Point an array of pointers to an array of arrays

Question

I am trying to point to an array of arrays (learning purposes). I use this line for that purpose:

int Arr[6][6];    
int (*ptr)[6][6];
ptr = &Arr;

Is this correct? (EDIT: I intend to use a pointer to a 6x6 array. Also, considering I wish to take input and print it out also, which kind of declaration is better? array of pointers or pointer to 6x6 array?)

Next,

(The "reading and printing of 2d array using pointer" code is a total mess, which is why I won't show it here.)

I want to use scanf and printf to take input through ptr (which should assign it to Arr), and then print both Arr and ptr to see the syntax (and if it works at all). Obviously both printed arrays should be the same.

What syntax should I use? It's extremely confusing and I hope this will help me learn better.

Reading and printing through pointer:

for (i = 0; i < 6; i++)

{
    printf_s("\nEnter values for row %d (Enter 0 for 1-1, 2-2 etc.:\n", i + 1);
    for (j = 0; j < 6; j++)
    {
        printf_s("Row %d Column %d", i + 1, j + 1);
        scanf_s("%d", &(*ptr)[i][j]); ........(ptr)[i][j] doesn't allocate to Arr correctly
    }
}
for (i = 0; i < 6; i++)
    {
        printf_s("\n");
        for (j = 0; j < 6; j++)
        {
            printf_s("%d\t", (*ptr)[i][j]); .........*(ptr)[i][j] is wrong because it dereferences the whole 2D array.
        }
    }

Credits: haccks (I changed the question a lot of times, so I'm pointing out mistakes I did during the same)

Why are we not allowed to see the declaration of Arr? Licensing issues? — Kerrek SB
– Kerrek SB, Commented Jan 3, 2014 at 22:31
array of pointers: T *arr[size];; array of pointers to array-of-arrays: T (*arr[size])[width][length]. — user529758
– user529758, Commented Jan 3, 2014 at 22:39
@user3138129; No..Arr is not just an int array. It is a 2D array of ints. — haccks
– haccks, Commented Jan 3, 2014 at 22:41

haccks · Accepted Answer · 2014-01-03 23:34:05Z

3

int (*ptr)[6][6]; declare ptr as a pointer to an array of 6 elements , each of which are array of 6 integers ( of type int (*)[6][6] ) . Now &Arr is also int (*)[6][6] type, your assignment is legal.
In your scanf_s argument, you are missing &. This should be

scanf_s("%d", &(*ptr)[i][j]);

Note that (*ptr)[i][j] is of int type. It is like Arr[i][j]. To store values to Arr[i][j] you need &, similarly & is used here to store values (*ptr)[i][j].

See the corrected program :

#include <stdio.h>
#include <stdlib.h>

int main(void)
{
    int Arr[6][6];
    int (*ptr)[6][6];
    ptr = &Arr;

    for (int i = 0; i < 6; i++)
    {
        printf("\nEnter values for row %d (Enter 0 for 1-1, 2-2 etc.:\n", i + 1);
        for (int j = 0; j < 6; j++)
        {
            printf("Row %d Column %d: ", i + 1, j + 1);
            scanf("%d", &(*ptr)[i][j]);

        }
    }
    for (int i = 0; i < 6; i++)
    {
        for (int j = 0; j < 6; j++)
            printf("%d\t", (*ptr)[i][j]);
        printf("\n");
    }

}

Output: http://ideone.com/q6BQ61

edited Jan 3, 2014 at 23:34

answered Jan 3, 2014 at 22:36

haccks

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9 Comments

Some programmer dude Over a year ago

Technically, if Arr is also an array of arrays of arrays of integers, e.g. int Arr[X][6][6], that would be a legal assignment as well.

haccks Over a year ago

@JoachimPileborg; In that case I think Arr would be the legal assignment.

user3138129 Over a year ago

And how do I take input to ptr?

haccks Over a year ago

You can do this by (*ptr)[i][j].

user3138129 Over a year ago

okay, I'll show what I tried in another edit in the question. It didn't work though..

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