 | This is a solution to exercise 10 on page 117 of C by Dissection using while loops. There is a similar solution with for loops. |
#include <stdio.h>
int main(void){
int n, sum = 0;
printf("Enter a positive or negative integer:\n");
scanf("%d", &n);
if (n < 0){
int i = 2 * n;
while (i <= n) sum += i++;
}
else {
int i = n;
while (i <= 2 * n) sum += i++;
}
printf("%d \n", sum);
return 0;
}
| This is a solution to exercise 10 on page 117 of C by Dissection using while loops. There is a similar solution with for loops. |
| printf("Enter a positive or negative integer:\n");
scanf("%d", &n); |
We prompt the user for input with printf, and read a number into the variable n with scanf. |
| if (n < 0){
int i = 2 * n;
while (i <= n) sum += i++;
}
else {
int i = n;
while (i <= 2 * n) sum += i++;
} |
The if control structure selects one out of two different 'directions', depending on the sign of n. Notice that we chose the else branch in case n is 0. This is arbitrary; The other branch whould also have been possible without changing the result of the program. |
| int i = 2 * n;
while (i <= n) sum += i++; |
We first declare i locally. In case n is negative we iterate i from 2*n to n using a while loop. 2*n is smaller than n because n is negative. Notice that the assignment sum += i++ both assigns sum and increments i. The incrementing of i is done after sum is assigned. It is crucical for the termination of the while loop that i is incremented. It can be argued that sum += i++ is too difficult to grasp. Therefore it could be replaced by the two assignments {sum = sum + i; i++}. |
| int i = n;
while (i <= 2 * n) sum += i++; |
We first declare i locally. Here we know that n is 0 or positive. The while loop iterates i from n to 2 * n. Notice again that the assignment sum += i++ both assigns sum and increments i. |
| printf("%d \n", sum); |
We print the final sum outside of both the if and the whiles. |