step4:
increment (decrement) loop counter (generally in the body of the loop)Arijit Sur, Dhiman Saha, Tanmoy Dey, Tirthankar Dasgupta, Chaitnya K. K., Piseke Dayakar, Gaurav Garg
Fundamental of WHILE and FOR loop:
Control Flow in the WHILE Loop:
step1: initialization of loop counter i=1; //Outside of Loop
step2: condition for entering loop while (condition)
{ //body of the loop
................
step3: body of the loop ....................
i = i + 1; (step 4)
} //End of WHILE loop
step4:
increment (decrement) loop counter (generally in the body of the loop)
Illustrative
Example:
Print first n consecutive integer numbers using WHILE loop.
#include <stdio.h>
void main()
{
int n,i;
printf("Enter value of n :\n");
scanf("%d",&n);
i=1; //Initialization while loop counter
while (i<n) //condition to enter the loop
{
printf("\n%d\n",i);
i=i+1; // Increment of loop counter
}
}
Control Flow in the FOR Loop:
for (initialization; condition; increment/decrement)
{
.................................
// Body of the loop
....................................
}
Illustrative Example:
Print first n consecutive integer numbers using FOR loop
#include <stdio.h>
void main()
{
int n, i;
printf("Enter value of n :\n");
scanf("%d",&n);
for (i=1;i<=n;i++)
{
printf("\n%d\n",i);
}
}
Tutorial Example:
1. Convert a given binary number into equivalent decimal number and vice versa.
// Use "cc -lm <programname.c>" for compilation
// This part is for converting binary to decimal.
#include <stdio.h>
#include <math.h>
void main()
{
long int bin_num;
int sum,pos;
printf("Enter a binary number (maximum 8 digits) :");
scanf("%ld",&bin_num);
sum=0;
pos=0;
while ((bin_num/10) > 0)
{
sum=sum+(bin_num%10)*pow(2,pos);
pos=pos+1;
bin_num=bin_num/10;
}
sum=sum+bin_num*pow(2,pos);
printf("The Decimal equivalent is = %d\n",sum);
}
// This part is for converting decimal to binary.
#include <stdio.h>
#include <math.h>
void main()
{
int dec_num;
int pos;
long int sum;
printf("Enter a Decimal number (maximum 512) :");
scanf("%d",&dec_num);
sum=0;
pos=0;
while ((dec_num/2) > 0)
{
sum=sum+(dec_num%2)*(long int)pow(10,pos);
pos=pos+1;
dec_num=dec_num/2;
}
sum=sum+dec_num*(long int)pow(10,pos);
printf("The Binary equivalent is = %d\n",sum);
}
2. Print all the triod numbers between 100 and 1000.
Definition: Triod Number: A number is a triod number if sum of cubic of their digits are
equal to the number.
For example: 153 is a triod number if 13 + 53 + 33 = 153.
#include <stdio.h>
#include <math.h>
void main()
{
int sum,i,num;
for (i=100;i<1000;i++)
{
sum=0;
num=i;
while ((num/10) > 0)
{
sum=sum+(int)pow((num%10),3);
num=num/10;
}
sum=sum+(int)pow(num,3);
if (sum==i)
printf("\n%d\n",i);
}
Sample Output:
153
370
371
407
Assignment 3:
a) Input a number (IP_num) and its base(IP_base). Input another base (OP_base) value. Convert the number to the second base.
For example: Input IP_num=184, IP_base=10, OP_base=8
Oput put will be = 270 (in base 8).
Hint: First converts the number (IP_num) from (IP_base) to the decimal (10) then convert into OP_base.
Test Run: I/P (s) O/P (s)
i) IP_num=184, IP_base=10, Op_base=8 : 270
ii)IP_num=1000, IP_base=2, Op_base=8 : 20
iii)IP_num=126, IP_base=6, Op_base=3 : 2000
iv)IP_num=12121, IP_base=3, Op_base=7 : 304
b) i) Print the sum of digits of a given positive decimal number (ip_NUM).
For example: ip_NUM=5348, O/P would be (5+3+4+8) = 20.
Test Run: I/P (s) O/P (s)
i) 3456 18
ii) 6759 27
iii) 100 1
iv) 5 5
ii) Print the sunny number of a given positive decimal number (ip_NUM).
Definition: Sunny Number : Sunny number is a single digit number which can be
obtained by repetitive summation of digits of a decimal number.
For example, for 5348, sum of digit= (5+3+4+8) = 20, sunny (5348)=2+0=2
again, for 3989, 9+9+8+3=29; 2+9=11; 1+1=2, sunny(3989) = 2
sunny(5)=5, sunny(53)=8 etc.
Test Run: I/P (s) O/P (s)
i) 2456 8
ii) 6759 9
iii) 100 1
iv) 5 5