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# Why 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;
}
```

What 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