#C program.
Example 6.1
A program to evaluate the equation
y = xn
when n is a non-negative integer, is given in Fig.6.2
The variable y is initialized to 1 and then multiplied by x, n times using the while loop. The loop control variable, count is initialized outside the loop and incremented inside the loop. When the value of count becomes greater than n, the control exists the loop.
EXAMPLE OF while STATEMENT
Program
main()
{
int count, n;
float x, y;
printf("Enter the values of x and n : ");
scanf("%f %d", &x, &n);
y = 1.0;
count = 1; /* Initialisation */
/* LOOP BEGINs */
while ( count <= n) /* Testing */
{
y = y*x;
count++; /* Incrementing */
}
/* END OF LOOP */
printf("\nx = %f; n = %d; x to power n = %f\n",x,n,y);
}
Output
Enter the values of x and n : 2.5 4
x = 2.500000; n = 4; x to power n = 39.062500
Enter the values of x and n : 0.5 4
x = 0.500000; n = 4; x to power n = 0.062500
Fig.6.2 Program to compute x to the power n using while loop
Example 6.2
A program to print the multiplication table from 1 x 1 to 12 x 10 as shown below is given in Fig. 6.3.
1 2 3 4 ......... 10
2 4 6 8 ......... 20
3 6 9 12 ......... 30
4 ......... 40
- -
- -
- -
12 . . . ........ 120
This program contains two do.... while loops in nested form. The outer loop is controlled by the variable row and executed 12 times. The inner loop is controlled by the variable column and is executed 10 times, each time the outer loop is executed. That is, the inner loop is executed a total of 120 times, each time printing a value in the table.
PRINTING OF MULTIPLICATION TABLE
Program:
#define COLMAX 10
#define ROWMAX 12
main()
{
int row,column, y;
row = 1;
printf(" MULTIPLICATION TABLE \n");
printf("-----------------------------------------\n");
do /*......OUTER LOOP BEGINS........*/
{
column = 1;
do /*.......INNER LOOP BEGINS.......*/
{
y = row * column;
printf("%4d", y);
column = column + 1;
}
while (column <= COLMAX); /*... INNER LOOP ENDS ...*/
printf("\n");
row = row + 1;
}
while (row <= ROWMAX);/*..... OUTER LOOP ENDS .....*/
printf("-----------------------------------------\n");
}
Output
MULTIPLICATION TABLE
-------------------------------------------------------
1 2 3 4 5 6 7 8 9 10
2 4 6 8 10 12 14 16 18 20
3 6 9 12 15 18 21 24 27 30
4 8 12 16 20 24 28 32 36 40
5 10 15 20 25 30 35 40 45 50
6 12 18 24 30 36 42 48 54 60
7 14 21 28 35 42 49 56 63 70
8 16 24 32 40 48 56 64 72 80
9 18 27 36 45 54 63 72 81 90
10 20 30 40 50 60 70 80 90 100
11 22 33 44 55 66 77 88 99 110
12 24 36 48 60 72 84 96 108 120
-------------------------------------------------------
Fig.6.3 Printing of a multiplication table using do...while loop
Example 6.3
The program in Fig.6.4 uses a for loop to print the "Powers of 2" table for the power 0 to 20, both positive and negative.
The program evaluates the value
p = 2 n
successively by multiplying 2 by itself n times.
1
q = 2-n = ----
p
Note that we have declared p as a long int and q as a double.
