How to create a pointer to a multidimentional array?

Question

Assume there is an N-dimentional array:

type list[S1][S2]...[SN]

How to create a pointer to this?

One could create another N-1-dimentional array of pointers by getting addresses of the lowest-level arrays, then getting addresses of the newly created arrays of pointers and so on, but this sounds like a complete mess. Typedef may come in handy, but that is not a fast way as well. What options are there to create a pointer to the whole array or to any level of this array?

"... a pointer to the whole array or to any level of this array?" For each item in the array? With a `for`-loop you can reach this simple... — SwissCodeMen, Oct 18 '20 at 21:37
@SwissCodeMen the problem is that except for the lowest-level arrays, each item in each array is a multidimentional array on its own — Kaiyaha, Oct 18 '20 at 21:39
*"How to create a pointer to this?"* - `type (*ptr)[S2][S3]....[SN] = list;` — WhozCraig, Oct 18 '20 at 21:42
***`typedef may come in handy, but that is not a fast way as well`*** typdef does not add any overhead so this statement is false by default — 0___________, Oct 18 '20 at 21:42
@WhozCraig it is not pointer to this. It is pointer to `type [S2]...[SN]` — 0___________, Oct 18 '20 at 21:43
@P__J__ By a _fast way_ I imply typing code, not compiling and executing it — Kaiyaha, Oct 18 '20 at 21:43
@P__J__ I have no idea what you're talking about. The OP asked how to create a proper pointer to `list`, and that's what it does. If the OP wants something else perhaps they should clarify what the real *problem* is. — WhozCraig, Oct 18 '20 at 21:44
@Kaiyaha if you're declaring a by-address formal declaration, then yes, you're correct. It would be `type (*ptr)[S1][S2]...[SN] = &list;` . Note the addressof operator applied to `list`. Otherwise it is as I showed prior. I sincerely doubt you want the latter. — WhozCraig, Oct 18 '20 at 21:49
@P__J__ Really? Then why deal with functions and classes? Let's just type more! — Kaiyaha, Oct 18 '20 at 21:50
actually C++ requires far more typing. Classes are not to "type" less — 0___________, Oct 18 '20 at 21:51
@WhozCraig Could you please explain your proposal as an answer? This is the solution I've been looking for — Kaiyaha, Oct 18 '20 at 21:54
@P__J__ Programmers like typing more, but somewhy they defined `integer` as `int` :) — Kaiyaha, Oct 18 '20 at 21:57
@Kaiyaha it is explained in the dupe. You need to learn the pointers — 0___________, Oct 18 '20 at 21:59
btw real programmers do not save on typing. They save on debugging using names which mean something (`static void configfs_adjust_dir_dirent_depth_after_populate(struct configfs_dirent *sd)`). — 0___________, Oct 18 '20 at 22:07
@P__J__ This all depends on a particular situation. And even here somewhy you name your struct pointer as `sd` — Kaiyaha, Oct 18 '20 at 22:22
@P__J__ Definitely the more senseful a name is the easier a programme is to debug. Still I doubt you would name your function as you did. Long constructions affect readability too — Kaiyaha, Oct 18 '20 at 22:24

0___________ · Accepted Answer · 2020-10-18T22:00:23.757

I would use typedef myself. Here you have an example: array of arrays;

#include <stdio.h>

typedef double arrd_t[2][3][4][5][6][7];
typedef arrd_t arrd_t_t[4][5][6][8][9][10];

arrd_t_t array;
arrd_t_t *ptr = &array;


double ret(arrd_t_t *ptr, size_t x,size_t y, size_t z, size_t a,size_t b, size_t c, size_t x1,size_t y1, size_t z1, size_t a1,size_t b1, size_t c1)
{
    return (*ptr)[x][y][z][a][b][c][x1][y1][z1][a1][b1][c1];
}

https://godbolt.org/z/G4oxsc

How to create a pointer to a multidimentional array?

1 Answers1