
getch();
}
Result:
input:
enter
the value of n:5
Output:
1 2 3
4 5
|
i=1
1<=5(T)
display 1
i=2
2<=5(T)
display 2
i=3
3<=5(T)
display 3
i=4
4<=5(T)
display 4
i=5
5<=5(T)
display 5
i=6
6<=5(F)
Exit
from the loop
|
i=1
1<=5(T)
display 1
i=2
2<=5(T)
display 2
i=3
3<=5(T)
display 3
i=4
4<=5(T)
display 4
i=5
5<=5(T)
display 5
i=6
6<=5(F)
Exit
from the loop
|
i=1
display 1
i=2
2<=5(T)
display 2
i=3
3<=5(T)
display 3
i=4
4<=5(T)
display 4
i=5
5<=5(T)
display 5
i=6
6<=5(F)
Exit
from the loop
|
Prime Number:
/*.C program to find the
given number is prime no or
not, using while loop */
|
/*.C program to find the
given number is prime no or
not, using for loop */
|
/*.C program to find the
given number is prime no or
not, using do-while loop */
|
#include<stdio.h>
#include<conio.h>
void
main()
{
int
i,n,count;
clrscr();
printf("enter
the value of n:");
scanf("%d",&n);
count=0;
i=1;
while(i<=n)
{
if(n%i==0)
{
printf("\n%d",i);
}
i++;
}
if(count==2)
{
printf("\n%d
is prime",n);
}
else
{
printf("\n%d
is not prime",n);
}
getch();
}
|
#include<stdio.h>
#include<conio.h>
void
main()
{
int
i,n,count;
clrscr();
printf("enter
the value of n:");
scanf("%d",&n);
count=0;
for(i=1;i<=n;i++)
{
if(n%i==0)
{
printf("\n%d",i);
}
}
if(count==2)
{
printf("\n%d
is prime",n);
}
else
{
printf("\n%d
is not prime",n);
}
getch();
}
|
#include<stdio.h>
#include<conio.h>
void
main()
{
int
i,n,count;
clrscr();
printf("enter
the value of n:");
scanf("%d",&n);
count=0;
i=1;
do
{
if(n%i==0)
{
printf("\n%d",i);
}
i++;
}
while(i<=n);
if(count==2)
{
printf("\n%d
is prime",n);
}
else
{
printf("\n%d
is not prime",n);
}
getch();
}
|
Result:
input:
enter
the value of n:5
Output:
1
5
no.of
factors=2
5
is prime
|
Result:
input:
enter
the value of n:5
Output:
1
5
no.of
factors=2
5
is prime
|
Result:
input:
enter
the value of n:5
Output:
1
5
no.of
factors=2
5
is prime
|
Step
1:
i=1
, count=0
1<=5(T)
If(5%1==0)
(T)
display
1
count=1;
Step 2:
i=2
, count=1
2<=5(T)
If(5%2==0)
(F)
Exit
condition
Step 3:
i=3
, count=1
3<=5(T)
If(5%3==0)
(F)
Exit
condition
Step 4:
i=4
, count=1
4<=5(T)
If(5%4==0)
(F)
Exit
condition
Step 5:
i=5
, count=1
5<=5(T)
If(5%5==0)
(T)
display
5
count=2;
i=6
6<=5(F)
Exit
from the loop
The
check :
If(count==2) i.e.
If(2==2) (T)
5 is prime
|
Step
1:
i=1
, count=0
1<=5(T)
If(5%1==0)
(T)
display
1
count=1;
Step 2:
i=2
, count=1
2<=5(T)
If(5%2==0)
(F)
Exit
condition
Step 3:
i=3
, count=1
3<=5(T)
If(5%3==0)
(F)
Exit
condition
Step 4:
i=4
, count=1
4<=5(T)
If(5%4==0)
(F)
Exit
condition
Step 5:
i=5
, count=1
5<=5(T)
If(5%5==0)
(T)
display
5
count=2;
i=6
6<=5(F)
Exit
from the loop
The
check :
If(count==2) i.e.
