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Answer 1

+5 A:

You have i++ in there. So consider this

Global.cam[i] = highx;
DebugPrint("cube i = " + cam[i]);
i++;

Otherwise, you're referencing the NEXT one every time.

glowcoder 2010-07-13 15:30:45

+1 beat me to it :-)

Péter Török 2010-07-13 15:32:40

The point is, by the time you look at cam[i], you have already incremented i, because the line above includes i++. In case that wasn't clear.

Carl Manaster 2010-07-13 15:34:21

Yep. Global.i++ is the same as something like `{old_i = Global.i; Global.i = Global.i+1; return old_i;}`

extraneon 2010-07-13 15:35:18

@Carl thanks for that clarification. That's exactly my point. Honestly, raise your hand if you've never been bit by this? :-)

glowcoder 2010-07-13 15:38:08

@glowcoder: My hand certainly isn't up!

Carl Manaster 2010-07-13 16:44:52

Answer 2

+1 A:

This is related to a previous answer of mine (link to the question; make sure to also check the first revision), so I'm partly responsible for the confusion.

Let's step back and consider the following snippet:

    char[] arr = new char[3];
    int index;

    index = 0;
    arr[index]   = 'A';
    arr[index+1] = 'B';
    arr[index+2] = 'C';
    System.out.println(arr); // "ABC"

    index = 0;
    arr[index++] = '1';
    arr[index++] = '2';
    arr[index++] = '3';
    System.out.println(arr); // "123"

In the first part of the snippet, index remains the same, and we manually write index+1, +2, etc as necessary. In the second snippet, we simply use index++. Both technique "work" in assigning elements to the array in increasing indices, but arguably the ++ is slightly "better", because the programmer doesn't have to manually do the increment. Moreover, the index++ version allows for instant reordering of the statements, whereas the explicit index+n version would need renumbering.

Depending on what needs to be done, there are better alternatives to either of the above, of course.

In the most ideal scenario, you can use the special array initializer syntax:

    char[] arr = { 'X', 'Y', 'Z' };
    System.out.println(arr); // "XYZ"

On the post-increment operator

Perhaps this snippet will shed some light:

    char[] arr = { '1', '2', '3' };
    int index = 0;

    System.out.println(index);        // "0"
    arr[index++] = 'X';
    System.out.println(index);        // "1"

    System.out.println(arr[index]);   // "2"
    System.out.println(arr[index-1]); // "X"
    System.out.println(arr[0]);       // "X"

The middle part is the key to understand the problem: after the assignment, index has been incremented by 1. This is why arr[index] is 2, which is the old value of arr[1]. The new value X is properly assigned to arr[0], or more generally, arr[index-1] immediately following the assignment.

References

JLS 15.14.2 Postfix Increment Operator ++

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