Additional Features of for Loop
The for loop in C has several capabilities that are not found in other loop constructs. For example, more than one variable can be initialized at a time in the for statement. The statements
p = 1;
for (n=0; n<17; ++n)
can be rewritten as
for (p=1, n=0; n<17; ++n)
USE OF for LOOP
Program:
main()
{
long int p;
int n;
double q;
printf("------------------------------------------\n");
printf(" 2 to power n n 2 to power -n\n");
printf("------------------------------------------\n");
p = 1;
for (n = 0; n < 21 ; ++n) /* LOOP BEGINS */
{
if (n == 0)
p = 1;
else
p = p * 2;
q = 1.0/(double)p ;
printf("%10ld %10d %20.12lf\n", p, n, q);
} /* LOOP ENDS */
printf("------------------------------------------\n");
}
Output -----------------------------------------------
2 to power n n 2 to power -n
-----------------------------------------------
1 0 1.000000000000
2 1 0.500000000000
4 2 0.250000000000
8 3 0.125000000000
16 4 0.062500000000
32 5 0.031250000000
64 6 0.015625000000
128 7 0.007812500000
256 8 0.003906250000
512 9 0.001953125000
1024 10 0.000976562500
2048 11 0.000488281250
4096 12 0.000244140625
8192 13 0.000122070313
16384 14 0.000061035156
32768 15 0.000030517578
65536 16 0.000015258789
131072 17 0.000007629395
262144 18 0.000003814697
524288 19 0.000001907349
1048576 20 0.000000953674
-----------------------------------------------
Fig.6.4 Program to print 'Power of 2' table using for loop
Example 6.4
A class of n students take an annual examination in m subjects. A program to read the marks obtained by each student in various subjects and to compute and print the total marks obtained by each of them is given in Fig.6.5.
The program uses two for loops, one for controlling the number of students and the other for controlling the number of subjects. Since both the number of students and the number of subjects are requested by the program, the program may be used for a class of any size and any number of subjects.
The outer loop includes three parts:
(1) reading of roll-numbers of students, one after another,
(2) inner loop, where the marks are read and totaled for each student, and
(3) printing of total marks and declaration of grades.
ILLUSTRATION OF NESTED LOOPS
Program
#define FIRST 360
#define SECOND 240
main()
{
int n, m, i, j,
roll_number, marks, total;
printf("Enter number of students and subjects\n");
scanf("%d %d", &n, &m);
printf("\n");
for (i = 1; i <= n ; ++i)
{
printf("Enter roll_number : ");
scanf("%d", &roll_number);
total = 0 ;
printf("\nEnter marks of %d subjects for ROLL NO %d\n",
m,roll_number);
for (j = 1; j <= m; j++)
{
scanf("%d", &marks);
total = total + marks;
}
printf("TOTAL MARKS = %d ", total);
if (total >= FIRST)
printf("( First Division )\n\n");
else if (total >= SECOND)
printf("( Second Division )\n\n");
else
printf("( *** F A I L *** )\n\n");
}
}
Output Enter number of students and subjects
3 6
Enter roll_number : 8701
Enter marks of 6 subjects for ROLL NO 8701
81 75 83 45 61 59
TOTAL MARKS = 404 ( First Division )
Enter roll_number : 8702
Enter marks of 6 subjects for ROLL NO 8702
51 49 55 47 65 41
TOTAL MARKS = 308 ( Second Division )
Enter roll_number : 8704
Enter marks of 6 subjects for ROLL NO 8704
40 19 31 47 39 25
TOTAL MARKS = 201 ( *** F A I L *** )
Fig.6.5 Illustration of nested for loops
Example 6.5
The program in Fig.6.8 illustrates the use of the break statement in a C program.
The program reads a list of positive values and calculates their average. The for loop is written to read 1000 values. However, if we want the program to calculate the average of any set of values less than 1000, then we must enter a 'negative' number after the last value in the list, to mark the end of input.
USE OF break IN A PROGRAM
Program
main()
{
int m;
float x, sum, average;
printf("This program computes the average of a
set of numbers\n");
printf("Enter values one after another\n");
printf("Enter a NEGATIVE number at the end.\n\n");
sum = 0;
for (m = 1 ; m < = 1000 ; ++m)
{
scanf("%f", &x);
if (x < 0)
break;
sum += x ;
}
average = sum/(float)(m-1);
printf("\n");
printf("Number of values = %d\n", m-1);
printf("Sum = %f\n", sum);
printf("Average = %f\n", average);
}
Output
This program computes the average of a set of numbers
Enter values one after another
Enter a NEGATIVE number at the end.
