Multi-dimensional array access using pointer in C

Question

Note that this has probably been answered before, but now in a way that I could understand it.

I'm trying to understand how pointers and arrays work together. I thought I understood both how and why 1D and 2D arrays could be accessed with pointers until I tried to wrap my head around 3D arrays (just to be clear, I understand 3D arrays, just not how to access them with pointers).

So let's say I have an array x[5] and I want to access the element x[1] using pointers. In that case, I can use *(&x[0]+1), because x is the same as &x[0] and *(x + 1) can be used to access a[1].

Now, for a 2D array, let's say it is declared as x[5][3].

To access x[1][2], I can use *(&x[0][0]+3*1+2). I can kind of understand this. We start at memory location x[0][0], then add 3 * 1 to get from x[0][0] to x[1][0], and then add 2 to get from x[1][0] to x[1][2].

The above makes sense to me until I get to a three dimensional array. Let's assume that x is declared as x[3][4][5], how would I get to, let's say element x[2][1][3] using the same logic(if possible)? Also, how can I apply this logic to higher dimensional arrays as well?

Any advice on this is greatly appreciated!

"x is the same as &x[0]" - NOOO. (sorry for shouting, but that has been asked and answered and discussed here way toooo often already!) - Do research! Hint: An array is not a pointer. — too honest for this site
– too honest for this site, Commented Apr 21, 2017 at 0:04
@Olaf ok, fair enough but 1. I got that information by actually doing research. 2. I don't keep up with all discussions going on here. 3. they return the same value. 4. My question was about accessing arrays using pointers, so while that particular statement may have been technically incorrect, it doesn't change what I want to know :) — Lobs001
– Lobs001, Commented Apr 21, 2017 at 0:12
@Olaf Also, can you point me to a good explanation of what the difference between the two statements is? I'm just curious...and the more you know the better :) — Lobs001
– Lobs001, Commented Apr 21, 2017 at 0:16
You may glean a lot from Array of pointers to an array of fixed size there are a LARGE number of very good and very RELEVANT answers there. (and loosely in your defense, it can be a challenge to pick out the relevant answers from 10,231 search results, but with diligence, it can be done) If you still have questions after digesting that answer, edit your question here and add specifics. — David C. Rankin
– David C. Rankin, Commented Apr 21, 2017 at 0:39
@DavidC.Rankin Thank you! I will look through it once I get home later today and update my q if there's more to add. — Lobs001
– Lobs001, Commented Apr 21, 2017 at 0:53

Macit · Accepted Answer · 2017-04-21 14:45:05Z

0

Supposing you have the following array:

//x[i][j][k]
int x[2][3][5] = { {{0,1,2,3,4},
                    {5,6,7,8,9},
                    {10,11,12,13,14}},

                    {{ 88,1,2,3,4 },
                    { 5,6,7,8,9},
                    { 10,11,12,77,14}} 
                };

Elements of 3D array are arranged in series in the memory like this:

{0,1,2,3,4},{5,6,7,8,9},{10,11,12,13,14},{ 88,1,2,3,4 },{ 5,6,7,8,9},{ 10,11,12,77,14}

Obviously, If you hava the address of the first element you can access the others by adding an offset. The address of the first element is **x, and the address of the next one is **x+1 and so forth. To access the elements of the x array look at the following example:

    int D3_width = sizeof(x)/sizeof(x[0]); //2
    int D2_width = sizeof(x[0]) / sizeof(x[0][0]); //3
    int D1_width = sizeof(x[0][0]) / sizeof(int); //5

    int i = 0, j = 2, k = 3;
    cout << *(**x + i*D2_width*D1_width + j*D1_width + k);

    output:13

You can use the same approach for other multi-dimensional arrays.

The formula is tested and works perfectly.

edited Apr 21, 2017 at 14:45

answered Apr 21, 2017 at 1:39

Macit

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2 Comments

Dmitri Over a year ago

@AjayBrahmakshatriya I think in **x, x decays to int (*)[3][5], which is dereferenced to int [3][5], which decays to int (*)[5], which is dereferenced to int [5], which decays to int *... so **x gives an int *, in a rather confusing way. I'd avoid it personally, but it seems to work.

Ajay Brahmakshatriya Over a year ago

@Dmitri Ah, I misread some parts. Will delete my comment. Thanks.

Patrick · Accepted Answer · 2017-04-21 00:21:39Z

-1

For a two-dimensional array you can skip through indexes of rows with pointer arithmetic as you were doing:

First index of first row + (length of row * desired Row)

If you are making your way up the dimensions, this expression will start to get rather large.

For 3D:

First index of first plane + (size of plane * desired plane)

Note that the plane is a 2-dimensional structure.

Now moving into 4D:

First index of first cube + (size of cube * desired cube )

Hope you get the idea.

answered Apr 21, 2017 at 0:21

Patrick

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