Exercise 1-15 - CBD p. 38

#include <stdio.h>


int main(void) {  
  
  int i = 1; 
  int sum = 0;
  int status = 0;

  printf("Variation 1\n");

  while (i < 10) {
    sum = sum + 2 + i;
    i = i + 1;
    printf("sum = %d, i = %d \n", sum, i);
  }

  i = 1; 
  sum = 0;

  printf("\nVariation 2\n");

  while (i < 10) {
    sum = sum/3 + i*i;
    i = i + 1;
    printf("sum = %d, i = %d \n", sum, i);
  }

  return status;    

} /* END OF MAIN */

	int i = 1; int sum = 0; int status = 0;
	Three integer variables are declared and initialized.

 

while (i < 10) { sum = sum + 2 + i; i = i + 1; printf("sum = %d, i = %d \n", sum, i); }

The following loop is executed as long as the value of i is less than 10. Each time the loop is executed, sum is set equal to the value it had at the beginning of the loop, plus the current value of i+2. i is then incremented by one, and the two values are printed out. At the end of the loop, the condition is checked again; since i starts with the value 1 and is increased by one each time, the loop will be executed 9 times.

 

	i = 1; sum = 0;
	Reset the variables (otherwise the second loop will not be executed)

 

	while (i < 10) { sum = sum/3 + i*i; i = i + 1; printf("sum = %d, i = %d \n", sum, i); }
	One more time, except this time sum is set to a third of the value it had, plus the square of i.

 

	return status;
	Terminate program. Output of the program: Variation 1 sum = 3, i = 2 sum = 7, i = 3 sum = 12, i = 4 sum = 18, i = 5 sum = 25, i = 6 sum = 33, i = 7 sum = 42, i = 8 sum = 52, i = 9 sum = 63, i = 10 Variation 2 sum = 1, i = 2 sum = 4, i = 3 sum = 10, i = 4 sum = 19, i = 5 sum = 31, i = 6 sum = 46, i = 7 sum = 64, i = 8 sum = 85, i = 9 sum = 109, i = 10