21 23 24 22 26 22 -1
Number of values = 6
Sum = 138.000000
Average = 23.000000
Fig.6.8 Use of break in a program
Example 6.6
A program to evaluate the series
1
------ = 1 + x + x2 + x3 + ..... + xn
1-x
for -1 < x < 1 with 0.01 per cent accuracy is given in Fig.6.9. The goto statement is used to exit the loop on achieving the desired accuracy.
We have used the for statement to perform the repeated addition of each of the terms in the series. Since it is an infinite series, the evaluation of the function is terminated when the term xn reaches the desired accuracy. The value of n that decides the number of loop operations is not known and therefore we have decided arbitrarily a value of 100, which may or may not result in the desired level of accuracy.
EXAMPLE OF exit WITH goto STATEMENT
Program
#define LOOP 100
#define ACCURACY 0.0001
main()
{
int n;
float x, term, sum;
printf("Input value of x : ");
scanf("%f", &x);
sum = 0 ;
for (term = 1, n = 1 ; n < = LOOP ; ++n)
{
sum += term ;
if (term < = ACCURACY)
goto output; /* EXIT FROM THE LOOP */
term *= x ;
}
printf("\nFINAL VALUE OF N IS NOT SUFFICIENT\n");
printf("TO ACHIEVE DESIRED ACCURACY\n");
goto end;
output:
printf("\nEXIT FROM LOOP\n");
printf("Sum = %f; No.of terms = %d\n", sum, n);
end:
; /* Null Statement */
}
Output
Input value of x : .21
EXIT FROM LOOP
Sum = 1.265800; No.of terms = 7
Input value of x : .75
EXIT FROM LOOP
Sum = 3.999774; No.of terms = 34
Input value of x : .99
FINAL VALUE OF N IS NOT SUFFICIENT
TO ACHIEVE DESIRED ACCURACY
Fig.6.9 Use of goto to exit from a loop
Example 6.7
The program in Fig.6.11 illustrates the use of continue statement.
The program evaluates the square root of a series of numbers and prints the results. The process stops when the number 9999 is typed in.
In case, the series contains any negative numbers, the process of evaluation of square root should be bypassed for such numbers because the square root of a negative number is not defined. The continue statement is used to achieve this. The program also prints a message saying that the number is negative and keeps an account of negative numbers.
The final output includes the number of positive values evaluated and the number of negative items encountered.
USE OF continue STATEMENT
Program:
#include
main()
{
int count, negative;
double number, sqroot;
printf("Enter 9999 to STOP\n");
count = 0 ;
negative = 0 ;
while (count < = 100)
{
printf("Enter a number : ");
scanf("%lf", &number);
if (number == 9999)
break; /* EXIT FROM THE LOOP */
if (number < 0)
{
printf("Number is negative\n\n");
negative++ ;
continue; /* SKIP REST OF THE LOOP */
}
sqroot = sqrt(number);
printf("Number = %lf\n Square root = %lf\n\n",
number, sqroot);
count++ ;
}
printf("Number of items done = %d\n", count);
printf("\n\nNegative items = %d\n", negative);
printf("END OF DATA\n");
}
Output
Enter 9999 to STOP
Enter a number : 25.0
Number = 25.000000
Square root = 5.000000
Enter a number : 40.5
Number = 40.500000
Square root = 6.363961
Enter a number : -9
Number is negative
Enter a number : 16
Number = 16.000000
Square root = 4.000000
Enter a number : -14.75
Number is negative
Enter a number : 80
Number = 80.000000
Square root = 8.944272
Enter a number : 9999
Number of items done = 4
Negative items = 2
END OF DATA
_______________________________________________________________
Fig.6.11 Use of continue statement
