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Yan Kozlovskiy
Courses Plus Student 13,773 PointsWhy are the answers 2, 6, 14 instead of 2, 6, 11?
#include <stdio.h>
int main() {
int many[] = {2, 4, 8};
int sum = 0;
for (int i = 0; i < 3; i++;) {
sum += many[i];
printf("sum %d\n", sum);
}
return 0;
}
2 Answers
Seab Anthony
Courses Plus Student 453 Points<p>Think of what happens in the segment sum += many[i]; Another way to write that is: sum = sum + many[i];</p> <p> 2 = 0 + 2................when i=0, many[i] is 2 so the sum is 2!</p> <p> 6 = 2 + 4................when i = 1, many[i] is 4 so add that to the current value of sum.</p> <p> 14 = 6 + 8................when i=3, many[i] is 8 so add that to current value of sum.</p> <p> Hence the printed values of sum are 2, 6, 14! </p>
Michael Hulet
47,913 PointsWhat you're doing here is creating an array of numbers (2, 4, 8), creating a variable to hold the sum as it is determined, then looping over each number in the array and adding it to the sum at that time. Each time the loop is run, the sum will be equal to all the numbers in the array before the one it's currently on, added together. Essentially, in its current state, it's equivalent to doing this:
int sum = 2 + 4 + 8;
Of course, the sum of 2 & 4 is 6, and the sum of 6 & 8 is 14. Thus, the final value of the sum is 14
jennygallagher
2,632 Pointsjennygallagher
2,632 PointsVery clear and concise, thank you!