If(2==2) (T)
5 is prime
|
Step
1:
i=1
, count=0
1<=5(T)
If(5%1==0)
(T)
display
1
count=1;
Step 2:
i=2
, count=1
2<=5(T)
If(5%2==0)
(F)
Exit
condition
Step 3:
i=3
, count=1
3<=5(T)
If(5%3==0)
(F)
Exit
condition
Step 4:
i=4
, count=1
4<=5(T)
If(5%4==0)
(F)
Exit
condition
Step 5:
i=5
, count=1
5<=5(T)
If(5%5==0)
(T)
display
5
count=2;
i=6
6<=5(F)
Exit
from the loop
The
check :
If(count==2) i.e.
If(2==2) (T)
5 is prime
|
Perfect Number:
Sum of the factors of the given number is
equals to twice of a given number is called Perfect Number.
/*.C program to find the
given number is perfect no or
not, using while loop */
|
/*.C program to find the
given number is perfect no or
not , using for loop */
|
/*.C program to find the
given number is perfect no or
not ,using do-while loop */
|
#include<stdio.h>
#include<conio.h>
void
main()
{
int
i,n,sum;
clrscr();
printf("enter
the value of n:");
scanf("%d",&n);
sum=0;
i=1;
while(i<=n)
{
if(n%i==0)
{
printf("\n%d",i);
sum=sum+i;
}
i++;
}
printf("\nsum
of factors=%d",sum);
if(sum==2*n)
{
printf("\n%d
is perfect Number",n);
}
else
{
printf("\n%d
is not perfect Number ",n);
}
getch();
}
|
#include<stdio.h>
#include<conio.h>
void
main()
{
int
i,n,sum;
clrscr();
printf("enter
the value of n:");
scanf("%d",&n);
sum=0;
for(i=1;
i<=n;i++)
{
if(n%i==0)
{
printf("\n%d",i);
sum=sum+i;
}
}
printf("\nsum
of factors=%d",sum);
if(sum==2*n)
{
printf("\n%d
is perfect Number",n);
}
else
{
printf("\n%d
is not perfect Number ",n);
}
getch();
}
|
#include<stdio.h>
#include<conio.h>
void
main()
{
int
i,n,sum;
clrscr();
printf("enter
the value of n:");
scanf("%d",&n);
sum=0;
i=1;
do
{
if(n%i==0)
{
printf("\n%d",i);
sum=sum+i;
}
i++;
}
while(i<=n);
printf("\nsum
of factors=%d",sum);
if(sum==2*n)
{
printf("\n%d
is perfect Number",n);
}
else
{
printf("\n%d
is not perfect Number ",n);
}
getch();
}
|
Result:
input:
enter
the value of n:6
Output:
1
2
3
6
sum
of factors=12
6
is perfect Number
|
Result:
input:
enter
the value of n:6
Output:
1
2
3
6
sum
of factors=12
6
is perfect Number
|
Result:
input:
enter
the value of n:6
Output:
1
2
3
6
sum
of factors=12
6
is perfect Number
|
Step
1:
i=1
, sum=0
1<=6(T)
If(6%1==0)
(T)
display
1
sum=1;
Step 2:
i=2
, sum=1
2<=6(T)
If(6%2==0)
(T)
display
2
sum=3;
Step 3:
i=3
, sum=3
3<=6(T)
If(6%3==0)
(T)
display
3
sum=6;
Step 4:
i=4
, sum=6
4<=6(T)
If(6%4==0)
(F)
Exit
condition
Step 5:
i=5
, sum=6
5<=6(T)
If(6%5==0)
(T)
Exit
condition
Step 6:
i=6
, sum=6
6<=6(T)
If(6%6==0)
(T)
display
6
sum=12;
Step
7:
i=7,
sum=12
7<=7(F)
Exit
from the loop
The
check :
If(sum==2*n) i.e.
