Program 26) This program illustrates the use of continue statement.
The program evaluates the square root of a series of number and prints the results. The process stops when the number 9999 in 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.
PROGRAM
#include <math.h>
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 ("%1f", &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 = %1f\n Square root =
%1f\n\n", number, sqroot) ;
count++ ;
}
printf (" Number of items done = %d\n", count) ;
printf ("\n\nNegative items = %\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
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