Why no out of bounds error on for loop (Java)

Question

I'm doing a problem on codingbat.com and am confused with why this solution to the problem does not give an index out of bounds error. Wouldn't the first for loop search for an index that is beyond the length of the passed array?

Here's the problem:

We'll say that a value is "everywhere" in an array if for every pair of adjacent elements in the array, at least one of the pair is that value. Return true if the given value is everywhere in the array.

Here's my working solution:

public boolean isEverywhere(int[] nums, int val) {
 boolean flag1 = true;
 boolean flag2 = true;
 for (int i = 0; i < nums.length; i += 2) {
    if (nums[i] != val) flag1 = false;
 }
 for (int i = 0; i < nums.length - 1; i += 2) {
    if (nums[i + 1] != val) flag2 = false;
 }
 return flag1 || flag2;
}

Thanks!

I'm not sure what you are asking, but you seem to misunderstand what `i < nums.length` does — OneCricketeer, Dec 28 '16 at 01:16
as you are only going as high as one less than `nums.length - 1`, then `nums[i + 1]` will be OK — Scary Wombat, Dec 28 '16 at 01:17
By the way, you only need one loop to check positions `i` and `i + 1` — OneCricketeer, Dec 28 '16 at 01:18

Elliott Frisch · Answer 1 · 2016-12-28T01:23:21.150

3

No. The test happens after i is incremented. So, once i is not less than the length of the nums array the loop stops. The second loop uses i + 1 at if (nums[i + 1] != val) which is why it needs to test that i is less than the length minus one.

Also you can make the method static (since it uses no instance state).

edited Dec 28 '16 at 01:23

answered Dec 28 '16 at 01:16

Elliott Frisch

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score 0 · Answer 2 · answered Dec 28 '16 at 01:58

0

It wouldn't because while i += 2 is written after i < nums.length, the loop still checks that i is less than the given length(nums.length) before going onto the body of the loop even after 2 has been added on.

answered Dec 28 '16 at 01:58

Ali Camilletti

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Why no out of bounds error on for loop (Java)

2 Answers2