If(12==12) (T)
12 is perfect
|
Step
1:
i=1
, sum=0
1<=6(T)
If(6%1==0)
(T)
display
1
sum=1;
Step 2:
i=2
, sum=1
2<=6(T)
If(6%2==0)
(T)
display
2
sum=3;
Step 3:
i=3
, sum=3
3<=6(T)
If(6%3==0)
(T)
display
3
sum=6;
Step 4:
i=4
, sum=6
4<=6(T)
If(6%4==0)
(F)
Exit
condition
Step 5:
i=5
, sum=6
5<=6(T)
If(6%5==0)
(T)
Exit
condition
Step 6:
i=6
, sum=6
6<=6(T)
If(6%6==0)
(T)
display
6
sum=12;
Step
7:
i=7,
sum=12
7<=7(F)
Exit
from the loop
The
check :
If(sum==2*n) i.e.
If(12==12) (T)
12 is perfect
|
Step
1:
i=1
, sum=0
1<=6(T)
If(6%1==0)
(T)
display
1
sum=1;
Step 2:
i=2
, sum=1
2<=6(T)
If(6%2==0)
(T)
display
2
sum=3;
Step 3:
i=3
, sum=3
3<=6(T)
If(6%3==0)
(T)
display
3
sum=6;
Step 4:
i=4
, sum=6
4<=6(T)
If(6%4==0)
(F)
Exit
condition
Step 5:
i=5
, sum=6
5<=6(T)
If(6%5==0)
(T)
Exit
condition
Step 6:
i=6
, sum=6
6<=6(T)
If(6%6==0)
(T)
display
6
sum=12;
Step
7:
i=7,
sum=12
7<=7(F)
Exit
from the loop
The
check :
If(sum==2*n) i.e.
If(12==12) (T)
12 is perfect
|
Infinite for Loop:
/*.C program to find the
Sum of the individual digits of
positive Integer using while loop */
|
/*.C program to find the
Sum of the individual digits of
positive Integer using for loop */
|
#include<stdio.h>
#include<conio.h>
void
main()
{
int n,r,sum;
clrscr();
printf("enter
the value of n:");
scanf("%d",&n);
sum=0;
while(n>0)
//while(n!=0)
{
r =n%10;
sum=sum+r;
n=n/10;
}
printf("\nsum
of the individual digits of a given Number=%d",sum);
getch();
}
|
#include<stdio.h>
#include<conio.h>
void
main()
{
int n,r,sum;
clrscr();
printf("enter
the value of n:");
scanf("%d",&n);
for(sum=0;n>0;)
{
r =n%10;
sum=sum+r;
n=n/10;
}
printf("\nsum
of the individual digits of a given Number=%d",sum);
getch();
}
|
Result:
input:
enter
the value of n:123
Output:
Sum=6
|
Result:
input:
enter
the value of n:123
Output:
Sum=6
|
Step
1:
n=123
, sum=0
123>0(T)
r=123%10=3
sum=sum+r i.e.
sum=0+3=3;
n=n/10 i.e
n=123/10=12
Step
2:
n=12
, sum=3
12>0(T)
r=12%10=2
sum=sum+r i.e.
sum=3+2=5;
n=n/10 i.e
n=12/10=1
Step
3:
n=1
, sum=5
1>0(T)
r=1%10=1
sum=sum+r i.e.
sum=5+1=6;
n=n/10 i.e
n=1/10=0
Step
4:
n=0
, sum=6
0>0(F)
Exit
from the loop
Display sum=6
|
Step
1:
n=123
, sum=0
123>0(T)
r=123%10=3
sum=sum+r i.e.
sum=0+3=3;
n=n/10 i.e
n=123/10=12
Step
2:
n=12
, sum=3
12>0(T)
r=12%10=2
sum=sum+r i.e.
sum=3+2=5;
n=n/10 i.e
n=12/10=1
Step
3:
n=1
, sum=5
1>0(T)
r=1%10=1
sum=sum+r i.e.
sum=5+1=6;
n=n/10 i.e
n=1/10=0
Step
4:
n=0
, sum=6
0>0(F)
Exit
from the loop
Display sum=6
|
Armstrong Number:
Sum of the cube of the individual digits of a
given number equals to given number is called
Armstrong Number.
