### 2D Arrays: Mistake?

Did I mess up somewhere?

I wanted to create a 2D array similar to...
[1 8 7 1]
[8 9 6 0]
[5 8 4 5]
[3 2 4 3]

I tried...
`int arr[4][4] = {1,8,7,1,8,9,6,0,5,8,4,5,3,2,4,3};`
and
`int arr[4][4] = {{1,8,7,1},{8,9,6,0},{5,8,4,5},{3,2,4,3}};`

I printed them and they came out as I wanted.

My question is, why does it print 8 when I do:
`cout << arr[0][4] << endl;`
It prints it as if it was in the same row, when it should be out of bounds. Did I miss a simple concept?
I'm only guessing here, but arr[0][4] is the element after arr[0][3], which is also indexed as arr[1][0], which is == 8
closed account (zb0S216C)
In terms of memory layout, a matrix is one long array with no padding between each element. It's effectively equivalent to this:

 ``12345`` ``````// This: int Array[3][3]; // ...is actually this: int Array[9];``````

Knowing this, we can easily compute the destination: Array + (Column-Index + Row-Index). I'm not 100% sure about why this is, but I've thought of a few possible reasons:

1. Book-keeping is easier and less stack space is required to keep track of each row.
2. Avoids memory fragmentation (if dynamically allocated).
3. Makes more effective use of the CPU & code caches.
4. Easier to optimise.

Wazzak
Last edited on
That is unfortunate... guess I will have to tell my code if it passes 3 on x, it is out of bound...

Sometimes I feel C++ is one of the best languages, then at other times, it is just so bad.

Thank you both for explaining this to me.
Which is why there are things like constants and pre-compiler directives.

Did you know that the only reason out of bounds would crash your program is if you run into memory that another program is using? Just becuase you do this:
 ``12`` ``````int array[4]; cout << array[4];``````

Doesn't mean your program will crash. It's up to you to make sure you stay in bounds. Sorry if you already know that...
Last edited on
@LowestOne
Let's say you have a 3x3 matrix on a piece of paper. If you search for the location 1,4, it is not possible because it is off the matrix. You can't do it on the because it doesn't work like that in C++. When I check to see if that location is NULL, it isn't. That is what I meant by out of bound.

I will check it differently now using constants.
 Let's say you have a 3x3 matrix on a piece of paper. If you search for the location 1,4, it is not possible because it is off the matrix.

Look at it this way: The operation is undefined. Just like the behavior you invoke in C++ when you access memory outside the bounds of an array.

 When I check to see if that location is NULL, it isn't. That is what I meant by out of bound.

You aren't allowed to check that location because it's out of bounds.

@cire
You actually can in C++...
 When I check to see if that location is NULL

That's impossible. An address is NULL only when it's 0. Since arrays are pointers to the first element of a sequence, the first element may be at address 0 (and crash the program with a segmentation fault if you try to write to it), but an element even one place away from the first can't have address 0

Though I don't think you can actually create an array starting at NULL
Last edited on
Topic archived. No new replies allowed.