/*.C program to find the
Given number is Arm strong no
or not, using while loop */
|
/*.C program to find the
Given number is Arm strong no
or not, using for loop */
|
#include<stdio.h>
#include<conio.h>
void
main()
{
int r,n,sum,temp;
clrscr();
printf("enter
the value of n:");
scanf("%d",&n);
temp=n;
sum=0;
while(n>0)
//while(n!=0)
{
r =n%10;
sum=sum+r*r*r;
n=n/10;
}
printf("\nsum=%d",sum);
if(sum==temp)
{
printf("\n%d
is Armstrong",temp);
}
else
{
printf("\n%d
is not Armstrong",temp);
}getch();
}
|
#include<stdio.h>
#include<conio.h>
void
main()
{
int r,n,sum,temp;
clrscr();
printf("enter
the value of n:");
scanf("%d",&n);
temp=n;
for(sum=0;n>0;)
{
r =n%10;
sum=sum+r*r*r;
n=n/10;
}
printf("\nsum=%d",sum);
if(sum==temp)
{
printf("\n%d
is Armstrong",temp);
}
else
{
printf("\n%d
is not Armstrong",temp);
}getch();
}
|
Result:
input:
enter
the value of n:153
Output:
153 is Armstrong
|
Result:
input:
enter
the value of n:153
Output:
153 is Armstrong
|
Step
1:
n=153
, sum=0
153>0(T)
r=153%10=3
sum=sum+r*r*r i.e.
sum=0+3*3*3=27;
n=n/10 i.e
n=153/10=12
Step
2:
n=15
, sum=27
15>0(T)
r=15%10=5
sum=sum+r*r*r i.e.
sum=27+125=152;
n=n/10 i.e
n=15/10=1
Step
3:
n=1
, sum=152
1>0(T)
r=1%10=1
sum=sum+r*r*r i.e.
sum=152+1=153;
n=n/10 i.e
n=1/10=0
Step
4:
n=0
, sum=153
0>0(F)
Exit
from the loop
The
check :
If(sum==temp) i.e.
If(153==153) (T)
153 is Armstrong
|
Step
1:
n=153
, sum=0
153>0(T)
r=153%10=3
sum=sum+r*r*r i.e.
sum=0+3*3*3=27;
n=n/10 i.e
n=153/10=12
Step
2:
n=15
, sum=27
15>0(T)
r=15%10=5
sum=sum+r*r*r i.e.
sum=27+125=152;
n=n/10 i.e
n=15/10=1
Step
3:
n=1
, sum=152
1>0(T)
r=1%10=1
sum=sum+r*r*r i.e.
sum=152+1=153;
n=n/10 i.e
n=1/10=0
Step
4:
n=0
, sum=153
0>0(F)
Exit
from the loop
The
check :
If(sum==temp) i.e.
If(153==153) (T)
153 is Armstrong
|
Nested loops:
Loop inside another loop is known as nested
loop.
Nested for loops:
for loop inside another for loop is known as
nested loop.
/*.C program to display Prime
no.s upto x Natural
numbers
*/
|
#include<stdio.h>
#include<conio.h>
void
main()
{
int
i,x,n,count;
clrscr();
printf("enter
the value of x:");
scanf("%d",&x);
printf("\nPrime
no's upto %d Natural Numbers:",x);
for(n=1;n<=x;n++)
{
count=0;
for(i=1;i<=n;i++)
{
if(n%i==0)
{
count++;
}
}
if(count==2)
{
printf("\t%d",n);
}
}
getch();
}
|
Result:
input:
enter
the value of x:100
Output:
Prime
no's upto 100 Natural Numbers:
2
3 5 7
11
13 17
19 23 29
31 37 41
43 47
53 59
61 67 71
73 79 83
89 97
|
/*.C program to display
Perfect no.s upto x
Natural
Numbers */
|
#include<stdio.h>
#include<conio.h>
void
main()
{
int
i,x,n,sum;
clrscr();
printf("enter
the value of x:");
scanf("%d",&x);
printf("\nPerfect
no's upto %d Natural Numbers:",x);
for(n=1;n<=x;n++)
{
sum=0;
for(i=1;i<=n;i++)
{
if(n%i==0)
{
sum=sum+i;
}
}
if(sum==2*n)
{
printf("\t%d",n);
}
}
getch();
}
|
Result:
input:
enter
the value of x:100
Output:
Perfect
no's upto 100 Natural Numbers: 6 28
|
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